Matroid theory studies the abstract notion of independence that underlies linear algebra, graph theory, matching theory, and optimization. A matroid is a finite combinatorial structure in which one can speak of independent sets, bases, rank, circuits, closure, and duality in a way that simultaneously generalizes linearly independent vectors and forests in a graph. This course develops matroids as a language for recognizing when greedy algorithms work, for comparing different notions of dependence, and for moving between combinatorial, algebraic, geometric, and algorithmic viewpoints. The main themes are abstraction, duality, structure, and optimization. Early chapters introduce the independence axioms and the basic examples, then build the standard toolkit of rank functions, closures, circuits, flats, and greedy bases. The course then turns to dual matroids, deletion and contraction, and minors, which provide the structural operations used throughout the subject. Representability over fields and graphic and cographic matroids connect the abstract theory back to matrices and graphs. The later chapters broaden the theory toward deeper applications and modern developments. Matroid intersection, union, transversal matroids, gammoids, and matching matroids show how matroids organize major combinatorial optimization problems. Characteristic and Tutte polynomials introduce enumerative invariants, while tropical geometry and valuated matroids connect matroid theory to piecewise-linear geometry. The final chapter surveys log-concavity and modern structure theory, showing how classical matroid ideas continue to shape current research in combinatorics. # Introduction Matroid theory begins from a repeated phenomenon: several different parts of mathematics use the word independent, and many of their strongest theorems have the same formal shape. Linear algebra studies independent columns of a matrix, graph theory studies forests inside a graph, and optimization studies when a greedy choice can be made without later regret. The aim of the course is to isolate the shared combinatorial structure behind these examples and then develop the consequences of that abstraction. This introductory chapter sets the scope of the course before the axioms begin in Chapter 1. It records the guiding examples, the central questions, and the main themes: equivalent axiom systems, duality, minors, representability, and greedy algorithms. The goal is not to prove the main theorems yet, but to explain why the subject has a natural internal logic. ## Independence as a Common Pattern What does it mean for a collection of objects to have no redundancy? In a [vector space](/page/Vector%20Space), redundancy means that some vector lies in the span of the others. In a graph, redundancy means that adding an edge has created a cycle. In a matching problem, redundancy can mean that too many chosen objects demand the same limited resource. The first theme of the course is that these examples behave alike under deletion of chosen objects and under an exchange principle. If a set has no redundancy, then any smaller set has no redundancy. If two non-redundant sets have different sizes, then the larger one contains an element that can be added to the smaller one without creating redundancy. These two principles will become the independent-set axioms. Before stating the formal definition in the next chapter, it is useful to name the kind of structure the course studies. [definition: Matroid Informal Model] A matroid is a finite combinatorial structure consisting of a ground set $E$ together with a specified collection of subsets of $E$ called independent sets, satisfying hereditary and exchange rules. [/definition] This is only a preview of the formal definition, but it already indicates the course's point of view. The elements of $E$ might be columns of a matrix, edges of a graph, or members of an abstract finite set. The matroid remembers which subsets are independent and forgets the extra coordinates, embeddings, labels, or drawings that produced those dependencies. [example: Three Sources of Independence] Let the ground set be the column labels of $A$, the edge set of $G$, or the finite set $E$ in the partitioned example. These give three independence rules: \begin{align*} I \text{ is independent exactly when the columns } \{A_i:i\in I\}\text{ are linearly independent over }k. \end{align*} \begin{align*} F \text{ is independent exactly when } F\text{ contains no cycle in }G. \end{align*} \begin{align*} I \text{ is independent exactly when } |I\cap E_j|\le b_j\text{ for every }j. \end{align*} Each rule is hereditary. For column sets, if $J\subset I$ and \begin{align*} \sum_{j\in J} c_jA_j=0, \end{align*} then extending the coefficients by $c_i=0$ for every $i\in I\setminus J$ gives \begin{align*} \sum_{i\in I} c_iA_i=0. \end{align*} Since the columns indexed by $I$ are linearly independent, every coefficient is zero, so every $c_j=0$. For forests, a cycle contained in $J\subset F$ would also be a cycle contained in $F$. For the partition rule, if $J\subset I$, then for each block $E_\ell$, \begin{align*} |J\cap E_\ell|\le |I\cap E_\ell|\le b_\ell. \end{align*} The same three examples also illustrate exchange. For vectors, if $I$ and $J$ are linearly independent and $|I|<|J|$, then \begin{align*} \dim \operatorname{span}\{A_i:i\in I\}=|I|. \end{align*} Also, \begin{align*} \dim \operatorname{span}\{A_j:j\in J\}=|J|. \end{align*} Since $|I|<|J|$, the span of the columns indexed by $J$ cannot be contained in the span of the columns indexed by $I$. Choose $e\in J$ with $A_e\notin \operatorname{span}\{A_i:i\in I\}$. If a relation \begin{align*} c_eA_e+\sum_{i\in I} c_iA_i=0 \end{align*} had $c_e\ne 0$, then \begin{align*} A_e=-c_e^{-1}\sum_{i\in I} c_iA_i, \end{align*} contradicting $A_e\notin \operatorname{span}\{A_i:i\in I\}$. Hence $c_e=0$, and then [linear independence](/page/Linear%20Independence) of $I$ forces every $c_i=0$, so $I\cup\{e\}$ is independent. For forests, let $F$ and $H$ be forests on the same vertex set $V$ with $|F|<|H|$. A forest on $V$ with edge set $S$ has $|V|-|S|$ connected components, so \begin{align*} |V|-|F|>|V|-|H|. \end{align*} Thus $F$ has more connected components than $H$. If every edge of $H\setminus F$ joined two vertices already connected in $F$, then adding all edges of $H$ could not reduce the number of connected components below that of $F$, contradicting the displayed inequality. Therefore some $e\in H\setminus F$ joins two different components of $F$, and adding such an edge cannot create a cycle. For the partition rule, suppose $I$ and $J$ satisfy all capacity inequalities and $|I|<|J|$. If every block had $|I\cap E_\ell|\ge |J\cap E_\ell|$, then summing over the partition would give $|I|\ge |J|$, a contradiction. Hence some block $E_\ell$ satisfies \begin{align*} |I\cap E_\ell|<|J\cap E_\ell|\le b_\ell. \end{align*} Choose $e\in (J\cap E_\ell)\setminus I$. Then \begin{align*} |(I\cup\{e\})\cap E_\ell|=|I\cap E_\ell|+1\le |J\cap E_\ell|\le b_\ell, \end{align*} and all other block counts are unchanged. Thus in all three settings, removing elements preserves independence, and a larger independent set can donate an element to a smaller one without creating redundancy. [/example] The example also warns us that matroids are not only disguised vector spaces or disguised graphs. Vector matroids and graphic matroids are motivating families, but partition and transversal matroids show that the language is broader. A recurring question will be which matroids come from linear algebra, which come from graphs, and which are genuinely outside both worlds. ## Bases and Greedy Choice Why should an abstraction of independence be useful for optimization? The practical reason is that many optimization problems ask for a largest or heaviest independent set. In linear algebra, a basis is a maximal independent set; in a connected graph, a spanning tree is a maximal forest; in a resource allocation problem, a maximal feasible selection uses the available capacities as far as possible. The course will make this precise by proving that the exchange axiom forces all maximal independent sets to have the same size. These maximal independent sets are the bases of the matroid. Once bases are available, the greedy algorithm becomes the first major bridge between structure and computation. [definition: Basis Informal Model] A basis of a matroid is a maximal independent subset of its ground set. [/definition] The word maximal is with respect to inclusion, not necessarily with respect to a numerical weight. This raises the first algorithmic test for the definition: if bases are the right abstraction of maximal non-redundant sets, then choosing the best available element at each step should succeed exactly in the matroidal setting. The next preview theorem states this criterion. [quotetheorem:6597] [citeproof:6597] This theorem explains why the course treats the exchange axiom as more than a technical condition. It is exactly the property that makes local choice compatible with global optimality. The hypotheses are doing real work: for example, take $E=\{a,b,c\}$ and let the feasible sets be all subsets of $\{a,b\}$ together with $\varnothing$ and $\{c\}$. This hereditary system is not a matroid, since $\{a,b\}$ is independent but $\{c\}$ cannot be enlarged by either $a$ or $b$; with weights $w(c)=3$ and $w(a)=w(b)=2$, greedy chooses $c$ while the basis $\{a,b\}$ has larger total weight. The theorem does not say that greedy solves every optimization problem on a matroid, only that linear weight maximization over bases is governed exactly by the exchange axiom. Chapter 3 will revisit the theorem after bases from Chapter 1, rank from Chapter 2, and contraction from Chapter 5 have been developed carefully. [example: Minimum Spanning Trees as Matroid Greedy] Let $G=(V,E)$ be connected, with edge weights $w:E\to \mathbb R$. In the graphic matroid, the independent sets are the forests. Its bases are the maximal forests, and in a connected graph those are exactly the spanning trees: if a forest $F$ is not spanning-connected, then two components of $(V,F)$ are joined by some edge of $G$, and adding an edge between two different components cannot create a cycle. Kruskal's algorithm orders edges by increasing $w$ and adds an edge exactly when the chosen edges remain a forest. To write this as the greedy algorithm for maximizing a matroid weight, choose \begin{align*} C=\max_{e\in E} w(e) \end{align*} and define \begin{align*} u(e)=C-w(e). \end{align*} Then $u(e)\ge 0$ for every $e\in E$. Also, for any two edges $e_1,e_2\in E$, the inequality $w(e_1)\le w(e_2)$ is equivalent to $C-w(e_1)\ge C-w(e_2)$, which is equivalent to $u(e_1)\ge u(e_2)$. Thus ordering edges by increasing $w$ is the same as ordering them by decreasing $u$. For any spanning tree $T$, the graph $(V,T)$ is connected and acyclic, so $|T|=|V|-1$. Therefore \begin{align*} u(T)=\sum_{e\in T}u(e) \end{align*} \begin{align*} =\sum_{e\in T}(C-w(e)) \end{align*} \begin{align*} =\sum_{e\in T}C-\sum_{e\in T}w(e) \end{align*} \begin{align*} =C|T|-w(T) \end{align*} \begin{align*} =C(|V|-1)-w(T). \end{align*} Hence, for spanning trees $T$ and $T'$, the inequality $u(T)\ge u(T')$ is equivalent to \begin{align*} C(|V|-1)-w(T)\ge C(|V|-1)-w(T'). \end{align*} Subtracting the common term $C(|V|-1)$ gives $-w(T)\ge -w(T')$, which is equivalent to $w(T)\le w(T')$. Maximizing the matroid weight $u$ over bases is therefore exactly the same problem as minimizing the original total edge weight $w$ over spanning trees. The exchange property is the matroid-level reason that Kruskal's locally admissible edge choices can be compared with an optimal spanning tree without losing optimality. [/example] ## Rank, Closure, and Circuits How can a theory defined by independent sets talk about dimension, span, and minimal dependence? Linear algebra suggests three derived ideas. The rank of a set measures the size of its largest independent part, the closure of a set records which elements are forced by it, and a circuit is a minimal dependent configuration. These notions are not extra decoration. They provide alternative axiom systems for the same objects, and each is better suited to different problems. Rank is natural for inequalities and polyhedra, circuits are natural for local dependence, and closure is natural for geometry. [definition: Rank Informal Model] For a matroid on a finite ground set $E$, the rank function is the map \begin{align*} r:2^E\to \mathbb Z_{\ge 0} \end{align*} defined by $r(A)=\max\{|I|: I\subseteq A \text{ and } I \text{ is independent}\}$ for each $A\subset E$. [/definition] Rank turns independence into a numerical function on all subsets of the ground set. The next theorem is needed because it gives the fundamental overlap inequality for this function, replacing the dimension formula from linear algebra by a statement that still makes sense for graphs, matchings, and abstract matroids. [quotetheorem:5808] [citeproof:5808] Submodularity will be one of the main computational signatures of matroids. The exchange axiom is again essential. In the hereditary system with feasible sets all subsets of $\{a,b\}$ together with $\varnothing$ and $\{c\}$, the associated maximum-size function satisfies $r(\{a,c\})=1$, $r(\{b,c\})=1$, $r(\{a,b,c\})=2$, and $r(\{c\})=1$, so submodularity would require $2\ge 3$, which fails. The theorem also does not claim the vector-space dimension formula survives as equality; the set-theoretic intersection $A\cap B$ may carry less span than the intersection of the two generated subspaces. It is also the language in which matroid polytopes and optimization theorems are usually stated. [example: Dimension Inequality] Let the vector matroid be represented by vectors $\{v_e:e\in E\}$ in a finite-dimensional vector space, and for $S\subset E$ write \begin{align*} \langle S\rangle=\operatorname{span}\{v_e:e\in S\}. \end{align*} The matroid rank of $S$ is then \begin{align*} r(S)=\dim \langle S\rangle. \end{align*} For two subsets $A,B\subset E$, set $U=\langle A\rangle$, $V=\langle B\rangle$, and $W=\langle A\cap B\rangle$. Every vector indexed by $A\cap B$ is indexed both by $A$ and by $B$, so \begin{align*} W\subset U\cap V. \end{align*} Taking dimensions gives \begin{align*} \dim W\le \dim(U\cap V). \end{align*} The span of the union is the sum of the two spans, because a linear combination of vectors indexed by $A\cup B$ is the sum of a linear combination of vectors indexed by $A$ and a linear combination of vectors indexed by $B$, and conversely every vector in $U+V$ has that form. Hence \begin{align*} \langle A\cup B\rangle=U+V. \end{align*} Using the vector-space dimension formula and then the inequality $\dim W\le \dim(U\cap V)$, we get \begin{align*} r(A)+r(B)=\dim U+\dim V. \end{align*} \begin{align*} \dim U+\dim V=\dim(U+V)+\dim(U\cap V). \end{align*} \begin{align*} \dim(U+V)+\dim(U\cap V)\ge \dim(U+V)+\dim W. \end{align*} Substituting $U+V=\langle A\cup B\rangle$ and $W=\langle A\cap B\rangle$ gives \begin{align*} \dim(U+V)+\dim W=\dim\langle A\cup B\rangle+\dim\langle A\cap B\rangle. \end{align*} Therefore \begin{align*} r(A)+r(B)\ge r(A\cup B)+r(A\cap B). \end{align*} The matroid rank inequality is thus the vector-space dimension formula with $\langle A\cap B\rangle$ replacing $U\cap V$; equality can fail exactly when $\langle A\cap B\rangle$ is a proper subspace of $\langle A\rangle\cap\langle B\rangle$. [/example] ## Duality and Minors What operations should preserve the class of matroids? For graphs, deleting an edge and contracting an edge are fundamental operations. For vector configurations, deleting a column and quotienting by a chosen vector play analogous roles. Matroid theory packages these operations as deletion and contraction, and their repeated use produces minors. Duality is the second structural operation. In a graph embedded in the plane, cycles in a planar dual graph correspond to cuts in the original graph. In linear algebra, orthogonal complements exchange independent and coindependent behaviour. Matroid duality abstracts these correspondences without requiring a planar drawing or an [inner product](/page/Inner%20Product). [definition: Minor Informal Model] A minor of a matroid is obtained by repeatedly deleting and contracting elements of its ground set. [/definition] Minors are the language of structure theory. Many important classes of matroids are described by excluding certain minors, just as planar graphs are characterised by excluding $K_5$ and $K_{3,3}$ as graph minors. The same structure theory also needs an operation that reverses independence and spanning, because excluded-minor arguments and planar graph examples repeatedly pass between cycles and cuts. This is the role of the dual matroid. [definition: Dual Matroid Informal Model] The dual matroid of a matroid on ground set $E$ is the matroid whose bases are the complements in $E$ of the bases of the original matroid. [/definition] This definition will require proof when introduced formally: the complements of bases must satisfy the basis axioms. Once established, duality turns deletion into contraction, loops into coloops, circuits into cocircuits, and rank into corank. The subject becomes much more symmetric after duality is available. [example: Planar Graph Duality] Let $G$ be a connected plane graph, and let $G^*$ be its planar dual. Write $e^*$ for the dual edge corresponding to $e\in E(G)$, so this gives a bijection $E(G)\leftrightarrow E(G^*)$. We show that if $T\subset E(G)$ is a spanning tree of $G$, then \begin{align*} T^*=\{e^*:e\in E(G)\setminus T\} \end{align*} is a spanning tree of $G^*$. Let $n=|V(G)|$, $m=|E(G)|$, and let $f$ be the number of faces of the plane embedding of $G$. The vertices of $G^*$ are the faces of $G$, so \begin{align*} |V(G^*)|=f. \end{align*} Since $T$ is a spanning tree of $G$, \begin{align*} |T|=n-1. \end{align*} Hence \begin{align*} |T^*|=|E(G)\setminus T| =m-|T| =m-(n-1) =m-n+1. \end{align*} By [Euler's formula](/theorems/2014) for a connected plane graph, \begin{align*} n-m+f=2. \end{align*} Rearranging gives \begin{align*} f=m-n+2, \end{align*} and therefore \begin{align*} |T^*|=m-n+1=f-1=|V(G^*)|-1. \end{align*} It remains to see that $T^*$ has no cycle. If $T^*$ contained a cycle $C^*$ in $G^*$, then the corresponding primal edges \begin{align*} C=\{e\in E(G):e^*\in C^*\} \end{align*} would form an edge cut in $G$ by planar cycle-cut duality. Since $C\subset E(G)\setminus T$, the tree $T$ would contain no edge of this cut. Removing the cut separates the vertex set of $G$ into two nonempty parts, so a spanning connected subgraph must use at least one edge of the cut; this contradicts that $T$ is connected. Thus $T^*$ is acyclic, and an acyclic graph on $|V(G^*)|$ vertices with $|V(G^*)|-1$ edges is connected, so $T^*$ is a spanning tree of $G^*$. Thus the bases of the cycle matroid of $G$ correspond exactly to complements of bases of the cycle matroid of $G^*$. This is the concrete graph-theoretic model for dual matroids: planar duality turns spanning trees into complementary spanning trees. [/example] ## Representability and Fields Which matroids can be realised by vectors over a field? This question is where the course connects combinatorics to linear algebra and algebraic geometry. A vector representation turns each element of the ground set into a column vector, with independence interpreted as linear independence. Representability depends on the field. Some matroids are representable over every field, some over fields of one characteristic but not another, and some over no field. This dependence is one reason matroid theory sits between discrete mathematics and algebra. [definition: Representable Matroid Informal Model] A matroid is representable over a field $k$ if its elements can be labelled by vectors over $k$ so that a subset is independent exactly when the corresponding vectors are linearly independent over $k$. [/definition] The formal version will allow parallel elements, loops, and projective rescaling. The informal definition still captures the essential point: a representation is not just a drawing but an exact encoding of the independence relation. [example: Uniform Matroids and Field Size] The uniform matroid $U_{2,n}$ has ground set of size $n$, and its independent subsets are exactly the subsets of size at most $2$. A representation over $\mathbb F_q$ therefore needs $n$ nonzero vectors in $\mathbb F_q^2$ such that no two are scalar multiples: if two chosen vectors were scalar multiples, that pair would be linearly dependent, while every pair in $U_{2,n}$ must be independent. The projective line over $\mathbb F_q$ is the set of one-dimensional subspaces of $\mathbb F_q^2$. There are \begin{align*} q^2-1 \end{align*} nonzero vectors in $\mathbb F_q^2$, since $\mathbb F_q^2$ has $q^2$ vectors and exactly one zero vector. Each one-dimensional subspace contains \begin{align*} q-1 \end{align*} nonzero vectors, namely the nonzero scalar multiples of any fixed nonzero vector in it. Hence the number of projective points is \begin{align*} \frac{q^2-1}{q-1} =\frac{(q-1)(q+1)}{q-1} =q+1. \end{align*} Thus a representation of $U_{2,n}$ over $\mathbb F_q$ can use at most one vector from each projective point, so it requires \begin{align*} n\le q+1. \end{align*} Conversely, if $n\le q+1$, choose $n$ distinct projective points and one nonzero representative vector from each. Distinct projective points are distinct one-dimensional subspaces, so no two representatives are scalar multiples, and therefore every pair is linearly independent in $\mathbb F_q^2$. This realizes exactly the independent sets of $U_{2,n}$. So $U_{2,n}$ is representable over $\mathbb F_q$ exactly when the projective line over $\mathbb F_q$ has enough points, showing that representability can depend on the size of the field. [/example] ## Polyhedra and Algorithms How does a finite independence structure become a polyhedral object? Given a matroid on $E$, each subset $I\subset E$ has an incidence vector in $\mathbb R^E$. Taking the convex hull of incidence vectors of independent sets or bases produces the independence polytope or base polytope. The rank inequalities define these polytopes in a compact and useful way. This viewpoint links the greedy theorem to linear programming: greedy optimization over matroids is equivalent to optimizing a linear function over a polytope whose facets are controlled by rank. [definition: Base Polytope Informal Model] The base polytope of a matroid on $E$ is the convex hull in $\mathbb R^E$ of the incidence vectors of its bases. [/definition] The geometry of the base polytope records the exchange structure of the matroid. To make that geometry useful for optimization, we need an inequality description rather than a list of all bases. The rank function supplies exactly those inequalities. [quotetheorem:6663] [citeproof:6663] This theorem is a preview of a later polyhedral chapter, where the proof will be stated with the required facts about total dual integrality or greedy optimization. Each part of the statement matters. The equality $x(E)=r(E)$ restricts attention to bases rather than arbitrary independent sets; the inequalities $x(A)\le r(A)$ exclude selecting too many elements from a dependent region; and nonnegativity prevents artificial signed solutions. Without the matroid rank axioms these inequalities need not describe the convex hull of feasible maximal sets: the non-matroid hereditary system above has a rank-like maximum-size function, but its exchange failure creates feasible maxima of different sizes and breaks the greedy separation argument. The theorem does not say that every submodular inequality system is a matroid base polytope, nor that the displayed description automatically gives integrality without the matroid structure behind $r$. It shows why rank submodularity is not merely a formal inequality: it is what allows the polytope to have the expected integral vertices. ## How the Course Will Develop What should the reader expect after this introduction? The course begins with the independent-set and basis axioms, then builds the rank, circuit, closure, and flat languages. It then studies duality, deletion, contraction, minors, representability, and the special behaviour of binary, regular, and graphic matroids. The later part of the course turns to structure and applications. Greedy optimization and matroid polytopes explain why matroids are central in combinatorial optimization. Representability and characteristic phenomena connect the subject to finite fields and algebraic geometry. The final structural results show that matroids are not a list of examples but a robust category of finite geometries with their own internal operations. [remark: Prerequisites and Style] The course assumes linear algebra, graph theory, elementary abstract algebra, basic polyhedral geometry, and familiarity with algorithms. Linear algebra supplies the model for rank and span, graph theory supplies cycles and cuts, and polyhedral geometry supplies the language of convex hulls and linear inequalities. The proofs will usually be combinatorial, but the examples will repeatedly translate the same statement into vector, graph, and optimization language. [/remark] By the end of the course, the central question will have changed. Instead of asking whether a theorem about vector spaces or graphs has a matroid analogue, we will often ask which matroid theorem explains the vector-space or graph-theoretic result we already knew. That reversal is the main reason matroid theory is an organizing language for independence and duality. The introduction has set up matroids as a common language behind familiar forms of independence. We now begin building that language from its most basic object: a finite set together with axioms for which subsets count as independent. # 1. Independence Axioms and First Examples These notes develop the first part of a course on matroid theory: the abstraction of independence from linear algebra, graph theory, and matching theory. The main goal is to learn how a finite set can carry an independence structure without choosing coordinates, an inner product, or a particular graph drawing. The only prerequisites used in this chapter are finite set notation, basic linear independence over a field, elementary graph terminology, and the language of matchings in a bipartite graph. Matroids begin with a question that appears in several parts of mathematics: what does it mean for a finite set of objects to be independent? In linear algebra the objects are vectors, in graph theory they are edges, and in matching theory they are elements that can be assigned to distinct witnesses. The first chapter isolates the common rules behind these examples and uses them to define matroids by independent sets. The guiding point is that matroids remember which subsets are independent and forget everything else. For a matrix, a matroid forgets coordinates and row operations but retains the column-dependence pattern. For a graph, it forgets the drawing of the graph but retains which edge sets contain cycles. ## Independence Axioms Which properties of linear independence are strong enough to support a useful theory, but weak enough to include graph forests and matching examples? Two features are indispensable: passing to a smaller set preserves independence, and larger independent sets can be enlarged by elements from even larger independent sets. These become the hereditary and exchange axioms. [definition: Matroid By Independent Sets] Let $E$ be a finite set. A matroid $M$ on ground set $E$ is a pair $M=(E,\mathcal I)$ where $\mathcal I \subseteq 2^E$ satisfies: 1. $\varnothing \in \mathcal I$. 2. If $I \in \mathcal I$ and $J \subset I$, then $J \in \mathcal I$. 3. If $I,J \in \mathcal I$ and $|I| < |J|$, then there exists $e \in J \setminus I$ such that $I \cup \{e\} \in \mathcal I$. [/definition] The members of $\mathcal I$ are called independent sets. A subset of $E$ which is not independent is called dependent. The second axiom is the hereditary property, while the third axiom is the exchange axiom. Before discussing maximal independent sets, it is useful to check that the motivating example from linear algebra really has these properties. [example: Linear Independence Satisfies Exchange] Let $V$ be a vector space over a field $k$, let $v_1,\dots,v_n\in V$, and take $E=\{1,\dots,n\}$. Declare $I\subseteq E$ to be independent exactly when the indexed family $(v_i)_{i\in I}$ is linearly independent. The empty set is independent because the only linear combination indexed by $\varnothing$ is the empty sum $0$, with no nonzero coefficient available. If $I$ is independent and $J\subseteq I$, then $J$ is independent: any relation \begin{align*} \sum_{j\in J} a_j v_j=0 \end{align*} extends to a relation on $I$ by setting $a_i=0$ for $i\in I\setminus J$. Since $I$ is independent, every coefficient in the extended relation is $0$, so every $a_j=0$. Now let $I,J\subseteq E$ be independent with $|I|<|J|$. We show that some $e\in J\setminus I$ has $I\cup\{e\}$ independent. Suppose not. For each $e\in J\setminus I$, the set $I\cup\{e\}$ is dependent, so there are coefficients, not all zero, with \begin{align*} b_e v_e+\sum_{i\in I} b_i v_i=0. \end{align*} The coefficient $b_e$ cannot be $0$, because then $\sum_{i\in I} b_i v_i=0$ would be a nontrivial relation among the independent family $(v_i)_{i\in I}$. Hence $b_e\ne 0$, and \begin{align*} v_e=-\sum_{i\in I} \frac{b_i}{b_e}v_i. \end{align*} Thus every $v_e$ with $e\in J\setminus I$ lies in $\operatorname{span}\{v_i:i\in I\}$, while every $v_e$ with $e\in J\cap I$ lies there already. Therefore \begin{align*} \operatorname{span}\{v_j:j\in J\}\subseteq \operatorname{span}\{v_i:i\in I\}. \end{align*} Since both indexed families are independent, their spans have dimensions $|J|$ and $|I|$, respectively. The inclusion gives \begin{align*} |J|=\dim \operatorname{span}\{v_j:j\in J\} \le \dim \operatorname{span}\{v_i:i\in I\} =|I|, \end{align*} contradicting $|I|<|J|$. Hence some $e\in J\setminus I$ makes $I\cup\{e\}$ independent, which is the exchange axiom. Thus every finite list of vectors determines a matroid. Repeated equal vectors are still distinct ground-set elements because the ground set records their indices, not just their values in $V$. [/example] The example shows why exchange is the correct abstraction of linear independence. Once independent sets can be enlarged in a controlled way, the natural next question is what happens when the enlargement process stops; those terminal independent sets are the matroidal analogue of vector-space bases. [definition: Basis] Let $M=(E,\mathcal I)$ be a matroid. A basis of $M$ is an independent set $B \in \mathcal I$ which is maximal under inclusion among the independent subsets of $E$. [/definition] A basis is not chosen in advance; it is produced by growing an independent set until no element can be added. In a general downward-closed family, different maximal sets can have different sizes. This motivates the first basis theorem: exchange should force all terminal independent sets to have a common cardinality. [quotetheorem:6598] [citeproof:6598] This common size is the rank of the whole matroid, a notion developed systematically in the next chapter. Each hypothesis has a role in the statement. The finite ground set ensures that maximal independent sets exist by repeatedly adding elements until no extension is possible; without finiteness, the family of finite subsets of an infinite set has no maximal member. The axiom $\varnothing\in\mathcal I$ starts the enlargement process; if $\mathcal I=\varnothing$, then the phrase "basis of $M$" has no object to refer to. Heredity is what makes maximality among independent sets a meaningful analogue of spanning: for instance, if $E=\{1,2\}$ and $\mathcal F=\{\varnothing,\{1\},\{1,2\}\}$ is replaced by the non-hereditary family $\{\varnothing,\{1,2\}\}$, then $\{1\}$ is not available as a smaller independent part of the proposed basis. The exchange axiom is the part that forces different maximal choices to have the same size. The theorem should not be read as a statement about arbitrary hereditary families. For example, on $E=\{1,2,3\}$ the family \begin{align*} \mathcal F=\{\varnothing,\{1\},\{2\},\{3\},\{1,2\}\} \end{align*} is hereditary, but its maximal members are $\{1,2\}$ and $\{3\}$, of sizes $2$ and $1$. This example explains why common size is a meaningful first test, but it also exposes a remaining gap: even when all maximal independent sets have the same size, a useful theory needs a way to compare two different maximal choices element by element. The next theorem supplies that missing control by showing that one basis can be transformed toward another by a valid single-element exchange. [quotetheorem:5809] [citeproof:5809] Basis exchange is the first structural theorem of the course. It says that bases are connected by single-element swaps, a principle that later becomes central in duality, greedy algorithms, and the geometry of the matroid base polytope. The proof uses more than the fact that bases have equal size: it uses heredity to know that $B_1\setminus\{e\}$ is still independent, exchange to insert an element from $B_2$, and maximality to conclude that the resulting independent set is again a basis. Each ingredient is needed. If heredity fails, deleting $e$ from a basis may leave the family of allowed sets, so there is no independent set to augment. If exchange fails, equal-size maximal sets can sit in separate components of the feasible family. If maximality is omitted, the swapped set may be independent of the right size relative to $B_1$ but need not be terminal in the ambient family. The theorem has two important limitations. First, equal-size maximal sets need not satisfy exchange: on $E=\{1,2,3,4\}$ the two sets $B_1=\{1,2\}$ and $B_2=\{3,4\}$ are equal-size maximal members of the collection $\{B_1,B_2\}$, but exchanging $1$ out of $B_1$ with either $3$ or $4$ gives a set not in the collection. Second, basis exchange guarantees that at least one valid replacement exists for each removed element; it does not say that every element of $B_2\setminus B_1$ is a valid replacement. Thus basis exchange is a genuine compatibility condition among the maximal sets, not merely a consequence of their common cardinality. [example: Bases In A Small Vector Matroid] Consider the three indexed vectors in $\mathbb R^2$ \begin{align*} v_1=(1,0), \quad v_2=(0,1), \quad v_3=(1,1). \end{align*} Each one-element set is independent: if $a v_i=(0,0)$, then $v_i\ne (0,0)$ forces $a=0$ by comparing a nonzero coordinate. Now check the two-element sets. For $\{1,2\}$, \begin{align*} a v_1+b v_2=a(1,0)+b(0,1)=(a,b). \end{align*} Thus $a v_1+b v_2=(0,0)$ implies $(a,b)=(0,0)$, so $a=0$ and $b=0$. Hence $\{1,2\}$ is independent. For $\{1,3\}$, \begin{align*} a v_1+b v_3=a(1,0)+b(1,1)=(a+b,b). \end{align*} If this equals $(0,0)$, then $b=0$ from the second coordinate, and then $a+b=0$ gives $a=0$. Hence $\{1,3\}$ is independent. For $\{2,3\}$, \begin{align*} a v_2+b v_3=a(0,1)+b(1,1)=(b,a+b). \end{align*} If this equals $(0,0)$, then $b=0$ from the first coordinate, and then $a+b=0$ gives $a=0$. Hence $\{2,3\}$ is independent. The full set is dependent because \begin{align*} v_1+v_2-v_3=(1,0)+(0,1)-(1,1)=(0,0), \end{align*} and the coefficients $1,1,-1$ are not all zero. Therefore the independent sets are $\varnothing$, the three one-element sets, and the three two-element sets. Since the only three-element subset is dependent, the maximal independent sets, hence the bases, are exactly \begin{align*} \{1,2\}, \quad \{1,3\}, \quad \{2,3\}. \end{align*} Basis exchange is visible in this list: for example, if $B_1=\{1,2\}$ and $B_2=\{1,3\}$, then removing $2$ from $B_1$ and adding $3$ gives $\{1,3\}$, which is again a basis. [/example] The example makes bases look like the main objects rather than auxiliary objects. This raises a reverse question: if someone gives only the collection of bases, can the independent sets and the matroid axioms be recovered from them? [quotetheorem:6599] [citeproof:6599] This theorem lets us move between independent-set language and basis language according to the problem, but the hypotheses are essential. Nonemptiness prevents the reconstruction from producing no bases at all. The no-containment condition rules out collections such as $\{\{1\},\{1,2\}\}$, for which the displayed exchange condition has too few cases to prevent incompatible sizes. Basis exchange then prevents equal-size but mutually isolated maximal sets from being declared bases: the collection $\{\{1,2\},\{3,4\}\}$ on $E=\{1,2,3,4\}$ reconstructs a hereditary family whose maximal independent sets are exactly those two pairs, but independent-set exchange fails because $\{1\}$ cannot be augmented by any element of $\{3,4\}$ to lie inside a listed base. The finiteness of $E$ is also part of the theorem, not a cosmetic assumption. In the finite setting every independent set obtained from the reconstruction can be extended to a listed basis by adding elements until no more can be added. For infinite ground sets, maximal extension may require an additional choice principle and the finite basis-exchange axiom no longer captures all possible infinite-base behaviour. Finally, the reconstruction remembers only the abstract independence structure; it does not recover a particular matrix, graph, or matching presentation. Thus an abstract matroid may be studied without choosing coordinates, a drawing, or a bipartite presentation, while representable, graphic, and transversal matroids are extra presentations of that same independence data when such presentations exist. Independent sets are natural for building examples, while bases are natural when comparing maximal objects of fixed size. In practice, checking that a proposed family is a matroid usually follows the same workflow: first identify the ground set, then prove that feasibility survives deleting elements, and finally prove exchange by starting with two feasible sets $I,J$ with $|I|<|J|$ and finding a specific element of $J\setminus I$ that can be added to $I$. The examples below show that the final step usually contains the real mathematics: Steinitz exchange for vectors, components for forests, quota counting for partition constraints, and alternating paths for matchings. ## Standard Sources Of Matroids Which familiar independence systems satisfy the exchange axiom, rather than merely being hereditary? The first examples answer this question from several directions. Each example has a different origin, but each reduces to the same pattern: feasible subsets can be shrunk, and any smaller feasible subset can be augmented from a larger one. [definition: Vector Matroid] Let $k$ be a field, let $A$ be an $m\times n$ matrix over $k$, and let $E=\{1,\dots,n\}$ index the columns of $A$. The vector matroid $M[A]$ has ground set $E$, and a subset $I\subseteq E$ is independent when the columns indexed by $I$ are linearly independent over $k$. [/definition] Vector matroids are the model example: they are the reason the axiom is called exchange. Row operations do not change the linear dependence relations among columns, so they do not change the associated matroid. [example: Column Matroid Over A Field] Over $\mathbb F_2$, take columns \begin{align*} c_1=(1,0,0), \quad c_2=(0,1,0), \quad c_3=(1,1,0), \quad c_4=(1,0,1). \end{align*} We compute two sample subsets of the vector matroid. First, \begin{align*} c_1+c_2+c_3=(1,0,0)+(0,1,0)+(1,1,0). \end{align*} Adding coordinates in $\mathbb F_2$ gives \begin{align*} (1,0,0)+(0,1,0)+(1,1,0)=(1+0+1,\;0+1+1,\;0+0+0). \end{align*} Since $1+1=0$ in $\mathbb F_2$, this becomes \begin{align*} (1+0+1,\;0+1+1,\;0+0+0)=(0,0,0). \end{align*} Thus $1c_1+1c_2+1c_3=(0,0,0)$, and the coefficients $1,1,1$ are not all zero. Hence $\{1,2,3\}$ is dependent. By contrast, suppose \begin{align*} a c_1+b c_2+d c_4=(0,0,0) \end{align*} with $a,b,d\in\mathbb F_2$. Expanding the left side gives \begin{align*} a(1,0,0)+b(0,1,0)+d(1,0,1)=(a,0,0)+(0,b,0)+(d,0,d). \end{align*} Adding coordinates gives \begin{align*} (a,0,0)+(0,b,0)+(d,0,d)=(a+d,\;b,\;d). \end{align*} Therefore \begin{align*} (a+d,\;b,\;d)=(0,0,0). \end{align*} Comparing the third coordinate gives $d=0$, comparing the second coordinate gives $b=0$, and then the first coordinate gives $a+d=a+0=a=0$. Therefore every relation among $c_1,c_2,c_4$ has all coefficients equal to zero, so $\{1,2,4\}$ is independent. This illustrates that the matroid records exactly the dependence pattern of the columns over the chosen field; changing the field can change which coefficient relations are available. [/example] The vector example explains the linear origin of matroids, but the course also needs a non-linear source where independence is still governed by exchange. Graphs provide this source: an edge set is feasible when it avoids cycles, so the next definition turns forests into independent sets. [definition: Graphic Matroid] Let $G=(V,E)$ be a finite graph. The graphic matroid $M(G)$ has ground set $E$, and a subset $F\subseteq E$ is independent when the subgraph $(V,F)$ is a forest. [/definition] The hereditary property says that a subgraph of a forest is a forest. For exchange, let $F_1,F_2$ be forests with $|F_1|<|F_2|$. A forest on the vertex set $V$ with edge set $F$ has $|V|-|F|$ connected components, so $(V,F_1)$ has more connected components than $(V,F_2)$. If every edge of $F_2\setminus F_1$ had both endpoints inside a single component of $(V,F_1)$, then every path in $(V,F_2)$ would also stay inside a component of $(V,F_1)$, forcing $(V,F_2)$ to have at least as many components as $(V,F_1)$. This contradicts $|F_1|<|F_2|$. Hence some edge $e\in F_2\setminus F_1$ joins two distinct components of $(V,F_1)$, and $F_1\cup\{e\}$ remains a forest. [example: Cycle Matroids Of A Triangle And A Square] Label the edges of the triangle $C_3$ by $e_1,e_2,e_3$. The empty set, each one-edge set, and each two-edge set are forests: with fewer than three edges in a triangle, there is no closed walk using distinct edges, so no cycle is present. The full edge set is dependent because the subgraph with edge set $\{e_1,e_2,e_3\}$ contains the cycle \begin{align*} e_1,\ e_2,\ e_3. \end{align*} Thus the independent sets of $M(C_3)$ are exactly the subsets of $\{e_1,e_2,e_3\}$ of size at most $2$. The only minimal dependent set is $\{e_1,e_2,e_3\}$, so this is the unique circuit. The bases are the maximal independent sets, namely the three two-edge subsets; each is a spanning tree of the triangle. Label the edges of the square $C_4$ by $f_1,f_2,f_3,f_4$ in cyclic order. Any set of at most three edges is a forest: deleting at least one edge from the four-edge cycle breaks the only cycle, leaving either a path, a shorter path together with an isolated vertex, or fewer edges. The full edge set is dependent because the subgraph with edge set $\{f_1,f_2,f_3,f_4\}$ contains the cycle \begin{align*} f_1,\ f_2,\ f_3,\ f_4. \end{align*} Therefore the independent sets of $M(C_4)$ are exactly the subsets of $\{f_1,f_2,f_3,f_4\}$ of size at most $3$, and the unique circuit is the full four-edge set. Its bases are the maximal independent sets, namely the four three-edge subsets; each is a spanning tree of the square. [/example] Graphic matroids retain the cycle structure of a graph but not all graph-theoretic information. To see that independence is controlled by cycles rather than by cardinality alone, it is useful to inspect a graph with several different three-edge shapes. [example: Cycle Matroid Of $K_4$] Let $K_4$ have vertex set $\{1,2,3,4\}$ and edge set \begin{align*} E=\{12,13,14,23,24,34\}. \end{align*} In the graphic matroid $M(K_4)$, an edge set is independent exactly when it forms a forest. A basis is therefore a maximal forest in $K_4$. Since a tree on four vertices has three edges, every spanning tree of $K_4$ gives a basis with three edges; for example \begin{align*} \{12,23,34\} \end{align*} is a path through all four vertices, so it is connected and has no cycle. Now compare three different three-edge sets. The set \begin{align*} T=\{12,23,13\} \end{align*} is dependent because its edges form the cycle \begin{align*} 1 \xrightarrow{12} 2 \xrightarrow{23} 3 \xrightarrow{13} 1. \end{align*} By contrast, \begin{align*} P=\{12,23,34\} \end{align*} is independent: its edges form the path $1-2-3-4$, and a path has no repeated vertex before its endpoint, so it contains no cycle. Similarly, \begin{align*} S=\{12,13,14\} \end{align*} is independent: every edge of $S$ is incident to the central vertex $1$, so any walk using distinct edges must move from a leaf to $1$ and then to another leaf, and cannot return to its starting leaf without reusing an edge. Thus $T$, $P$, and $S$ all have three edges, but $T$ is dependent while $P$ and $S$ are independent. In $M(K_4)$, independence is controlled by the presence of cycles, not by cardinality alone. [/example] The preceding examples come with external structure: vectors or graphs. The next family strips away that structure and keeps only the rule that an independent set has bounded size, giving a symmetric test case for the axioms. [definition: Uniform Matroid] Let $0\le r\le n$. The uniform matroid $U_{r,n}$ has ground set $E$ with $|E|=n$, and a subset $I\subseteq E$ is independent when $|I|\le r$. [/definition] Every $r$-element subset is a basis of $U_{r,n}$. Uniform matroids show that matroids can have very high symmetry, since all elements play the same role. [example: The Uniform Matroid $U_{2,4}$] Let $E=\{1,2,3,4\}$. By the definition of $U_{2,4}$, a subset $I\subseteq E$ is independent exactly when $|I|\le 2$. Thus the independent sets are $\varnothing$, the four one-element sets, and the six two-element sets \begin{align*} \{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}. \end{align*} Every subset of size $3$ or $4$ is dependent because its cardinality is greater than $2$. The bases are the maximal independent sets. A one-element independent set $\{i\}$ is not maximal: since $E$ has four elements, there is some $j\in E\setminus\{i\}$, and then \begin{align*} |\{i,j\}|=2, \end{align*} so $\{i,j\}$ is still independent. A two-element independent set cannot be enlarged while staying independent, since adding any new element produces a three-element set and \begin{align*} 3>2. \end{align*} Therefore the bases are exactly the six two-element subsets listed above. Basis exchange is visible from the list. For example, take $B_1=\{1,2\}$ and $B_2=\{3,4\}$. If we remove $1$ from $B_1$, then replacing it by $3$ gives \begin{align*} (B_1\setminus\{1\})\cup\{3\}=\{2,3\}, \end{align*} and replacing it by $4$ gives \begin{align*} (B_1\setminus\{1\})\cup\{4\}=\{2,4\}. \end{align*} Both resulting sets have size $2$, hence both are bases of $U_{2,4}$. This shows concretely that in a uniform matroid, independence depends only on cardinality. [/example] Uniform matroids impose one global quota. Many combinatorial constraints impose several local quotas at once, so the next construction divides the ground set into blocks and bounds how many elements may be chosen from each block. [definition: Partition Matroid] Let $E$ be a finite set partitioned as $E=E_1\sqcup\cdots\sqcup E_m$, and let $r_i$ be integers with $0\le r_i\le |E_i|$. The partition matroid has ground set $E$, and $I\subseteq E$ is independent when $|I\cap E_i|\le r_i$ for every $i\in\{1,\dots,m\}$. [/definition] Partition matroids are an important bridge from pure independence to optimization. They express constraints such as choosing at most two items from one category and at most one from another. [example: A Partition Constraint] Suppose the ground set is partitioned as \begin{align*} E=E_{\text{red}}\sqcup E_{\text{blue}}\sqcup E_{\text{green}}, \end{align*} with capacities \begin{align*} r_{\text{red}}=1,\qquad r_{\text{blue}}=2,\qquad r_{\text{green}}=1. \end{align*} By the definition of the partition matroid, a subset $I\subseteq E$ is independent exactly when \begin{align*} |I\cap E_{\text{red}}|\le 1,\qquad |I\cap E_{\text{blue}}|\le 2,\qquad |I\cap E_{\text{green}}|\le 1. \end{align*} If $I$ contains two red elements, then \begin{align*} |I\cap E_{\text{red}}|\ge 2>1=r_{\text{red}}, \end{align*} so $I$ violates the red capacity and is dependent, even if $I$ has few elements overall. On the other hand, if $J$ contains one red element, two blue elements, and one green element, then \begin{align*} |J\cap E_{\text{red}}|=1\le 1,\qquad |J\cap E_{\text{blue}}|=2\le 2,\qquad |J\cap E_{\text{green}}|=1\le 1. \end{align*} Thus $J$ satisfies all three capacity inequalities and is independent. This example shows that partition matroid independence is checked block by block: total size alone does not decide independence. [/example] The partition example shows independence controlled by local capacity limits. A different combinatorial problem asks whether chosen elements can be assigned to distinct witnesses, and this motivates the definition of transversal matroids. [definition: Transversal Matroid] Let $E$ be a finite set, and let $A_1,\dots,A_m\subseteq E$. The transversal matroid defined by the family $(A_1,\dots,A_m)$ has ground set $E$, and a subset $I\subseteq E$ is independent if there exists an injection $\varphi:I\to\{1,\dots,m\}$ such that $x\in A_{\varphi(x)}$ for every $x\in I$. [/definition] Equivalently, put a bipartite graph between left vertices $E$ and right vertices $1,\dots,m$, joining $x$ to $j$ when $x\in A_j$. Independent sets are those left-vertex sets that can be matched into the right side. This formulation makes the hereditary property immediate, since deleting left vertices from a matched set leaves a smaller matching. It also explains why the exchange axiom is subtler here than for partition matroids: adding a new left vertex may require reassigning several already matched elements along an alternating path, rather than checking a single quota. [example: A Small Transversal Matroid] Let $E=\{a,b,c\}$ and take \begin{align*} A_1=\{a,b\},\qquad A_2=\{b,c\}. \end{align*} We check independence by writing explicit injections from the chosen subset into $\{1,2\}$, with each element sent to a set that contains it. First, $\{a,c\}$ is independent. Define \begin{align*} \varphi:\{a,c\}\to \{1,2\},\qquad \varphi(a)=1,\qquad \varphi(c)=2. \end{align*} The two images $1$ and $2$ are distinct, so $\varphi$ is injective. Also \begin{align*} a\in A_{\varphi(a)}=A_1=\{a,b\},\qquad c\in A_{\varphi(c)}=A_2=\{b,c\}. \end{align*} Thus $\{a,c\}$ is independent in the transversal matroid. The set $\{a,b\}$ is independent as well. Define \begin{align*} \psi:\{a,b\}\to \{1,2\},\qquad \psi(a)=1,\qquad \psi(b)=2. \end{align*} Again the images are distinct, and \begin{align*} a\in A_{\psi(a)}=A_1=\{a,b\},\qquad b\in A_{\psi(b)}=A_2=\{b,c\}. \end{align*} So $\{a,b\}$ satisfies the matching condition. By contrast, $\{a,b,c\}$ is dependent. Any injection \begin{align*} \theta:\{a,b,c\}\to \{1,2\} \end{align*} would have to assign three distinct elements of $\{1,2\}$ as the values $\theta(a),\theta(b),\theta(c)$. But $\{1,2\}$ has only two elements, so among the three values $\theta(a),\theta(b),\theta(c)$ at least two must be equal. Hence no such injection exists, and $\{a,b,c\}$ is not independent. This example shows the matching nature of transversal independence: two chosen elements can be assigned to distinct available positions, but all three elements cannot be assigned when only two positions exist. [/example] The proof that transversal matroids satisfy exchange is a matching argument. Let $I$ and $J$ be matchable left-vertex sets with $|I|<|J|$, and fix matchings that cover them. Starting from the vertices of $J\setminus I$, follow alternating paths that use unmatched edges from left to right and matched edges from right to left relative to the matching of $I$. If such a path reaches a right vertex not used by the matching of $I$, reversing the path augments the matching and covers $I\cup\{x\}$ for the starting element $x\in J\setminus I$. If no starting element reached an unused right vertex, let $R$ be the set of right vertices reachable by such alternating paths and let $L$ be the reachable left vertices. Every vertex of $R$ is already matched to a vertex of $L\cap I$, and every neighbour of $L$ lies in $R$; otherwise the alternating search could continue. The matching covering $J$ must assign the vertices of $J\cap L$ to distinct neighbours inside $R$, but $J\setminus I$ contributes more starting vertices than the matching of $I$ can absorb inside this trapped region. This gives the Hall-type contradiction. Thus some element of $J\setminus I$ can be added to $I$ while preserving matchability. ## Isomorphism And Loss Of Coordinates When should two matroids be considered the same? Since a matroid is a ground set together with an independence structure, the appropriate notion of sameness is a relabelling of the ground set that preserves independent subsets. This lets us separate the combinatorial structure from the labels used to present it. [definition: Matroid Isomorphism] Let $M_1=(E_1,\mathcal I_1)$ and $M_2=(E_2,\mathcal I_2)$ be matroids. An isomorphism $\psi:M_1\to M_2$ is a bijection $\psi:E_1\to E_2$ such that, for every $I\subseteq E_1$, \begin{align*} I\in\mathcal I_1 \iff \psi(I)\in\mathcal I_2. \end{align*} [/definition] An isomorphism preserves dependence, bases, and every invariant that can be stated only in terms of independent subsets. It does not preserve coordinates, vector lengths, angles, or a particular drawing of a graph. [example: Row-Equivalent Matrices Give The Same Matroid] Let $A$ be an $m\times n$ matrix over a field $k$, with columns $a_1,\dots,a_n$, and let $P$ be an invertible $m\times m$ matrix. The columns of $PA$ are $Pa_1,\dots,Pa_n$, because left multiplication by $P$ applies $P$ to each column of $A$. Fix a subset $I\subseteq \{1,\dots,n\}$ and coefficients $(c_i)_{i\in I}$. By linearity of matrix multiplication, \begin{align*} \sum_{i\in I} c_i(Pa_i)=P\left(\sum_{i\in I} c_i a_i\right). \end{align*} If the columns indexed by $I$ in $A$ satisfy the relation \begin{align*} \sum_{i\in I} c_i a_i=0, \end{align*} then \begin{align*} \sum_{i\in I} c_i(Pa_i)=P\left(\sum_{i\in I} c_i a_i\right)=P0=0. \end{align*} Thus every zero relation among the columns of $A$ gives the same zero relation among the corresponding columns of $PA$. Conversely, suppose the corresponding columns of $PA$ satisfy \begin{align*} \sum_{i\in I} c_i(Pa_i)=0. \end{align*} Using the displayed linearity identity gives \begin{align*} P\left(\sum_{i\in I} c_i a_i\right)=0. \end{align*} Multiplying both sides by $P^{-1}$ gives \begin{align*} P^{-1}P\left(\sum_{i\in I} c_i a_i\right)=P^{-1}0. \end{align*} Since $P^{-1}P$ is the identity matrix and $P^{-1}0=0$, this becomes \begin{align*} \sum_{i\in I} c_i a_i=0. \end{align*} Therefore the same coefficient tuples give zero relations among the columns indexed by $I$ in $A$ and in $PA$. Hence $(a_i)_{i\in I}$ is linearly independent exactly when $(Pa_i)_{i\in I}$ is linearly independent, so $A$ and $PA$ have the same column matroid. Elementary row operations are left multiplication by invertible matrices, so they change the matrix representation but not the matroid. [/example] This example illustrates a central slogan: a matroid represented by vectors remembers the linear dependence relations among labelled columns, but it forgets the coordinates used to write those columns. The next example shows the same phenomenon in graph language. [example: Relabelling A Graphic Matroid] Let $G$ be the square graph with edge set \begin{align*} E=\{e_1,e_2,e_3,e_4\} \end{align*} in cyclic order, so the only cycle using edges of $G$ is the full four-edge cycle. Thus a subset $F\subseteq E$ is independent in the graphic matroid exactly when \begin{align*} F\ne \{e_1,e_2,e_3,e_4\}. \end{align*} Indeed, if $F=E$, then $F$ contains the cycle $e_1,e_2,e_3,e_4$; if $F\ne E$, then at least one edge of the square is missing, so the remaining graph is a path, a disjoint union of paths, or has fewer edges, and hence contains no cycle. Now relabel the edges by \begin{align*} \psi(e_1)=a,\qquad \psi(e_2)=b,\qquad \psi(e_3)=c,\qquad \psi(e_4)=d. \end{align*} The map $\psi:E\to \{a,b,c,d\}$ is a bijection. For any subset $F\subseteq E$, \begin{align*} F=E \quad\Longleftrightarrow\quad \psi(F)=\{a,b,c,d\}, \end{align*} because $\psi$ sends all four distinct edges of $E$ to all four distinct new labels. Therefore \begin{align*} F\ne E \quad\Longleftrightarrow\quad \psi(F)\ne \{a,b,c,d\}. \end{align*} So $F$ is independent in the original square matroid exactly when $\psi(F)$ is independent in the relabelled square matroid. Hence the two graphic matroids are isomorphic: the four-edge cyclic dependence is unchanged, while only the edge names have changed. [/example] Matroid isomorphism also clarifies why different mathematical presentations can define the same object. A rank-$2$ vector configuration of four points in general position and the uniform matroid $U_{2,4}$ have the same independent subsets after relabelling, even though one description uses coordinates and the other uses only cardinality. [remark: What Matroids Remember] A matroid remembers which subsets are independent, which subsets are bases, and which minimal dependent subsets will later be called circuits. It forgets metric and coordinate information. This is why matroid theory can compare matrices, graphs, and matchings inside one language. [/remark] The chapter has established the independent-set axioms, proved the first basis theorems, and built a stock of examples that will recur throughout the course. Chapter 2 refines the independence and basis language developed here into rank, circuits, closure, and flats, giving several equivalent ways to look at the same structure. Independent sets and bases give the first workable definition of a matroid, but they are only one viewpoint. The next chapter translates the same structure into numerical, geometric, and dependence-based language through rank, closure, circuits, and flats. # 2. Rank, Closure, Circuits, and Flats Rank is the numerical language behind the independence axioms from the previous chapter. The prerequisites for this chapter are the independent-set axioms and bases from Chapter 1, together with the vector and graphic examples introduced there. Instead of asking only whether a set is independent, we ask for the size of the largest independent part it contains. This chapter develops rank, circuits, closure, and flats as four interchangeable ways of describing the same matroid structure, and it explains how familiar geometric and graph-theoretic notions reappear without coordinates. ## Rank and Submodularity How much independent information is contained in a subset $A \subset E$? In a vector space this question asks for the dimension of the span of the chosen vectors, while in a graph it asks for the size of a largest forest inside a chosen edge set. Matroid rank abstracts both counts. [definition: Rank Function Of A Matroid] Let $M = (E, \mathcal I)$ be a matroid. The rank function of $M$ is the function $r: 2^E \to \mathbb N \cup \{0\}$ defined by \begin{align*} r(A) = \max\{|I| : I \subseteq A,\ I \in \mathcal I\} \end{align*} for every $A \subset E$. [/definition] The set $A$ is independent exactly when $r(A)=|A|$, so rank remembers the independent sets. It also gives a way to measure dependence locally: adding an element either raises rank by one or adds no new independent information. [example: Rank In A Vector Matroid] Let $M$ be the vector matroid on a finite list of vectors $(v_e)_{e\in E}$ in a vector space $V$ over a field $k$, and fix $A\subset E$. By definition of a vector matroid, a subset $I\subseteq A$ is independent in $M$ exactly when the sublist $(v_e)_{e\in I}$ is linearly independent in $V$. Hence the rank definition gives \begin{align*} r(A)=\max\{|I|:I\subseteq A,\ (v_e)_{e\in I}\text{ is linearly independent}\}. \end{align*} Every linearly independent sublist indexed by $A$ lies inside $\operatorname{span}\{v_e:e\in A\}$, so its size is at most $\dim_k\operatorname{span}\{v_e:e\in A\}$. Conversely, because the finite list $(v_e)_{e\in A}$ spans $\operatorname{span}\{v_e:e\in A\}$, the standard basis-extraction argument from linear algebra produces a sublist indexed by some $I_0\subset A$ that is a basis of this span. Therefore \begin{align*} |I_0|=\dim_k\operatorname{span}\{v_e:e\in A\}. \end{align*} The maximum size of a linearly independent sublist indexed by $A$ is exactly this dimension, so \begin{align*} r(A)=\dim_k \operatorname{span}\{v_e:e\in A\}. \end{align*} Thus rank in the vector matroid records the dimension of the indexed span, while discarding the actual coordinates of the vectors. [/example] The vector example shows rank coming from geometry, but many matroid arguments need a neutral family where the only feature of a subset is its size. Such examples test whether a theorem is genuinely matroidal rather than secretly using coordinates or graph connectivity. This motivates the uniform matroids, where dependence begins exactly when a chosen cardinality threshold is exceeded. [definition: Uniform Matroid] Let $E$ be a finite set with $|E|=n$, and let $0\le k\le n$. The uniform matroid $U_{k,n}$ on $E$ is the matroid whose independent sets are exactly the subsets $I\subset E$ with $|I|\le k$. [/definition] In $U_{k,n}$, every subset of size $k$ is a basis and every subset of size $k+1$ is a circuit. The notation remembers the rank $k$ and the size $n$ of the ground set, but not the labels of the elements. The vector example suggests that rank should behave like dimension, but a matroid need not come from an ambient vector space. We therefore need intrinsic laws for rank, expressed only in terms of subsets of $E$. The most important such law controls how much rank can be counted twice when two subsets overlap. [quotetheorem:6600] [citeproof:6600] Submodularity is the main new inequality in this chapter. It says that the rank gained by adding $B$ to $A$ is at most the rank gained by adding $B$ before $A$ is present; shared information is not counted twice. The hypotheses are not cosmetic. If $A$ and $B$ are not subsets of the same ground set, the term $A\cap B$ has no intrinsic meaning; if a proposed rank function only satisfies the size, monotonicity, and one-element bounds, the later example on $E=\{a,b,c\}$ shows that submodularity can still fail. The theorem also does not say that every dimension identity survives as an equality: matroid rank keeps only the dimension of spans of indexed subsets, not the actual subspaces inside an ambient vector space. This inequality is the rank form of exchange, and it is the tool that makes the next axiom system possible. [example: Submodularity For Indexed Vector Sets] Let $(v_e)_{e\in E}$ be a finite list of vectors in a vector space, and for $A,B\subset E$ write $U_A=\operatorname{span}\{v_e:e\in A\}$ and $U_B=\operatorname{span}\{v_e:e\in B\}$. In the associated vector matroid, rank is dimension of the indexed span, so $r(A)=\dim U_A$ and $r(B)=\dim U_B$. Also, \begin{align*} r(A\cup B)=\dim \operatorname{span}\{v_e:e\in A\cup B\}=\dim(U_A+U_B). \end{align*} The intersection term is \begin{align*} r(A\cap B)=\dim \operatorname{span}\{v_e:e\in A\cap B\}. \end{align*} Substituting these four identities into rank submodularity from *Rank Axioms From Independence* gives \begin{align*} \dim U_A+\dim U_B\ge \dim(U_A+U_B)+\dim \operatorname{span}\{v_e:e\in A\cap B\}. \end{align*} The last term is the span of the common indexed vectors, not necessarily $U_A\cap U_B$. For instance, let $E=\{a,b\}$, take $A=\{a\}$ and $B=\{b\}$, and suppose $v_a=v_b\ne 0$. Then $A\cap B=\varnothing$, so \begin{align*} \operatorname{span}\{v_e:e\in A\cap B\}=\operatorname{span}(\varnothing)=0. \end{align*} On the other hand, \begin{align*} U_A=\operatorname{span}\{v_a\}. \end{align*} Since $v_b=v_a$, we also have \begin{align*} U_B=\operatorname{span}\{v_b\}=\operatorname{span}\{v_a\}. \end{align*} Therefore \begin{align*} U_A\cap U_B=\operatorname{span}\{v_a\}, \end{align*} which is a one-dimensional line. Thus the usual identity $\dim U+\dim W=\dim(U+W)+\dim(U\cap W)$ is a statement about subspaces themselves, while matroid submodularity is the indexed-set shadow that remains after the actual subspace intersection has been forgotten. [/example] The indexed-vector example shows why these rank laws are the right abstraction of dimension. The next question is whether they merely follow from matroids or whether they can replace the independence axioms altogether. This replacement is useful in constructions where it is simpler to prescribe numerical ranks than to list every independent subset. [quotetheorem:6601] [citeproof:6601] This equivalence tells us that rank is not extra structure on a matroid. It is a complete encoding of the same object, particularly well suited to inequalities and optimization. Each rank law has a role: the size bound prevents artificial ranks larger than cardinality, monotonicity prevents rank from decreasing when elements are added, the one-element bound prevents a single element from contributing more than one unit, and submodularity supplies exchange. The last role is visible on $E=\{a,b,c\}$ with \begin{align*} r(\varnothing)=0,\qquad r(\{a,b\})=r(E)=2,\qquad r(A)=1\text{ for every other nonempty }A\subset E. \end{align*} This function satisfies the first three rank laws, but not submodularity for $A=\{a,c\}$ and $B=\{b,c\}$, since $1+1<2+1$. The sets with $r(I)=|I|$ are $\varnothing$, the singletons, and $\{a,b\}$; the singleton $\{c\}$ cannot be augmented by any element of $\{a,b\}$ toward the larger independent set $\{a,b\}$. The theorem is therefore an equivalence for functions satisfying all four laws, not a licence to check only local cardinality bounds. Later constructions often define a candidate rank first, so this theorem gives the checklist for proving that the construction is matroidal. ## Circuits And Elimination If rank measures how large an independent part can be, what is the smallest possible reason for dependence? In linear algebra this is a minimal linear relation among vectors. In a graph it is a cycle. Matroids call these minimal obstructions circuits. [definition: Circuit] Let $M=(E,\mathcal I)$ be a matroid. A circuit of $M$ is a subset $C\subset E$ such that $C\notin \mathcal I$ and every proper subset of $C$ belongs to $\mathcal I$. [/definition] A dependent set contains a circuit by repeatedly deleting elements while dependence remains. Thus circuits are the local pieces from which all dependence is detected. [example: Circuits In Graphic Matroids] Let $G$ be a finite graph and let $M(G)$ be its graphic matroid on the edge set $E(G)$, so an edge set is independent exactly when it is a forest. We show that the circuits of $M(G)$ are exactly the edge sets of simple cycles in $G$. First let $C$ be the edge set of a simple cycle. The subgraph with edge set $C$ is not a forest, since it contains that cycle, so $C$ is dependent in $M(G)$. If $e\in C$, then deleting $e$ breaks the simple cycle into a simple path through the same vertices. A path has no cycle, so $C\setminus\{e\}$ is a forest. More generally, every proper subset $C'\subsetneq C$ is contained in $C\setminus\{e\}$ for some $e\in C\setminus C'$, and every subgraph of a forest is a forest. Thus every proper subset of $C$ is independent, so $C$ is a circuit. Conversely, let $D\subset E(G)$ be a circuit of $M(G)$. Since $D$ is dependent, the subgraph with edge set $D$ is not a forest, so it contains at least one simple cycle with edge set $C\subset D$. The set $C$ is dependent by the first paragraph. If $C\subsetneq D$, then $D$ would have a proper dependent subset, contradicting the minimal dependence required of a circuit. Therefore $C=D$. Hence every circuit is the edge set of a simple cycle, and no other edge set is a circuit. [/example] The graphic example makes circuits concrete, but it also raises the structural question of which families of minimal dependent sets can arise. Minimal dependence must be stable under combining two dependences that share an element, because otherwise the family would not support exchange. This need for a compatibility law motivates the next theorem, the circuit version of the exchange principle. [quotetheorem:6602] [citeproof:6602] Circuit elimination is the matroid abstraction of cancelling a common term in two linear dependences, or splicing two graph cycles through a shared edge. The condition that the circuits are distinct is necessary: taking the same circuit twice would ask for a circuit inside $C\setminus\{e\}$, but every proper subset of a circuit is independent. The shared element is also essential, since there is no common dependence to cancel when two circuits are disjoint. For a nontrivial failure, take the family $\{\{1,2,3\},\{1,4,5\}\}$ on $E=\{1,2,3,4,5\}$; the two proposed circuits share $1$, but elimination would require a circuit contained in $\{2,3,4,5\}$, and the family contains none. This theorem becomes the defining replacement for the exchange axiom in the circuit language. [example: Eliminating Cycles In A Graph] Let $C_1$ and $C_2$ be distinct simple cycles in a graph, and let $e$ be an edge common to both. Write $D$ for the symmetric difference of their edge sets: \begin{align*} D=(C_1\setminus C_2)\cup(C_2\setminus C_1). \end{align*} Since $e\in C_1\cap C_2$, the edge $e$ is in neither $C_1\setminus C_2$ nor $C_2\setminus C_1$, so \begin{align*} D\subset (C_1\cup C_2)\setminus\{e\}. \end{align*} Because $C_1\ne C_2$, there is at least one edge belonging to exactly one of them, so $D\ne\varnothing$. At every vertex $x$, the number of incident edges from a simple cycle is either $0$ or $2$. Hence, modulo $2$, \begin{align*} \deg_D(x)\equiv \deg_{C_1}(x)+\deg_{C_2}(x)\equiv 0+0\equiv 0. \end{align*} Thus every vertex of the subgraph with edge set $D$ has even degree. Since $D$ is nonempty and finite, start with an edge of $D$ and follow unused incident edges; the even-degree condition prevents the walk from getting stuck before it returns to a previously visited vertex. The segment between the first repeated vertex and its repetition is a simple cycle whose edge set $C_3$ satisfies \begin{align*} C_3\subset D\subset (C_1\cup C_2)\setminus\{e\}. \end{align*} In the graphic matroid, this simple cycle is exactly the circuit whose existence is guaranteed abstractly by *Circuit Elimination*. [/example] The cycle example shows the operation that circuit elimination captures in graphs. We now reverse the viewpoint and ask whether this operation, together with minimality, is enough to reconstruct a matroid. The answer is yes, and it gives a circuit-only axiom system. [quotetheorem:6603] [citeproof:6603] This theorem is often the most efficient way to prove that a proposed family of minimal obstructions is matroidal. The non-containment hypothesis is necessary because circuits are meant to be minimal dependent sets: on $E=\{1,2,3\}$, the proposed family $\{\{1,2\},\{1,2,3\}\}$ cannot be the circuit family of any matroid because the larger set contains a smaller dependence. Circuit elimination is an additional condition, not a consequence of non-containment alone. For instance, on $E=\{1,2,3,4,5\}$ the family $\{\{1,2,3\},\{1,4,5\}\}$ has no containment, but elimination at $1$ would require a circuit contained in $\{2,3,4,5\}$ and none is present. The theorem does not say that any list of plausible minimal obstructions defines a matroid; it says that non-containment plus this precise elimination law is exactly the missing structure. ## Closure And Spanning When should an element be regarded as already forced by a set $A$? In vector spaces, this is membership in the linear span of $A$. In graphs, an edge is forced by a forest when its endpoints are already connected by the chosen edges. Matroid closure abstracts this idea through rank. [definition: Closure] Let $M$ be a matroid on a finite set $E$ with rank function $r$. The closure operator of $M$ is the function $\operatorname{cl}:2^E\to 2^E$ defined for every $A\subset E$ by \begin{align*} \operatorname{cl}(A)=\{e\in E : r(A\cup\{e\})=r(A)\}. \end{align*} [/definition] Elements in $\operatorname{cl}(A)$ add no new rank to $A$. This includes every element of $A$, but may also include external elements dependent on $A$. [example: Closure In A Graphic Matroid] Let $H_A=(V(G),A)$, and write $c(A)$ for the number of connected components of $H_A$. In a graphic matroid, independent sets are forests. If the components of $H_A$ have vertex sets $V_1,\dots,V_m$, then any forest contained in $A$ has at most $|V_i|-1$ edges inside $V_i$ for each $i$, and choosing a spanning tree in each component gives exactly that many edges. Hence \begin{align*} r(A)=\sum_{i=1}^m (|V_i|-1)=\sum_{i=1}^m |V_i|-m=|V(G)|-c(A). \end{align*} Let $e$ have endpoints $x$ and $y$. If $x$ and $y$ lie in the same connected component of $H_A$, then adding $e$ does not change the connected components, so \begin{align*} c(A\cup\{e\})=c(A) \end{align*} and therefore \begin{align*} r(A\cup\{e\})=|V(G)|-c(A\cup\{e\})=|V(G)|-c(A)=r(A). \end{align*} Thus $e\in\operatorname{cl}(A)$. Conversely, if $x$ and $y$ lie in different connected components of $H_A$, then adding $e$ merges exactly those two components and leaves all other components unchanged. Hence \begin{align*} c(A\cup\{e\})=c(A)-1, \end{align*} so \begin{align*} r(A\cup\{e\})=|V(G)|-c(A\cup\{e\})=|V(G)|-(c(A)-1)=r(A)+1. \end{align*} Therefore $r(A\cup\{e\})\ne r(A)$, so $e\notin\operatorname{cl}(A)$. Thus closure in a graphic matroid records exactly which edges have endpoints already connected by the chosen edge set. [/example] The graphic example identifies closure by connectivity, while the circuit language identifies dependence by minimal cycles. These two descriptions should agree: being forced by $A$ should mean that $e$ completes a minimal dependence with elements of $A$. The next theorem makes this agreement precise and gives a practical test for closure. [quotetheorem:6604] [citeproof:6604] This characterisation turns closure into a dependence detector. The alternative $e\in A$ is necessary because closure is extensive even when no circuit is involved; for instance, a coloop belongs to its own closure but is not contained in any circuit. The circuit condition is the non-tautological part: outside $A$, closure means that $e$ completes a minimal dependence with elements already in $A$. Both hypotheses matter. If $e\notin A$ is omitted from the circuit alternative, a loop $e$ is handled by the circuit $\{e\}\subset A\cup\{e\}$, while an element already in $A$ need not lie in any circuit. Conversely, if a circuit $C$ containing $e$ is not contained in $A\cup\{e\}$, it says nothing about whether $A$ already forces $e$: in the uniform matroid $U_{2,4}$, the circuit $\{e,f,g\}$ does not put $e$ in the closure of $\{f\}$. This theorem is a test for membership in one closure, not a statement that every circuit meeting $A$ forces all its remaining elements from $A$. [definition: Spanning Set] Let $M$ be a matroid on $E$. A subset $A\subset E$ spans a subset $B\subset E$ if $B\subset \operatorname{cl}(A)$. The set $A$ is spanning in $M$ if $\operatorname{cl}(A)=E$. [/definition] Spanning is the global form of closure, while independence says that no element of the set is redundant. Since bases were introduced as maximal independent sets, we should check that the familiar linear algebra characterisation survives. The next theorem confirms that a basis is exactly an independent set that spans the entire ground set. [quotetheorem:6605] [citeproof:6605] This theorem lets us pass between the three basic descriptions of a basis: maximal independent set, minimal spanning set, and set of full rank with no redundancy. Both hypotheses are doing work. Independence is required: in the uniform matroid $U_{2,3}$ on $\{1,2,3\}$, the whole set spans but is dependent, so it is not a basis. Spanning is required: in the same matroid, $\{1\}$ is independent but has rank $1<2$ and can be enlarged. The conclusion is also about bases of the whole matroid, not about bases of a chosen subset; an independent set can span its own closure without spanning $E$. Finally, the theorem does not say that every spanning set contains a unique basis; $U_{2,3}$ has three different bases inside the spanning set $\{1,2,3\}$. The result will be used repeatedly when arguments switch between adding elements to an independent set and proving that a set already reaches the whole ground set. ## Flats, Hyperplanes, And Special Elements Which subsets are closed under all forced consequences? In projective geometry these are subspaces; in a graph they correspond to edge sets containing every edge inside each connected component they already determine. Matroid flats formalise this closure-complete condition. [definition: Flat] Let $M$ be a matroid on $E$. A flat of $M$ is a subset $F\subset E$ such that \begin{align*} \operatorname{cl}(F)=F. \end{align*} [/definition] A flat is a set to which no outside element can be added without increasing rank. Hence flats are the natural closed sets of matroid geometry. [example: Flats In A Projective Plane Configuration] Represent the projective plane as the one-dimensional subspaces of a three-dimensional vector space $W$ over $k$, and choose a nonzero representative vector $v_e\in W$ for each point $e\in E$. For a projective point $p$, let \begin{align*} E_p=\{e\in E:[v_e]=p\}. \end{align*} If $e\in E_p$, then every $f\in E_p$ satisfies $v_f=\lambda v_e$ for some $\lambda\in k^\times$, so \begin{align*} \operatorname{span}\{v_e,v_f\}=kv_e \end{align*} and adding $f$ to $\{e\}$ does not raise rank. If $g\notin E_p$, then $v_g$ is not a scalar multiple of $v_e$, so \begin{align*} \dim_k\operatorname{span}\{v_e,v_g\}=2>\dim_k kv_e=1. \end{align*} Thus the rank-one flat through $e$ is exactly $E_p$; it is a singleton when no other representative in $E$ determines the same projective point. Now let $L$ be a projective line, corresponding to a two-dimensional subspace $U\subset W$, and put \begin{align*} E_L=\{e\in E:v_e\in U\}. \end{align*} If $E_L$ contains two distinct projective points, then their representative vectors are not scalar multiples, so $\dim_k\operatorname{span}\{v_e:e\in E_L\}=2$. For $f\in E_L$, adding $f$ keeps the span inside $U$, hence does not increase rank. For $g\notin E_L$, we have $v_g\notin U$, so \begin{align*} \dim_k(U+kv_g)=\dim_k U+1=3, \end{align*} and adding $g$ raises rank from $2$ to $3$. Therefore the rank-two flats are exactly the subsets of $E$ lying on a common projective line and containing every represented point of $E$ on that line. Conversely, if $F$ is a rank-two flat, then $U_F=\operatorname{span}\{v_e:e\in F\}$ has dimension $2$, so its projectivization is a line, and flatness forces $F=\{e\in E:v_e\in U_F\}$. If the represented points span the whole projective plane, then \begin{align*} \dim_k\operatorname{span}\{v_e:e\in E\}=3. \end{align*} Any flat of rank $3$ already spans the same three-dimensional space as $E$, so every element of $E$ lies in its closure; since a flat equals its closure, that flat must be $E$. Thus $E$ is the unique flat of full rank. [/example] Among proper flats, the largest ones play the role of codimension-one subspaces. They are important because they are the closed sets just below the whole ground set. [definition: Hyperplane] Let $M$ be a matroid on $E$ with rank $r(E)$. A hyperplane is a flat $H\subsetneq E$ such that \begin{align*} r(H)=r(E)-1. \end{align*} [/definition] Hyperplanes are maximal proper flats. They will later become the complements of cocircuits under matroid duality. [example: Hyperplanes In A Graphic Matroid] Let $G$ be a connected graph with vertex set $V$ and edge set $E$, and let $H\subset E$. In the graphic matroid $M(G)$, the rank of an edge set is the size of a largest forest it contains. If the spanning subgraph $(V,H)$ has connected components with vertex sets $V_1,\dots,V_m$, then a forest in $H$ has at most $|V_i|-1$ edges inside the component $V_i$, and choosing a spanning tree in each component attains this bound. Therefore \begin{align*} r(H)=\sum_{i=1}^m(|V_i|-1). \end{align*} Since the sets $V_1,\dots,V_m$ partition $V$, this becomes \begin{align*} r(H)=|V|-m. \end{align*} Writing $c(H)=m$ for the number of connected components of $(V,H)$, we have \begin{align*} r(H)=|V|-c(H). \end{align*} Because $G$ is connected, $c(E)=1$, so \begin{align*} r(E)=|V|-1. \end{align*} Thus $H$ has rank one less than $r(E)$ exactly when \begin{align*} |V|-c(H)=r(H)=r(E)-1=|V|-2. \end{align*} Subtracting $|V|$ from both sides gives $-c(H)=-2$, hence \begin{align*} c(H)=2. \end{align*} So a hyperplane in $M(G)$ is a flat whose spanning subgraph has exactly two connected components. Now use the graphic closure test: an edge lies in $\operatorname{cl}(H)$ exactly when its endpoints are already connected in $(V,H)$. Since a flat satisfies $\operatorname{cl}(H)=H$, no edge of $E\setminus H$ can have both endpoints in the same component of $(V,H)$. Because $(V,H)$ has exactly two components, every edge of $E\setminus H$ runs between those two components. Therefore deleting $E\setminus H$ from $G$ leaves exactly those two components, so $E\setminus H$ is a cut. This cut is inclusion-minimal. If $D'\subsetneq E\setminus H$, choose an edge \begin{align*} e\in (E\setminus H)\setminus D'. \end{align*} The edge $e$ joins the two components of $(V,H)$, while the edges of $H$ already connect all vertices inside each component. Hence the graph with edge set $E\setminus D'$ is connected, so deleting the smaller set $D'$ does not disconnect $G$. Conversely, let $D\subset E$ be an inclusion-minimal cut and put $H=E\setminus D$. Then $(V,H)$ is disconnected. Minimality forces $(V,H)$ to have exactly two connected components: if it had three or more components, adding back one edge of $D$ could merge at most two of them, so deleting $D\setminus\{e\}$ would still leave the graph disconnected, contradicting minimality. Also, every edge of $D$ must join the two components of $(V,H)$; if some edge of $D$ had both endpoints in one component, adding it back would not connect the two components, so deleting $D\setminus\{e\}$ would still disconnect $G$. Therefore no edge outside $H$ has endpoints already connected in $(V,H)$, so no edge outside $H$ lies in $\operatorname{cl}(H)$. Since every element of $H$ lies in $\operatorname{cl}(H)$, this gives \begin{align*} \operatorname{cl}(H)=H. \end{align*} Thus $H$ is a flat. Since $c(H)=2$, \begin{align*} r(H)=|V|-2=r(E)-1. \end{align*} Therefore $H$ is a hyperplane. Hence the hyperplanes of $M(G)$ are exactly the complements of inclusion-minimal cuts of $G$. [/example] The hyperplane example shows that closure language detects global separations. The same rank language also detects element-level degeneracies, and without names for them every later example would have to restate the same rank tests. The first rank-zero exception is especially important: it is an element already forced by the empty set, so it can never participate in an independent set and must be separated from ordinary elements before discussing bases, closure, or parallelism. This motivates the following name. [definition: Loop] Let $M$ be a matroid on $E$. An element $e\in E$ is a loop if \begin{align*} r(\{e\})=0. \end{align*} [/definition] A loop is an element that is dependent by itself, so it is the smallest possible circuit. Once this zero-rank behaviour has been isolated, the next extreme to name is an element that is forced into every full-rank independent set. Such an element behaves like a bridge in a graph. [definition: Coloop] Let $M$ be a matroid on $E$. An element $e\in E$ is a coloop if $e$ belongs to every basis of $M$. [/definition] A coloop is indispensable, whereas a loop is unusable. To analyze matroids with repeated representations, we also need a name for the case where two distinct elements make the same one-dimensional contribution to rank. This definition is necessary for treating parallel graph edges and repeated projective points uniformly. [definition: Parallel Elements] Let $M$ be a matroid on $E$. Distinct non-loop elements $e,f\in E$ are parallel if \begin{align*} r(\{e,f\})=1. \end{align*} [/definition] Parallel elements form a two-element circuit. This is the matroid version of two nonzero scalar multiples in a vector representation, or two parallel edges joining the same pair of vertices in a graph. [example: Loops And Parallel Edges In Graphs] Let $G$ be a graph that may have loops and parallel edges, and let $M(G)$ be its graphic matroid. If $e$ is a graph loop, then the one-edge set $\{e\}$ contains a cycle, namely the loop itself. Therefore $\{e\}$ is dependent in $M(G)$, so the largest independent subset of $\{e\}$ is $\varnothing$. Hence \begin{align*} r(\{e\})=|\varnothing|=0, \end{align*} and $e$ is a matroid loop. Now let $e$ and $f$ be distinct non-loop edges with the same two endpoints. The set $\{e\}$ is a forest, because a single non-loop edge does not contain a cycle, so \begin{align*} r(\{e\})=1. \end{align*} Similarly, $r(\{f\})=1$. But the two edges $e$ and $f$ together form a cycle of length two: traverse $e$ from one endpoint to the other and return along $f$. Thus $\{e,f\}$ is dependent, while each one-element subset is independent. The largest independent subsets of $\{e,f\}$ therefore have size $1$, so \begin{align*} r(\{e,f\})=1. \end{align*} Since $e$ and $f$ are distinct non-loop elements and $r(\{e,f\})=1$, they are parallel matroid elements. Finally, let $e$ be a bridge of $G$. Removing $e$ increases the number of connected components, so no forest that omits $e$ can connect the two sides separated by $e$. A basis of $M(G)$ is a maximal forest in $G$, and maximality forces it to include an edge connecting those two sides; the only such edge available across this separation is the bridge $e$. Hence every basis of $M(G)$ contains $e$, so $e$ is a coloop. [/example] The hierarchy of closure, flats, and special elements packages local dependence into a geometry. Since closure has its own exchange law, we can ask whether closure operators can define matroids directly. The next theorem answers this and completes the passage from rank and circuits to span. [quotetheorem:6606] [citeproof:6606] Closure operators are somewhat redundant because they record the closure of every subset. The exchange axiom is the essential hypothesis: an arbitrary extensive, monotone, idempotent closure operator may describe a closure system but fail to define matroid independence. A concrete failure occurs on $E=\{e,f\}$ with \begin{align*} \operatorname{cl}(\varnothing)=\varnothing,\qquad \operatorname{cl}(\{e\})=\{e,f\},\qquad \operatorname{cl}(\{f\})=\{f\},\qquad \operatorname{cl}(E)=E. \end{align*} This operator is extensive, monotone, and idempotent, but $f\in\operatorname{cl}(\{e\})\setminus\operatorname{cl}(\varnothing)$ while $e\notin\operatorname{cl}(\{f\})$, so exchange fails. The theorem also does not say that a closure operator is determined by its values on singletons; different matroids can have the same singleton closures and different higher-rank flats. To remove redundant data, we now keep only the subsets that are already closed. The next theorem explains exactly which families of closed sets are rich enough to reconstruct the missing closure operator by intersection and still retain exchange. [quotetheorem:6607] [citeproof:6607] The flat axioms show how little data is needed to recover the whole matroid: it is enough to know the closed sets and how their covers divide the remaining elements. The intersection hypothesis is necessary because closures should be obtainable as smallest closed supersets; on $E=\{1,2,3\}$, the family $\{E,\{1,2\},\{2,3\}\}$ contains $E$ but is not closed under intersections, so the proposed closure of $\{2\}$ would be $\{2\}$, a set not in the family. The partition condition is the flat version of exchange. For example, $\{\varnothing,\{f\},E\}$ on $E=\{e,f\}$ is closed under intersections, but the minimal flat properly containing $\varnothing$ is only $\{f\}$, which does not partition $E$; the missing part records the asymmetric closure behaviour where $e$ forces $f$ but $f$ does not force $e$. The theorem does not classify flats by rank or give a representation over a field; it reconstructs an abstract matroid from a family of closed sets satisfying exactly these incidence rules. Together, rank, circuits, closure, and flats give four languages for the same structure: rank is best for inequalities, circuits for minimal dependence, closure for span, and flats for geometry. The four languages of Chapter 2 make matroids flexible enough to support both structural proofs and geometric intuition. We now use that flexibility to explain why greedy optimization works exactly in the matroidal setting. # 3. The Greedy Algorithm and Matroid Optimization The previous chapters developed independence, bases, rank, circuits, and closure as structural language for matroids. This chapter turns that structure into an algorithmic principle: the greedy algorithm is not merely a convenient heuristic, but a diagnostic test for matroids themselves. The basic problem is to maximize an additive weight function over independent sets or bases, and the main question is when the local instruction "take the heaviest feasible element" is guaranteed to produce a global optimum. The answer is precise: matroids are exactly the finite independence systems on which every nonnegative linear weight function can be optimized by this greedy rule. ## Weighted Bases and Greedy Choice Suppose a finite set $E$ carries weights $w:E\to \mathbb R$. The optimization problem asks for an independent set, or more often a basis, with maximum total weight. In a vector space this resembles selecting a maximum-weight linearly independent set of columns, while in a graph it resembles selecting a maximum-weight forest or, after changing signs, a minimum spanning tree. The question is why the local rule "take the best available element that keeps independence" ever produces a global optimum. The first restriction is that the objective must be additive. If the value of choosing two elements includes a bonus or penalty for choosing them together, then a decision made from the individual weights need not reflect the value of the pair. Greedy matroid optimization is therefore a theorem about linear objectives on sets, not about arbitrary utility functions on subsets. [definition: Weight Of A Set] Let $E$ be a finite set and let $w:E\to \mathbb R$ be a weight function. For $A\subset E$, define \begin{align*} w_{\mathcal P}: \mathcal P(E)&\to \mathbb R, & w_{\mathcal P}(A)&=\sum_{e\in A} w(e). \end{align*} [/definition] When no confusion can arise, write $w(A)$ for $w_{\mathcal P}(A)$. The additive form of the objective is important: the algorithm makes decisions element by element, so the objective must decompose element by element. To turn this objective into a procedure, we need a rule that reads the weights in descending order while consulting only the independence oracle already provided by the matroid. [definition: Greedy Algorithm For Independent Sets] Let $M=(E,\mathcal I)$ be a matroid on a finite ground set and let $w:E\to \mathbb R$ be a weight function. For an ordering $e_1,\dots,e_n$ of $E$ satisfying $w(e_1)\ge w(e_2)\ge \cdots \ge w(e_n)$, the greedy algorithm is the procedure \begin{align*} \operatorname{Greedy}:\{(M,w,e_1,\dots,e_n): e_1,\dots,e_n\text{ orders }E\text{ by nonincreasing }w\}\to \mathcal I \end{align*} defined by the following rule. Starting with $G_0=\varnothing$, define $G_i=G_{i-1}\cup\{e_i\}$ if $G_{i-1}\cup\{e_i\}\in\mathcal I$, and define $G_i=G_{i-1}$ otherwise. The output is $\operatorname{Greedy}(M,w,e_1,\dots,e_n)=G_n\in\mathcal I$. [/definition] Because the procedure scans the whole ground set and accepts every feasible element, it always returns a maximal independent set of the matroid. In a matroid, every maximal independent set is a basis, so this full-scan version always returns a basis, even when some accepted elements have negative weight. For independent-set optimization with arbitrary signs, the relevant convention is to ignore elements of negative weight, or equivalently to run the same full-scan procedure on the restriction to $\{e\in E:w(e)\ge 0\}$; when weights are nonnegative this distinction disappears. Ties may lead to different greedy bases, so correctness means that every possible tie-breaking order gives an optimum, or equivalently that the theorem is independent of harmless choices among equal weights. [example: Greedy Basis In A Uniform Matroid] In the uniform matroid $U_{r,n}$ on $E=\{1,\dots,n\}$, the independent sets are exactly the subsets of size at most $r$, so the bases are exactly the subsets of size $r$. Order the elements as $e_1,\dots,e_n$ with $w(e_1)\ge w(e_2)\ge \cdots \ge w(e_n)$. During the full scan, greedy accepts an element exactly when fewer than $r$ elements have already been accepted, because every set of size at most $r$ is independent and every set of size $r+1$ is dependent. Hence the greedy basis is \begin{align*} G=\{e_1,\dots,e_r\}. \end{align*} Let $B$ be any basis. If $B\ne G$, choose $e_j\in B\setminus G$ and $e_i\in G\setminus B$. Since $e_i\in G$ and $e_j\notin G$, we have $i\le r<j$, so $w(e_i)\ge w(e_j)$. The exchanged set \begin{align*} B'=(B\setminus\{e_j\})\cup\{e_i\} \end{align*} has size $r$, so it is again a basis of $U_{r,n}$. Its weight is \begin{align*} w(B')=w(B)-w(e_j)+w(e_i). \end{align*} Equivalently, \begin{align*} w(B')=w(B)+\bigl(w(e_i)-w(e_j)\bigr). \end{align*} Since $w(e_i)-w(e_j)\ge 0$, we get \begin{align*} w(B')\ge w(B). \end{align*} Repeating this exchange replaces all elements of $B\setminus G$ by elements of $G\setminus B$ without decreasing weight, so $w(G)\ge w(B)$ for every basis $B$. For maximum-weight independent sets with arbitrary signs, one may choose any subset of size at most $r$. If $w(e)<0$ and $e\in I$, then \begin{align*} w(I\setminus\{e\})=w(I)-w(e)>w(I), \end{align*} so no maximum-weight independent set contains a negative-weight element. Thus an optimal independent set consists of the positive-weight elements among $e_1,\dots,e_r$, with zero-weight elements optional because they do not change the total weight. This is the uniform-matroid exchange principle in its simplest form: a lower-weight selected element can always be replaced by a higher-weight unselected element while preserving independence. [/example] The uniform example suggests the general proof strategy, but a theorem is needed because most matroids do not allow us to read optimality class by class or coordinate by coordinate. A greedy basis should be compared to an arbitrary optimal basis by exchanging elements one at a time, preserving the fact that the greedy element has at least as much weight as the element it replaces. [quotetheorem:5810] [citeproof:5810] This theorem is the first major algorithmic payoff of the exchange axiom. It says that global optimality is certified by a local feasibility test, provided the feasible sets are the independent sets of a matroid. The two weight hypotheses should not be conflated. For independent sets, nonnegative weights matter because adding an element of negative weight can only make the objective worse, so a maximum independent set need not be maximal. For bases, all feasible sets have the same matroid rank, so arbitrary weights are allowed: negative entries merely make some basis elements undesirable, but they cannot be avoided by choosing a smaller feasible set. The exchange hypothesis is not cosmetic. In the hereditary system on $E=\{a,b,c\}$ with feasible sets $\varnothing,\{a\},\{b\},\{c\},\{b,c\}$, the set $\{a\}$ blocks both $b$ and $c$ even though the larger feasible set $\{b,c\}$ exists. With weights $w(a)=3$ and $w(b)=w(c)=2$, greedy takes $a$ and obtains weight $3$, while $\{b,c\}$ has weight $4$. Thus the theorem does not extend to arbitrary downward-closed constraints, and it does not say that every greedy-looking algorithm is correct; it applies to additive weights and to feasibility given by a matroid independence oracle. [example: Minimum Spanning Tree As Greedy Optimization] Let $G=(V,E)$ be a connected graph, and let $M(G)$ be its graphic matroid, so independent sets are forests and bases are spanning trees. Give each edge $e$ a cost $c(e)$, choose a constant $C$ with $C>c(e)$ for every $e\in E$, and define \begin{align*} w(e)=C-c(e). \end{align*} If $T$ is a spanning tree, then $|T|=|V|-1$. Therefore \begin{align*} w(T)=\sum_{e\in T}w(e)=\sum_{e\in T}(C-c(e)). \end{align*} Separating the two sums gives \begin{align*} w(T)=\sum_{e\in T}C-\sum_{e\in T}c(e). \end{align*} Since $\sum_{e\in T}C=C|T|=C(|V|-1)$ and $c(T)=\sum_{e\in T}c(e)$, this becomes \begin{align*} w(T)=C(|V|-1)-c(T). \end{align*} Thus, for spanning trees $T_1$ and $T_2$, the inequality $w(T_1)\ge w(T_2)$ is equivalent to \begin{align*} C(|V|-1)-c(T_1)\ge C(|V|-1)-c(T_2). \end{align*} Subtracting the common term $C(|V|-1)$ gives \begin{align*} -c(T_1)\ge -c(T_2). \end{align*} Multiplying by $-1$ reverses the inequality, so \begin{align*} c(T_1)\le c(T_2). \end{align*} Therefore maximizing $w(T)$ over bases of $M(G)$ is exactly the same as minimizing $c(T)$ over spanning trees of $G$. The greedy order for $w$ is the nondecreasing-cost order for $c$. Indeed, for edges $e$ and $f$, the inequality $w(e)\ge w(f)$ means \begin{align*} C-c(e)\ge C-c(f). \end{align*} Subtracting $C$ gives \begin{align*} -c(e)\ge -c(f), \end{align*} and multiplying by $-1$ gives \begin{align*} c(e)\le c(f). \end{align*} During the scan, an edge is accepted exactly when adding it to the current forest remains independent in the graphic matroid, which is exactly the condition that the edge does not create a cycle. Hence decreasing-$w$ greedy is Kruskal's algorithm: scan edges in nondecreasing cost and add precisely the cycle-free ones. By *Rado-Edmonds Greedy Theorem*, the resulting basis has maximum $w$-weight, and the calculation above shows that this basis is a spanning tree of minimum total cost. [/example] The same theorem also explains a scheduling rule for partition constraints. The point is not that all scheduling problems are matroidal, but that the ones with independent capacity constraints over disjoint classes are. [example: Partition Matroid Scheduling] Suppose jobs are partitioned into disjoint classes $E_1,\dots,E_m$, and at most $b_j$ jobs may be chosen from class $E_j$. Thus the feasible sets are \begin{align*} \mathcal I=\{I\subset E: |I\cap E_j|\le b_j\text{ for }1\le j\le m\}. \end{align*} The constraint is checked class by class: adding a job from $E_j$ changes only the count $|I\cap E_j|$. For each class, order its jobs as $E_j=\{e_{j,1},\dots,e_{j,n_j}\}$ with \begin{align*} w(e_{j,1})\ge w(e_{j,2})\ge \cdots \ge w(e_{j,n_j}). \end{align*} Put $q_j=\min\{b_j,n_j\}$. During the global greedy scan, a job in $E_j$ is accepted exactly while fewer than $b_j$ jobs from $E_j$ have already been accepted, and jobs from other classes do not affect that count. Hence the greedy output $G$ satisfies \begin{align*} G\cap E_j=\{e_{j,1},\dots,e_{j,q_j}\}. \end{align*} Let $I\in\mathcal I$ be any feasible set. Write $I\cap E_j=\{e_{j,t_1},\dots,e_{j,t_k}\}$ with $t_1<\cdots<t_k$. Since $I$ is feasible, $k\le b_j$, and therefore $k\le q_j$ unless $q_j=n_j$, in which case the same inequality is automatic because $k\le n_j$. For each $1\le \ell\le k$, the increasing order gives $t_\ell\ge \ell$, so \begin{align*} w(e_{j,t_\ell})\le w(e_{j,\ell}). \end{align*} Summing these $k$ inequalities gives \begin{align*} w(I\cap E_j)\le \sum_{\ell=1}^k w(e_{j,\ell}). \end{align*} Because all profits are nonnegative and $k\le q_j$, adding the remaining terms cannot decrease the sum: \begin{align*} \sum_{\ell=1}^k w(e_{j,\ell})\le \sum_{\ell=1}^{q_j} w(e_{j,\ell}). \end{align*} By the description of $G\cap E_j$, this last sum is $w(G\cap E_j)$, so \begin{align*} w(I\cap E_j)\le w(G\cap E_j). \end{align*} The classes are disjoint, hence weights add over the partition: \begin{align*} w(I)=\sum_{j=1}^m w(I\cap E_j). \end{align*} Using the classwise inequalities, \begin{align*} w(I)\le \sum_{j=1}^m w(G\cap E_j). \end{align*} Finally, \begin{align*} \sum_{j=1}^m w(G\cap E_j)=w(G). \end{align*} Thus $w(I)\le w(G)$ for every feasible set $I$, so the greedy set has maximum profit. Equivalently, each class contributes its most profitable jobs up to its capacity, while the single global scan interleaves those classwise choices by profit. [/example] ## Greedy Correctness Characterizes Matroids The previous section proved that matroids make greedy optimization correct. The converse asks whether this property is special: if an independence system allows greedy optimization for every weight function, must it satisfy exchange? The answer is yes, and it is one of the cleanest ways to recognize the matroid axioms from an optimization problem. [definition: Independence System] An independence system is a pair $(E,\mathcal I)$ where $E$ is finite, $\varnothing\in\mathcal I$, and whenever $I\in\mathcal I$ and $J\subset I$, we have $J\in\mathcal I$. [/definition] An independence system has the hereditary axiom but may fail exchange. The next example isolates the failure and shows how greedy can be misled by a locally attractive element. [example: Greedy Failure Without Exchange] Let $E=\{a,b,c\}$ and let \begin{align*} \mathcal I=\{\varnothing,\{a\},\{b\},\{c\},\{b,c\}\}. \end{align*} This is an independence system: $\varnothing\in\mathcal I$, and every subset of each listed feasible set is again one of the listed feasible sets. It is not a matroid, because $\{a\},\{b,c\}\in\mathcal I$ and $|\{a\}|<|\{b,c\}|$, but the two possible exchanges are not feasible: \begin{align*} \{a\}\cup\{b\}=\{a,b\}\notin\mathcal I. \end{align*} Also, \begin{align*} \{a\}\cup\{c\}=\{a,c\}\notin\mathcal I. \end{align*} Assign weights \begin{align*} w(a)=3,\qquad w(b)=2,\qquad w(c)=2. \end{align*} Since $w(a)>w(b)=w(c)$, greedy scans $a$ first. Starting from $\varnothing$, it accepts $a$ because \begin{align*} \varnothing\cup\{a\}=\{a\}\in\mathcal I. \end{align*} After that, it tests $b$ and $c$. The element $b$ is rejected because \begin{align*} \{a\}\cup\{b\}=\{a,b\}\notin\mathcal I. \end{align*} The element $c$ is rejected for the same reason: \begin{align*} \{a\}\cup\{c\}=\{a,c\}\notin\mathcal I. \end{align*} Thus the greedy output is $\{a\}$, with total weight \begin{align*} w(\{a\})=\sum_{e\in\{a\}}w(e)=w(a)=3. \end{align*} However, $\{b,c\}\in\mathcal I$, and its weight is \begin{align*} w(\{b,c\})=\sum_{e\in\{b,c\}}w(e)=w(b)+w(c)=2+2=4. \end{align*} Since \begin{align*} 4>3, \end{align*} the feasible set $\{b,c\}$ has larger weight than the greedy output. The exchange failure lets the locally best element $a$ block both lower-weight elements, even though together they form a better feasible set. [/example] The example used a hand-built bad weight function, so the next problem is to decide whether every failure of exchange can be detected in this way. This is exactly the converse direction needed to make greedy correctness a characterization rather than only a consequence of the matroid axioms. [quotetheorem:6608] [citeproof:6608] This proof is useful because it turns a structural axiom into an adversarial input for an algorithm. If exchange fails, weights can make greedy fill a smaller maximal independent set inside some subset $S$, while a larger independent set in the same $S$ remains available for comparison. The role of nonnegative weights is essential for the characterization in the independent-set form: the counterexample is built from positive and zero weights, so failure cannot be dismissed as an artefact of negative costs. The quantifiers in the theorem are also part of the content. Correctness for one weight vector does not imply matroidality: in the system $\{\varnothing,\{a\},\{b\},\{c\},\{b,c\}\}$, weights $w(a)=100$ and $w(b)=w(c)=1$ make greedy optimal even though exchange fails. Correctness for one tie-breaking rule is also too weak, because an implementation may happen to scan $b,c$ before $a$ on the weights $w(a)=w(b)=w(c)=1$ and return the larger feasible set. Finally, correctness for a restricted objective class, such as weights that always heavily favour $a$, says only that those particular objectives avoid the obstruction. Matroidality is detected by greedy correctness for every nonnegative linear weight function and every admissible ordering needed by the algorithmic statement. [remark: Tie Breaking And Characterization] The characterization is insensitive to harmless choices about tie-breaking once the quantifier over all weights is included. If a particular implementation fixes a deterministic tie-breaking rule, the converse uses weights with a small perturbation to force the bad choices into the prescribed order. Thus matroidality is detected by robust greedy correctness, not by a lucky ordering on one instance. [/remark] The theorem also gives a practical warning. When a constraint family resembles independence but lacks exchange, a greedy rule may still succeed for special weights, but there is no matroid-level guarantee covering all linear objectives. [example: A Cardinality Constraint With A Hidden Conflict] Consider three tasks $a,b,c$ with $b$ compatible with $c$, while $a$ is incompatible with both $b$ and $c$. The feasible sets are \begin{align*} \mathcal I=\{\varnothing,\{a\},\{b\},\{c\},\{b,c\}\}. \end{align*} This is a cardinality-looking constraint with a hidden conflict: although two tasks can sometimes be chosen, choosing $a$ prevents choosing either of the other two. First assign profits \begin{align*} w(a)=100,\qquad w(b)=1,\qquad w(c)=1. \end{align*} Since $100>1$, greedy scans $a$ before $b$ and $c$. Starting from $\varnothing$, it accepts $a$ because \begin{align*} \varnothing\cup\{a\}=\{a\}\in\mathcal I. \end{align*} It then rejects $b$ and $c$, since \begin{align*} \{a\}\cup\{b\}=\{a,b\}\notin\mathcal I \end{align*} and \begin{align*} \{a\}\cup\{c\}=\{a,c\}\notin\mathcal I. \end{align*} Thus the greedy output is $\{a\}$, with profit \begin{align*} w(\{a\})=w(a)=100. \end{align*} The only feasible set of size $2$ is $\{b,c\}$, and its profit is \begin{align*} w(\{b,c\})=w(b)+w(c)=1+1=2. \end{align*} Since $100>2$, the greedy set is optimal for this profit function. Now assign profits \begin{align*} w(a)=3,\qquad w(b)=2,\qquad w(c)=2. \end{align*} Again greedy scans $a$ first because $3>2$, accepts $a$, and then rejects $b$ and $c$ for the same infeasibility equations \begin{align*} \{a,b\}\notin\mathcal I,\qquad \{a,c\}\notin\mathcal I. \end{align*} So the greedy output again has profit \begin{align*} w(\{a\})=3. \end{align*} But the feasible set $\{b,c\}$ has profit \begin{align*} w(\{b,c\})=w(b)+w(c)=2+2=4. \end{align*} Since $4>3$, greedy is not optimal for this profit function. The same feasible-set structure can make greedy succeed for one weight vector and fail for another, so the structure of the constraint, not its superficial cardinality flavor, determines whether greedy has a theorem behind it. [/example] ## Matroid Polytopes And Linear Optimization Greedy optimization can be rephrased geometrically. Instead of optimizing over independent sets directly, represent each set by its incidence vector in $\mathbb R^E$ and optimize a linear functional over the convex hull of those vectors. The question becomes: which inequalities describe this polytope, and why does the greedy algorithm solve every linear objective over it? [definition: Incidence Vector] Let $E$ be a finite set. The incidence-vector map is \begin{align*} \mathbf{1}_{(\cdot)}:\mathcal P(E)&\to\mathbb R^E, & A&\mapsto \mathbf{1}_A, \end{align*} where $(\mathbf{1}_A)_e=1$ for $e\in A$ and $(\mathbf{1}_A)_e=0$ for $e\notin A$. [/definition] Incidence vectors translate set systems into $0$-$1$ geometry. However, optimizing over a list of incidence vectors still hides the geometry of all feasible mixtures. The next construction records the whole feasible region by taking the convex hull, so that weighted optimization becomes ordinary linear optimization over a polytope. [definition: Independent Set Polytope] Let $M=(E,\mathcal I)$ be a matroid. The independent-set polytope of $M$ is \begin{align*} P_I(M)=\operatorname{conv}\{\mathbf{1}_I:I\in\mathcal I\}\subset\mathbb R^E. \end{align*} [/definition] The convex-hull definition is exact but not yet useful as an inequality description, because it may involve exponentially many independent sets. Rank gives a compact conceptual description: a vector cannot place more total mass on a subset than the rank of that subset. This is not a formal consequence of monotonicity alone. For a general hereditary set system, the inequalities suggested by a rank-like function can admit fractional extreme points, so the matroid exchange axiom is the integrality mechanism behind the clean description. [quotetheorem:5830] [citeproof:5830] The theorem explains why the rank function is the correct polyhedral substitute for independence. It also shows that the many rank inequalities are not accidental: each subset $A$ imposes the statement that no independent set can use more than $r(A)$ elements from $A$. Each displayed hypothesis has a separate role. The inequalities $x_e\ge 0$ prevent the right-hand side from extending in negative coordinate directions; the rank inequalities encode all subset capacity restrictions; and the assumption that $r$ is a matroid rank function supplies the exchange and submodularity properties that force integrality. The theorem does not say that every rank-like capacity system has an integral polytope, nor does it replace the convex-hull definition by a short list of inequalities in general. The hypotheses are doing real work. For the hereditary non-matroid system on $E=\{a,b,c,d\}$ with maximal feasible sets \begin{align*} \{a,b\},\quad \{a,c\},\quad \{b,c\},\quad \{d\}, \end{align*} with all subsets of these sets also feasible, exchange fails because the independent sets $\{d\}$ and $\{a,b\}$ cannot be exchanged. Let $\rho(A)$ denote the maximum size of a feasible subset contained in $A$. The vector $x\in\mathbb R^E$ with $x_a=x_b=x_c=x_d=1/2$ satisfies $x(A)\le \rho(A)$ for every $A\subset E$: if $A$ has feasible rank $1$ then $|A|\le 2$, while if $A$ has feasible rank $2$ then $|A|\le 4$. But $x$ is not in the convex hull of feasible-set incidence vectors. Indeed, any convex combination with $x_d=1/2$ puts half its mass on the singleton $\{d\}$, and the remaining half can contribute at most $1$ in total to the coordinates $a,b,c$, whereas $x_a+x_b+x_c=3/2$. Thus the matroid exchange axiom is exactly what prevents these fractional impostors. Nor does the theorem by itself make optimization polynomial-time, since it lists one rank inequality for every subset of $E$; algorithmic use requires a separation oracle, a rank oracle, or a more compact special description. Its importance is that it identifies the exact convex hull, which is the starting point for later matroid intersection and submodular optimization arguments where greedy must be replaced by stronger polyhedral methods. [example: Uniform Matroid Polytope] For $U_{r,n}$, a subset is independent exactly when it has size at most $r$, so its rank is \begin{align*} r(A)=\min\{|A|,r\}. \end{align*} We compute the rank-inequality description from *Edmonds Matroid Polytope Theorem*. It gives \begin{align*} P_I(U_{r,n}) =\{x\in\mathbb R^n:x_i\ge 0\text{ for all }i,\ \sum_{i\in A}x_i\le \min\{|A|,r\}\text{ for all }A\subseteq \{1,\dots,n\}\}. \end{align*} Taking $A=\{i\}$ gives \begin{align*} x_i\le r(\{i\})=\min\{1,r\}\le 1, \end{align*} so every point in this polytope satisfies $0\le x_i\le 1$. Taking $A=\{1,\dots,n\}$ gives \begin{align*} \sum_{i=1}^n x_i\le r(\{1,\dots,n\})=\min\{n,r\}=r, \end{align*} because $0\le r\le n$ for $U_{r,n}$. Hence \begin{align*} P_I(U_{r,n})\subseteq \{x\in\mathbb R^n:0\le x_i\le 1\text{ for all }i,\ \sum_{i=1}^n x_i\le r\}. \end{align*} Conversely, suppose $x\in\mathbb R^n$ satisfies $0\le x_i\le 1$ for all $i$ and $\sum_{i=1}^n x_i\le r$. For any $A\subseteq\{1,\dots,n\}$, the upper bounds $x_i\le 1$ give \begin{align*} \sum_{i\in A}x_i\le \sum_{i\in A}1=|A|. \end{align*} Since $x_i\ge 0$, we also have \begin{align*} \sum_{i\in A}x_i\le \sum_{i=1}^n x_i\le r. \end{align*} Combining the two inequalities gives \begin{align*} \sum_{i\in A}x_i\le \min\{|A|,r\}=r(A). \end{align*} Thus all subset rank inequalities hold, and therefore \begin{align*} P_I(U_{r,n})=\{x\in\mathbb R^n:0\le x_i\le 1\text{ for all }i,\ \sum_{i=1}^n x_i\le r\}. \end{align*} The singleton constraints and the total-rank constraint already force every other subset constraint, so this uniform example shows how Edmonds' full rank-inequality description can be conceptually exact while containing many redundant inequalities. [/example] The uniform polytope includes incidence vectors of independent sets of every size up to $r$. Many optimization problems in matroid theory fix the size at $r(E)$, so the next object cuts out the face corresponding only to bases. [definition: Base Polytope] Let $M=(E,\mathcal I)$ be a matroid with set of bases $\mathcal B$. The base polytope of $M$ is \begin{align*} P_B(M)=\operatorname{conv}\{\mathbf{1}_B:B\in\mathcal B\}\subset\mathbb R^E. \end{align*} [/definition] The base polytope is the natural object for optimization problems where every feasible solution has the same cardinality, such as spanning trees. The definition by convex hull again leaves open the computational question: which linear inequalities describe this face? Since bases are the independent sets of maximum size, the expected answer is the independent-set description plus one global equation fixing total mass. [quotetheorem:6663] [citeproof:6663] The equality $\sum_{e\in E}x_e=r(E)$ is the essential extra condition distinguishing bases from arbitrary independent sets. Without it the displayed inequalities describe the independent-set polytope, so lower-dimensional feasible solutions remain: in $U_{2,3}$, the vector $\mathbf{1}_{\{1\}}$ satisfies all rank inequalities but represents an independent set of size $1$, not a basis. Thus the base polytope is not just a rank-capacity region; it is the top-rank face of that region. The matroid hypotheses are also needed here, not only the total-rank equality. Return to the hereditary non-matroid system on $\{a,b,c,d\}$ whose maximal feasible sets are $\{a,b\}$, $\{a,c\}$, $\{b,c\}$, and $\{d\}$. Its maximum feasible sets of size $2$ have incidence vectors supported entirely on $\{a,b,c\}$, so their convex hull has $x_d=0$. However, the vector $(1/2,1/2,1/2,1/2)$ satisfies the same rank inequalities and the equality $x(E)=2$, while it has $x_d=1/2$. The failure is that the rank-inequality framework is no longer integral once exchange is absent; the equation selecting top rank cannot repair a non-matroid rank function. The theorem is therefore a description of the base polytope once a matroid is already present, not a general recipe for arbitrary maximum-cardinality feasible sets. It also does not claim that every displayed inequality cuts out a facet, since loops, parallel elements, separations, and special matroids can make many inequalities redundant. Finally, the description may still be exponentially large, because the subset inequalities remain indexed by all $A\subset E$ unless a special representation gives a smaller equivalent system or a separation method. This description is a bridge to later optimization methods. For a graphic matroid it becomes the spanning-tree polytope, where rank inequalities can be translated into cut or connectivity constraints. For matroid intersection, one optimizes over points satisfying the independent-set inequalities of two matroids at once, and the loss of a single greedy rule is compensated by a stronger polyhedral algorithm. For linear programming over bases, the theorem says that optimizing a linear functional over $P_B(M)$ is still the same problem as choosing a maximum-weight basis; the polytope supplies certificates and dual variables even when the computational method is no longer the elementary greedy scan. The description can still be large because it retains all subset rank inequalities, and in applications one often replaces them by equivalent cut, connectivity, or separation conditions adapted to the particular matroid. The base polytope description often has a useful dual form obtained by applying the rank inequality to complements. For spanning trees, it becomes the familiar cut and partition style constraints expressing connectivity. [example: Spanning Tree Polytope From A Graphic Matroid] Let $G=(V,E)$ be a connected graph and let $M(G)$ be its graphic matroid, so independent sets are forests. For $A\subset E$, write $c(A)$ for the number of connected components of the spanning subgraph $(V,A)$. If those components have vertex sets $V_1,\dots,V_{c(A)}$, then a maximal forest in $(V,A)$ has $|V_s|-1$ edges inside the component with vertex set $V_s$. Summing over the components gives \begin{align*} \sum_{s=1}^{c(A)}(|V_s|-1)=\sum_{s=1}^{c(A)}|V_s|-\sum_{s=1}^{c(A)}1. \end{align*} Since the component vertex sets partition $V$, this becomes \begin{align*} \sum_{s=1}^{c(A)}(|V_s|-1)=|V|-c(A). \end{align*} Thus the graphic matroid rank is \begin{align*} r(A)=|V|-c(A). \end{align*} Because $G$ is connected, $c(E)=1$, so \begin{align*} r(E)=|V|-c(E)=|V|-1. \end{align*} By *Base Polytope Inequality Description*, the base polytope of $M(G)$ is the set of all $x\in\mathbb R^E$ satisfying $x_e\ge 0$ for every $e\in E$, \begin{align*} \sum_{e\in E}x_e=|V|-1, \end{align*} and, for every $A\subset E$, \begin{align*} \sum_{e\in A}x_e\le |V|-c(A). \end{align*} Equivalently, \begin{align*} P_B(M(G))=\operatorname{conv}\{\mathbf 1_B:B\text{ is a basis of }M(G)\}. \end{align*} The bases of $M(G)$ are maximal forests of rank $|V|-1$. A forest on $V$ with $|V|-1$ edges has $|V|-(|V|-1)=1$ connected component, so it is a spanning tree. Conversely, every spanning tree is a forest with $|V|-1$ edges, hence a basis of the graphic matroid. Therefore \begin{align*} P_B(M(G))=\operatorname{conv}\{\mathbf 1_T:T\text{ is a spanning tree of }G\}. \end{align*} If $c:E\to\mathbb R$ is an edge-cost function and $T$ is a spanning tree, then the linear objective at the incidence vector of $T$ is \begin{align*} c\cdot \mathbf 1_T=\sum_{e\in E}c(e)(\mathbf 1_T)_e. \end{align*} Since $(\mathbf 1_T)_e=1$ for $e\in T$ and $(\mathbf 1_T)_e=0$ for $e\notin T$, this equals \begin{align*} c\cdot \mathbf 1_T=\sum_{e\in T}c(e). \end{align*} Thus minimizing $c\cdot x$ over the base polytope is exactly the minimum-spanning-tree problem. The vertices are spanning-tree incidence vectors, and the graphic-matroid greedy algorithm scans edges by increasing cost and accepts exactly the edges that do not create a cycle, which is Kruskal's algorithm. [/example] ## Greedy Algorithms As Polyhedral Certificates The greedy theorem and the polytope theorem are two versions of the same phenomenon. The algorithm gives an optimal vertex for each weight vector, while the rank inequalities certify that no fractional point can do better. Understanding this equivalence is useful later when matroid intersection and submodular optimization replace a single matroid by more complicated feasible regions. [explanation: Threshold Decomposition Of A Weight Vector] Assume $E=\{e_1,\dots,e_n\}$ is ordered so that $w(e_1)\ge\cdots\ge w(e_n)\ge 0$. Put $w(e_{n+1})=0$ and $A_i=\{e_1,\dots,e_i\}$. Then $w=\sum_{i=1}^n (w(e_i)-w(e_{i+1}))\mathbf{1}_{A_i}$ as a vector in $\mathbb R^E$. Hence for any $x\in P_I(M)$, \begin{align*} w\cdot x\le \sum_{i=1}^n (w(e_i)-w(e_{i+1}))r(A_i). \end{align*} After the first $i$ steps, the greedy algorithm has produced a maximal independent subset of $A_i$: every element of $A_i$ has already been scanned, and any rejected element was infeasible with the current greedy prefix. In a matroid, every maximal independent subset of $A_i$ is a basis of the restriction $M|A_i$, so it has size $r(A_i)$. Hence the greedy independent set $G$ has exactly $r(A_i)$ selected elements inside $A_i$ after the first $i$ steps, and therefore \begin{align*} w(G)=\sum_{i=1}^n (w(e_i)-w(e_{i+1}))r(A_i). \end{align*} This equality is the bridge from the rank-inequality upper bound to greedy optimality, and it is the polyhedral shadow of the exchange argument. [/explanation] The threshold decomposition is also a model for how rank constraints interact with linear objectives. Rather than checking all independent sets, the proof checks nested subsets determined by the weight vector. [remark: Why Submodularity Appears] Rank submodularity is the hidden convexity behind the polytope theorem. It ensures that the set function $r$ behaves enough like a concave capacity function for the inequalities $x(A)\le r(A)$ to define an integral polytope. Without submodularity, the same-looking inequalities can have fractional vertices that do not correspond to feasible sets. [/remark] The next two chapters use these results structurally: Chapter 4 applies greedy language to complementary bases under duality, and Chapter 5 explains why deletion and contraction preserve the relevant matroidal optimization setting. Later structural theorems will often be read in two languages at once: exchange axioms for proofs, and polyhedral descriptions for optimization. Greedy optimization shows that bases can be selected and compared through weights and exchange. Duality now turns this picture around, replacing bases by their complements and revealing that many matroid statements come naturally in paired forms. # 4. Duality This chapter continues the course's development of matroid structure by introducing the dual matroid. The prerequisite material is the basis axiom system from Chapter 1 and the rank, circuits, flats, and closure language from Chapter 2; the minor operations of deletion and contraction are previewed here and developed systematically in Chapter 5. Duality turns bases into complementary bases, so it converts questions about spanning into questions about independence and converts graph cycles into graph cuts. The main goals are to prove that the dual construction is again a matroid, compute its rank and local obstructions, and explain why planar graph duality is an instance of the same abstract operation. ## Dual Bases and the Dual Matroid If bases are the maximal independent sets of a matroid, the first question is whether their complements carry their own exchange theory. For a matroid $M$ on ground set $E$, a basis $B$ records a maximal amount of independence. Its complement $E \setminus B$ records everything not needed by that basis, and duality asks whether these complements can serve as the bases of another matroid on the same ground set. [definition: Dual Bases] Let $M$ be a matroid on a finite ground set $E$ with set of bases $\mathcal{B}(M)$. The [dual basis](/theorems/414) family is \begin{align*} \mathcal{B}^*(M) = \{E \setminus B : B \in \mathcal{B}(M)\}. \end{align*} [/definition] The definition is economical, but it hides the main issue: the basis axioms require exchange. The next theorem is the construction theorem for the dual matroid, because it proves that complementary bases are not only a family of sets but the bases of a genuine matroid. [quotetheorem:6609] [citeproof:6609] The construction theorem supplies the missing exchange axiom, so the complementary basis family can now be promoted from notation to an object. The matroid-basis hypothesis is essential: on $E=\{1,2,3,4\}$, the family $\{\{1,2\},\{3,4\}\}$ has sets of equal size but fails basis exchange, and its complementary family is the same failing family. The finite-ground-set hypothesis also matters because the construction compares complements and cardinalities inside a fixed finite $E$; for infinite matroids, bases need not be controlled by the finite basis axiom used here. We name the object because the rest of the chapter will compute its rank, circuits, flats, and minors. [definition: Dual Matroid] Let $M$ be a matroid on a finite ground set $E$. The dual matroid $M^*$ is the matroid on $E$ whose bases are \begin{align*} \mathcal{B}(M^*) = \{E \setminus B : B \in \mathcal{B}(M)\}. \end{align*} [/definition] This definition raises an immediate question: does duality lose information, or can the original matroid be recovered from the dual? Since the construction uses complements of bases, the needed consistency result is that applying the same construction a second time returns the starting matroid. [quotetheorem:6662] [citeproof:6662] Double duality says that no information is lost: duality is an involution on matroids with fixed ground set. The conclusion depends on keeping the same ground set, since complements are taken inside $E$ at both stages. A boundary case shows the point: if a rank-one matroid on $\{a,b\}$ is identified with a relabelled copy on $\{x,y\}$ before taking the second dual, the complement operation no longer returns the original named bases on $\{a,b\}$. Thus the theorem is an equality for matroids on the same labelled ground set, while after relabelling it should be read only up to the chosen isomorphism. This result will justify proving only one half of several later dual identities, because applying the same statement to $M^*$ recovers the reverse direction. [example: Dual of a Uniform Matroid] Let $M=U_{r,n}$ on ground set $E$ with $|E|=n$. By definition of the uniform matroid, a subset $B\subset E$ is a basis of $M$ exactly when $|B|=r$. The bases of the dual are the complements of these bases, so \begin{align*} \mathcal{B}(M^*) &=\{E\setminus B : B\subset E,\ |B|=r\}. \end{align*} For such a set $B$, \begin{align*} |E\setminus B|=|E|-|B|=n-r. \end{align*} Conversely, if $C\subset E$ has $|C|=n-r$, then \begin{align*} |E\setminus C|=|E|-|C|=n-(n-r)=r, \end{align*} so $E\setminus C$ is a basis of $U_{r,n}$ and \begin{align*} C=E\setminus(E\setminus C) \end{align*} is a basis of $M^*$. Therefore the bases of $M^*$ are exactly the $(n-r)$-subsets of $E$, which are the bases of $U_{n-r,n}$. Hence \begin{align*} U_{r,n}^* = U_{n-r,n}. \end{align*} Duality therefore exchanges rank $r$ with corank $n-r$ while preserving the full symmetry of the ground set. In particular, $U_{0,n}^*=U_{n,n}$, so $n$ loops become $n$ coloops, and $U_{n,n}^*=U_{0,n}$, so $n$ coloops become $n$ loops. [/example] ## Rank in the Dual The basis definition gives the dual matroid, but it is not a convenient way to compute ranks of arbitrary subsets. The main problem is to express the rank $r^*(A)$ in terms of the original rank function $r$ of $M$. The answer measures how many elements of $A$ can be kept outside a basis of $M$. [quotetheorem:6610] [citeproof:6610] This formula makes duality computable without listing bases. The finite-ground-set hypothesis is again used through the cardinality term $|A|$, and the formula is most useful when the original rank function is already known. The formula does not say that dual rank is determined by $r(A)$ alone: in $U_{1,2}$ with $E=\{a,b\}$, both singletons have original rank $1$, but the value for $A$ is computed using the rank of the complement $E\setminus A$. More sharply, outside the matroid setting the expression need not be a rank function; if a set function on $\{a,b\}$ has $r(E)=2$ and $r(\{a\})=r(\{b\})=0$, then the displayed expression gives negative dual rank on a singleton. Conceptually, $r(E)-r(E\setminus A)$ is the amount of original rank that must be supplied by elements of $A$, so the remaining elements of $A$ are available for independence in the dual. [example: Loops and Coloops Under Duality] Let $e\in E$. Suppose first that $e$ is a loop of $M$. Then no basis $B$ of $M$ contains $e$, so $e\in E\setminus B$ for every basis $B$ of $M$. Since the bases of $M^*$ are exactly the sets $E\setminus B$ with $B$ a basis of $M$, the element $e$ lies in every basis of $M^*$. Hence $e$ is a coloop of $M^*$. The rank formula gives the same conclusion. Because $e$ is a loop, removing $e$ does not lower the rank of $E$, so $r(E\setminus\{e\})=r(E)$. By *Dual Rank Formula* and $r(\varnothing)=0$, \begin{align*} r^*(E)=|E|-r(E)+r(\varnothing)=|E|-r(E). \end{align*} Also, \begin{align*} r^*(E\setminus\{e\})=|E\setminus\{e\}|-r(E)+r(\{e\})=(|E|-1)-r(E)+0=|E|-1-r(E). \end{align*} Therefore, \begin{align*} r^*(E)-r^*(E\setminus\{e\})=(|E|-r(E))-(|E|-1-r(E))=1. \end{align*} Thus deleting $e$ from the whole ground set lowers dual rank by $1$, which is the rank characterization of a coloop. Now suppose that $e$ is a coloop of $M$. Then every basis $B$ of $M$ contains $e$, so no complement $E\setminus B$ contains $e$. Hence $e$ lies in no basis of $M^*$, and therefore $e$ is a loop of $M^*$. Equivalently, since $e$ is a coloop, $r(E\setminus\{e\})=r(E)-1$, so *Dual Rank Formula* gives \begin{align*} r^*(\{e\})=|\{e\}|-r(E)+r(E\setminus\{e\})=1-r(E)+(r(E)-1)=0. \end{align*} A singleton of rank $0$ is a loop, so duality exchanges loops and coloops. [/example] The rank formula also explains why duality reverses spanning and independence information. A subset of the dual has full dual rank exactly when its complement has the least possible original rank defect. [remark: Rank and Corank] For the whole ground set, \begin{align*} r^*(E)=|E|-r(E). \end{align*} The number $|E|-r(E)$ is called the corank or nullity of $M$. Thus the rank of the dual is the corank of the original matroid. [/remark] ## Circuits, Cocircuits, and Flats Rank tells us how large independent sets can be, but the local obstructions to independence are circuits. Duality introduces the corresponding local obstructions in the dual matroid, and these are important enough to receive their own name. [definition: Cocircuit] Let $M$ be a matroid on a finite ground set $E$. A cocircuit of $M$ is a circuit of the dual matroid $M^*$. [/definition] Cocircuits are not an additional structure; they are the circuit structure of $M^*$ read back on the ground set of $M$. The problem is to detect them inside $M$ without first constructing $M^*$, and the right comparison is with the largest proper closed sets in the original matroid. [quotetheorem:6611] [citeproof:6611] This theorem is often the most efficient way to find cocircuits: instead of looking for circuits in the dual, find hyperplanes in the original matroid and take complements. The equivalence uses both maximality and closedness. In $U_{3,5}$, a singleton is a proper flat of rank $1$, but it is not a hyperplane; its complement has four elements, whereas the cocircuits are complements of rank-$2$ hyperplanes and have size $3$. Thus non-maximal proper flats give complements that are too large to be minimally dependent in the dual. It also explains the graph-theoretic cut examples later in the chapter, where hyperplanes encode all edges except a minimal cut. The next example checks the theorem in the uniform case, where hyperplanes are especially explicit. [example: Cocircuits in a Uniform Matroid] Let $1\le r\le n$ and consider $U_{r,n}$ on a ground set $E$ with $|E|=n$. For any $A\subseteq E$, the rank in $U_{r,n}$ is \begin{align*} r_{U_{r,n}}(A)=\min(|A|,r). \end{align*} If $|A|\le r-2$, then choose $e\in E\setminus A$; adjoining $e$ gives \begin{align*} r_{U_{r,n}}(A\cup\{e\})=\min(|A|+1,r)=|A|+1>|A|=r_{U_{r,n}}(A), \end{align*} so $A$ is not a maximal proper flat. If $|A|=r-1$, then \begin{align*} r_{U_{r,n}}(A)=r-1<r=r_{U_{r,n}}(E), \end{align*} and for every $e\in E\setminus A$, \begin{align*} r_{U_{r,n}}(A\cup\{e\})=\min(r,r)=r. \end{align*} Thus the hyperplanes of $U_{r,n}$ are exactly the subsets of size $r-1$. By *[Cocircuits as Complements of Hyperplanes](/theorems/6611)*, the cocircuits of $U_{r,n}$ are exactly the sets $E\setminus H$ where $|H|=r-1$. For such an $H$, \begin{align*} |E\setminus H|=|E|-|H|=n-(r-1)=n-r+1. \end{align*} Conversely, if $C\subseteq E$ has $|C|=n-r+1$, then \begin{align*} |E\setminus C|=|E|-|C|=n-(n-r+1)=r-1, \end{align*} so $E\setminus C$ is a hyperplane and $C$ is a cocircuit. Therefore the cocircuits of $U_{r,n}$ are precisely the subsets of size $n-r+1$, matching the circuits of the dual uniform matroid $U_{n-r,n}$. The boundary cases fit the same interpretation when read carefully. In $U_{0,n}$, every singleton has rank $0$, so every element is a loop; the closure of $\varnothing$ is all of $E$, so the only flat is $E$ and there are no hyperplanes or cocircuits. In $U_{n,n}$, the hyperplanes are the subsets of size $n-1$, and their complements are the singletons, agreeing with the fact that $U_{n,n}^*=U_{0,n}$. [/example] The uniform example shows the simplest cocircuit calculation, but more complicated matroids require a language that keeps track of closure and circuit content at the same time. This motivates cyclic flats, which are the flats that have no element forced as a coloop inside the restriction. [definition: Cyclic Flat] Let $M$ be a matroid on a finite ground set $E$. A cyclic flat of $M$ is a flat $F\subset E$ such that the restriction $M|F$ has no coloops. [/definition] The condition says that every element of $F$ lies in some circuit contained in $F$. The question is whether this mixed closure-and-circuit condition is preserved by duality, and complements are forced by the way hyperplanes and cocircuits were paired above. [quotetheorem:6612] [citeproof:6612] Cyclic flats package both closure and circuit structure, so they are a useful meeting point between the rank and circuit descriptions of duality. The theorem is stronger than the hyperplane-cocircuit correspondence because it applies to flats of all ranks, not just maximal proper flats. It also shows a limitation of looking only at flats: ordinary flats are not simply carried to ordinary flats by complement unless the no-coloop condition is included. This [complement rule](/theorems/4970) will be useful whenever a matroid is encoded by its lattice of cyclic flats. ## Deletion and Contraction Under Duality Minors are formed by deleting elements and contracting elements. Since duality reverses the roles of independence and spanning, the natural question is how these two minor operations transform. The answer is that deletion and contraction are dual operations. [quotetheorem:6613] [citeproof:6613] This identity is one of the reasons duality is structural rather than cosmetic. The proof also shows why contraction is the correct dual operation rather than another deletion: the rank correction term $r_{M^*}(A)$ records exactly the elements made invisible by contracting $A$. A small boundary case prevents a common misreading. In $U_{1,2}$ on $\{a,b\}$ with $A=\{a\}$, the matroid $(M\setminus A)^*$ makes $b$ a loop, while $M^*\setminus A$ makes $b$ a coloop. Thus the dual of deletion is contraction, not deletion in general. The statement is limited to minors on the same named subset $A$, so the ground set after either operation is $E\setminus A$. [example: Contracting a Coloop and Deleting a Loop] Let $E$ be the ground set of $M$, and suppose $e$ is a coloop. Then every basis $B$ of $M$ contains $e$. A basis of $M^*$ has the form $E\setminus B$ with $B$ a basis of $M$, so \begin{align*} e\in B \quad \Longrightarrow \quad e\notin E\setminus B. \end{align*} Thus no basis of $M^*$ contains $e$, which means that $e$ is a loop of $M^*$. The minor identity is exactly the second part of *Deletion Contraction Duality*: \begin{align*} (M/e)^*=M^*\setminus e. \end{align*} On bases, this says the same thing explicitly. Since $e$ is in every basis of $M$, the bases of $M/e$ are \begin{align*} \{B\setminus\{e\}: B\in\mathcal{B}(M)\}. \end{align*} Taking dual bases inside $E\setminus\{e\}$ gives \begin{align*} (E\setminus\{e\})\setminus(B\setminus\{e\}) &=E\setminus B. \end{align*} These are precisely the bases of $M^*$, and none of them contain $e$, so they are also precisely the bases of $M^*\setminus e$. Hence contracting a coloop in $M$ corresponds to deleting the resulting loop in $M^*$. For a graphic matroid, a bridge belongs to every spanning forest basis, so it is a coloop of the cycle matroid. The dual statement says that this bridge becomes a loop in the cographic matroid, and deleting that loop is the dual operation to contracting the bridge. [/example] ## Graphic and Cographic Matroids Graphic matroids turn graph cycles into matroid circuits. Duality asks for the matroid whose circuits are the cocircuits of the graph, so the relevant graph-theoretic objects are minimal cuts rather than cycles. [definition: Bond] Let $G=(V,E)$ be a finite graph. A bond of $G$ is a minimal nonempty edge cut $\delta(S)$, where $\varnothing\subsetneq S\subsetneq V$ and \begin{align*} \delta(S)=\{uv\in E : u\in S,\ v\in V\setminus S\}. \end{align*} [/definition] Bonds are minimal cuts in the same way that cycles are minimal dependent edge sets in the cycle matroid. The bridge from graph theory back to matroids is to prove that these cuts are precisely the cocircuits of the cycle matroid. [quotetheorem:6614] [citeproof:6614] The bonds theorem converts the abstract word cocircuit into the concrete language of cuts. Minimality is essential. In a 4-cycle with cyclically ordered edges $e_1,e_2,e_3,e_4$, the set $\{e_1,e_2,e_3\}$ contains the bond $\{e_1,e_3\}$, but the three-edge set is not itself a bond because it properly contains a smaller cut that already disconnects the graph. Correspondingly, it is dependent in the dual matroid but not minimally dependent. The graph hypothesis also matters: for a cycle matroid, dependence is controlled by graph cycles, so dual dependence is controlled by minimal edge cuts. The dual of a graphic matroid is called cographic; it need not be graphic for an arbitrary graph, so the definition names the matroid before asking whether a graph realizes it. [definition: Cographic Matroid] Let $G$ be a finite graph. The cographic matroid of $G$ is \begin{align*} M^*(G)=M(G)^*. \end{align*} [/definition] The same dual operation appears outside graph theory. If a matroid is represented over a field $k$ by the columns of a matrix $A$, then a representation of the dual can be obtained from linear relations orthogonal to the row space of $A$. In coding-theoretic language, passing from a generator matrix to a parity-check matrix exchanges a linear code with its dual code, and the associated representable matroids undergo the same basis-complement duality. This is why cocircuits can be read as minimal parity checks in representable examples. A graph whose cycle matroid is $M(G)^*$ may not exist, but planarity supplies such a graph through the classical planar dual. The question is whether the cycle matroid of the planar dual graph realizes the cographic matroid of the original graph. [quotetheorem:6615] [citeproof:6615] The planar theorem turns an abstract dual matroid into an actual graph in a familiar case. The plane embedding is part of the hypothesis, since the dual graph is built from faces of that embedding rather than from the abstract graph alone. A non-planar graph such as $K_5$ has a cycle matroid and hence a cographic dual matroid, but there is no plane embedding from which to build a face-dual graph realizing that dual by ordinary graph cycles. Even for a planar abstract graph, changing the embedding can change the face structure used to define $G^\dagger$, so the theorem is about a connected plane graph, not only an abstract planar graph. The following example gives a small computation of that correspondence. [illustration:planar-square-dual] [example: A Square and Its Planar Dual] Let the cyclically ordered edges of the square be $E=\{e_1,e_2,e_3,e_4\}$. In the cycle matroid $M(G)$, bases are spanning trees of $G$. Removing $e_i$ from the 4-cycle leaves a path through all four vertices, so each $E\setminus\{e_i\}$ is a spanning tree. Conversely, a spanning tree on four vertices has three edges, and the only three-edge subsets of $E$ are the sets $E\setminus\{e_i\}$. Hence \begin{align*} \mathcal{B}(M(G))=\{E\setminus\{e_i\}:1\le i\le 4\}. \end{align*} Taking complementary bases in $E$ gives \begin{align*} \mathcal{B}(M(G)^*)=\{\{e_i\}:1\le i\le 4\}. \end{align*} The plane dual $G^\dagger$ has two vertices, one for the inside face and one for the outside face, and four parallel edges $e_1^\dagger,e_2^\dagger,e_3^\dagger,e_4^\dagger$. A single dual edge connects the two vertices and contains no cycle, so it is a spanning tree of $G^\dagger$. Any two distinct dual edges form a 2-edge cycle, so no spanning tree can contain two of them. Therefore \begin{align*} \mathcal{B}(M(G^\dagger))=\{\{e_i^\dagger\}:1\le i\le 4\}. \end{align*} Under the natural edge bijection $e_i\leftrightarrow e_i^\dagger$, the basis family of $M(G)^*$ matches the basis family of $M(G^\dagger)$, so \begin{align*} M(G)^*\cong M(G^\dagger). \end{align*} Equivalently, the cocircuits of the square are its two-edge cuts, and these become exactly the 2-edge cycles in the parallel-edge dual graph. [/example] Planar graph duality is therefore a special case of matroid duality, not a separate phenomenon. The matroid formulation keeps the same exchange, rank, and minor identities even when no planar drawing or dual graph is present. Duality showed that planar graph duality is only one instance of a broader matroidal operation. The next step is to understand how matroids behave under deletion and contraction, the basic ways of passing to smaller structures while preserving the theory. # 5. Deletion, Contraction, and Minors This chapter develops the basic operations used to pass from a matroid to smaller matroids. The prerequisites are the independent-set axioms and bases from Chapter 1, the rank and circuit language from Chapter 2, and dual matroids from Chapter 4. The main goals are to define deletion and contraction, prove that they preserve the matroid axioms, organise their iterates as minors, and connect this calculus to decomposition through direct sums, separations, and two-sums. The guiding point is that "removing an element" has two different meanings in matroid theory. Deletion forgets an element and preserves all independence information on the remaining ground set. Contraction first treats the element as already used, then records the residual independence; this is the operation that matches graph edge contraction and vector-space quotienting. ## Removing Elements While Preserving Independence What should it mean to remove an element from a matroid without changing the independence relations among the remaining elements? For deletion the answer is direct: forget the element and keep exactly the independent sets that avoided it. Contraction is subtler, because in a vector configuration or graph we often want to remove an element after using it to simplify the surrounding structure. [definition: Deletion] Let $M=(E,\mathcal I)$ be a matroid and let $e\in E$. The deletion of $e$ from $M$ is the matroid $M\setminus e$ on ground set $E\setminus\{e\}$ with independent sets \begin{align*} \mathcal I(M\setminus e)=\{I\subset E\setminus\{e\}: I\in\mathcal I(M)\}. \end{align*} [/definition] Restricting the independent sets could in principle break augmentation: two surviving independent sets still need exchange witnesses that also avoid $e$. The point to check is that deletion has not removed the element needed to satisfy the matroid axiom. [quotetheorem:6616] [citeproof:6616] Deletion is the least delicate minor operation: both matroid axioms survive because every set under discussion already avoided $e$. No hypothesis on $e$ is needed; in the one-element loop matroid $U_{0,1}$, deletion gives the empty matroid, and in the one-element coloop matroid $U_{1,1}$, deletion again gives the empty matroid. The theorem does not say that deletion preserves rank, bases, loops, or coloops globally: deleting the coloop from $U_{1,1}$ lowers the rank from $1$ to $0$. This contrasts with contraction, where the attempted rule "$I$ is independent after contraction when $I\cup\{e\}$ was independent before contraction" fails at the empty set if $e$ is a loop. That failure is why the next definition first treats nonloops and why loops need a separate convention. [definition: Contraction Of A Nonloop] Let $M=(E,\mathcal I)$ be a matroid and let $e\in E$ be a nonloop. The contraction of $e$ from $M$ is the matroid $M/e$ on ground set $E\setminus\{e\}$ with independent sets \begin{align*} \mathcal I(M/e)=\{I\subset E\setminus\{e\}: I\cup\{e\}\in\mathcal I(M)\}. \end{align*} [/definition] For contraction, augmentation has a new obstruction: exchanges in $M$ are tested after $e$ has already been adjoined, so a witness must survive after $e$ is removed from the ground set. The nonloop hypothesis is exactly what makes the empty set and the exchange step compatible. [quotetheorem:6617] [citeproof:6617] The nonloop hypothesis is used at the first step of the proof: it guarantees that $\varnothing$ is independent in $M/e$. If $e$ were a loop, for instance in $U_{0,1}$, the same independent-set formula would give no independent sets at all, because $\{e\}$ is dependent; this would not be a matroid. The theorem only proves that a single nonloop contraction is a matroid; it does not by itself prove that repeated contractions are order-independent, nor does it explain how contraction behaves under duality. Those questions need the rank formula and the later duality theorem. The exchange argument also records the intuition behind contraction: every independent set in $M/e$ is measured after reserving one unit of rank for $e$. To keep contraction available for every element, the loop case is defined by a convention compatible with this rank bookkeeping. [definition: Contraction Of A Loop] Let $M$ be a matroid and let $e\in E(M)$ be a loop. The contraction of $e$ from $M$ is \begin{align*} M/e=M\setminus e. \end{align*} [/definition] Can deletion and contraction be computed by a single rank formula? Without such a formula, even simple questions about the order of two contractions would require tracking independent sets after each intermediate operation, and this becomes ambiguous when an element turns into a loop after the first step. The next theorem is needed because later minor arguments use ranks rather than repeatedly unpacking independent-set definitions. [quotetheorem:6618] [citeproof:6618] The deletion formula says that no rank information inside $E\setminus\{e\}$ changes when $e$ is forgotten. The contraction formula says something different: ranks are measured after adjoining $e$, then subtracting the rank contributed by $e$ itself. The subtraction term is necessary: in $U_{1,2}$ with parallel elements $e$ and $f$, the set $\{f\}$ has rank $1$ in $M$, but after contracting $e$ it has rank $r_M(\{e,f\})-r_M(\{e\})=1-1=0$, so $f$ becomes a loop. The theorem does not say that deletion and contraction preserve the rank of the whole matroid, nor does it classify which elements become loops or coloops after the operation. It supplies the calculation needed to answer those questions in concrete cases. This formula is the main technical device used below, because it keeps track of elements whose status changes after earlier deletions or contractions. [example: Deleting A Column From A Vector Matroid] Let the columns of $A$ be indexed by $E$, and let $e\in E$. Write $A\setminus e$ for the matrix obtained by removing the column indexed by $e$, so the columns of $A\setminus e$ are indexed by $E\setminus\{e\}$. For any $I\subset E\setminus\{e\}$, the subfamily of columns indexed by $I$ in $A\setminus e$ is exactly the same subfamily of vectors as the columns indexed by $I$ in $A$. Therefore \begin{align*} I\in\mathcal I(M(A\setminus e)) \Longleftrightarrow \{A_i:i\in I\}\text{ is linearly independent}. \end{align*} By the definition of the vector matroid $M(A)$, the right-hand condition is equivalent to \begin{align*} I\in\mathcal I(M(A)). \end{align*} Since $I$ was already required to avoid $e$, this gives \begin{align*} \mathcal I(M(A\setminus e))=\{I\subset E\setminus\{e\}:I\in\mathcal I(M(A))\}. \end{align*} By the definition of deletion, the right-hand side is $\mathcal I(M(A)\setminus e)$. Hence \begin{align*} M(A\setminus e)=M(A)\setminus e. \end{align*} Thus deleting a column from a representing matrix changes only which columns are available for testing independence; it does not change any linear relation among the columns that remain. [/example] The vector example shows that deletion is a restriction of the represented configuration. The graphic example shows why contraction is not the same as deletion: an edge can be used to identify structure before it disappears. [example: Contracting An Edge In A Graphic Matroid] Let $G$ be a graph, let $e=uv$ be a nonloop edge, and identify the edges of $G/e$ with $E(G)\setminus\{e\}$. For $F\subset E(G)\setminus\{e\}$, the definition of contraction in a nonloop gives \begin{align*} F\in\mathcal I(M(G)/e) \Longleftrightarrow F\cup\{e\}\in\mathcal I(M(G)). \end{align*} By the definition of the graphic matroid, this is equivalent to \begin{align*} F\cup\{e\}\text{ contains no cycle in }G. \end{align*} We now compare this with acyclicity in $G/e$. If $F$ contains a cycle in $G/e$, then lifting that cycle back to $G$ gives either a cycle using only edges of $F$, or a $u$-$v$ path using only edges of $F$; in the second case that path together with $e$ is a cycle in $G$. Hence $F\cup\{e\}$ contains a cycle in $G$. Conversely, if $F\cup\{e\}$ contains a cycle in $G$, then either that cycle avoids $e$, giving the same cycle in $G/e$, or it uses $e$, in which case deleting $e$ from the cycle gives a $u$-$v$ path whose image after contracting $e$ is a cycle in $G/e$. Therefore \begin{align*} F\cup\{e\}\text{ is a forest in }G \Longleftrightarrow F\text{ is a forest in }G/e. \end{align*} Thus the independent sets of $M(G)/e$ are exactly the independent sets of $M(G/e)$ under the natural edge identification, so \begin{align*} M(G)/e\cong M(G/e). \end{align*} Contracting a nonloop edge in the graphic matroid therefore records precisely the usual graph operation of contracting that edge. [/example] ## Minors And Their Calculus Once deletion and contraction are available, the next question is how many different smaller matroids can be extracted from a given one. The answer is controlled by minors: they are obtained by deleting some elements and contracting others, and the order of these operations does not affect the result except for relabelling. [definition: Minor] A matroid $N$ is a minor of a matroid $M$ if there exist disjoint subsets $C,D\subset E(M)$ such that \begin{align*} N\cong M/C\setminus D. \end{align*} [/definition] Here $C$ is the set of contracted elements and $D$ is the set of deleted elements. The remaining ground set is $E(M)\setminus(C\cup D)$, but this notation would be ambiguous if deleting one element before contracting another changed the resulting rank function. [quotetheorem:6619] [citeproof:6619] This commutation result is the reason minors form a stable class of smaller configurations. The hypothesis that $e\ne f$ is essential: deletion and contraction are not two operations that can be applied to the same element in either order, because the element is absent after the first operation. The loop convention is also doing real work; after contracting $e$, a formerly ordinary element $f$ can become a loop, for example when $e$ and $f$ are parallel in $U_{1,2}$. The theorem gives equality only after the common remaining ground set has been identified by the identity map; it does not classify all minors of $M$, and it does not say that two unrelated minors with the same size are isomorphic. What it permits is the notation $M/C\setminus D$, shifting attention away from operation order toward which elements are contracted and which are deleted. [definition: Minor-Closed Class] A class $\mathcal C$ of matroids is minor-closed if, whenever $M\in\mathcal C$ and $N$ is a minor of $M$, then $N\in\mathcal C$. [/definition] Minor-closed classes are central because many natural matroid families are described by operations that survive deletion and contraction. Uniform and projective examples give the first concrete tests for how minors behave. [example: Minors Of A Uniform Matroid] Let $M=U_{2,4}$ on a ground set $E$ with $|E|=4$, and choose $e\in E$. For deletion, take any set $I\subset E\setminus\{e\}$. By the definition of deletion, $I$ is independent in $M\setminus e$ exactly when $I$ is independent in $M$. Since the independent sets of $U_{2,4}$ are precisely the subsets of $E$ of size at most $2$, this means \begin{align*} I\in\mathcal I(M\setminus e)\Longleftrightarrow |I|\le 2. \end{align*} The remaining ground set $E\setminus\{e\}$ has $3$ elements, so its subsets of size at most $2$ are exactly the independent sets of $U_{2,3}$. Hence \begin{align*} U_{2,4}\setminus e\cong U_{2,3}. \end{align*} The element $e$ is a nonloop, because $\{e\}$ has size $1\le 2$ and is therefore independent in $U_{2,4}$. For contraction, take $I\subset E\setminus\{e\}$. By the definition of contraction of a nonloop, \begin{align*} I\in\mathcal I(M/e)\Longleftrightarrow I\cup\{e\}\in\mathcal I(U_{2,4}). \end{align*} Again using the definition of $U_{2,4}$, the right-hand condition is \begin{align*} I\cup\{e\}\in\mathcal I(U_{2,4})\Longleftrightarrow |I\cup\{e\}|\le 2. \end{align*} Because $I\subset E\setminus\{e\}$, the element $e$ is not in $I$, so \begin{align*} |I\cup\{e\}|=|I|+1. \end{align*} Therefore \begin{align*} |I\cup\{e\}|\le 2\Longleftrightarrow |I|+1\le 2. \end{align*} Subtracting $1$ from both sides gives \begin{align*} |I|+1\le 2\Longleftrightarrow |I|\le 1. \end{align*} Thus \begin{align*} I\in\mathcal I(M/e)\Longleftrightarrow |I|\le 1. \end{align*} On the remaining $3$-element ground set, these are exactly the independent sets of $U_{1,3}$, so \begin{align*} U_{2,4}/e\cong U_{1,3}. \end{align*} Deletion keeps the rank bound $2$ on a smaller ground set, while contraction lowers the rank bound by one because every independent set is tested after adjoining the contracted element. [/example] Uniform matroids show the rank shift caused by contraction in its simplest form. The Fano matroid shows the same operation in a finite projective geometry, where quotienting by a point creates parallel classes. [example: Minors Of The Fano Matroid] Let $F_7=M(A)$, where the columns of $A$ are the seven nonzero vectors of $V=\mathbb F_2^3$. Since the only nonzero scalar in $\mathbb F_2$ is $1$, these seven nonzero vectors are exactly the points of the Fano plane. Deleting a point $p$ means removing the corresponding column, so $F_7\setminus p$ is represented by the six remaining nonzero vectors. For contraction, choose coordinates so that the contracted point is \begin{align*} p=(1,0,0). \end{align*} The quotient map \begin{align*} q:\mathbb F_2^3\to \mathbb F_2^3/\langle p\rangle \end{align*} may be identified with \begin{align*} q(x,y,z)=(y,z). \end{align*} For a set $I$ of remaining columns, the definition of contraction gives \begin{align*} I\in\mathcal I(F_7/p) \Longleftrightarrow I\cup\{p\}\text{ is linearly independent in }\mathbb F_2^3. \end{align*} This is the same as independence of the quotient images. Indeed, if \begin{align*} \sum_{i\in I}\alpha_i q(v_i)=0, \end{align*} then by linearity of $q$, \begin{align*} q\left(\sum_{i\in I}\alpha_i v_i\right)=0. \end{align*} Thus \begin{align*} \sum_{i\in I}\alpha_i v_i\in \ker(q)=\langle p\rangle, \end{align*} so there is some $\beta\in\mathbb F_2$ such that \begin{align*} \sum_{i\in I}\alpha_i v_i+\beta p=0. \end{align*} If $I\cup\{p\}$ is linearly independent, then every coefficient $\alpha_i$ and $\beta$ is zero. Conversely, any relation \begin{align*} \sum_{i\in I}\alpha_i v_i+\beta p=0 \end{align*} maps under $q$ to \begin{align*} \sum_{i\in I}\alpha_i q(v_i)+\beta q(p)=0, \end{align*} and since $q(p)=0$, this becomes \begin{align*} \sum_{i\in I}\alpha_i q(v_i)=0. \end{align*} Therefore $F_7/p$ is represented by the six quotient images of the remaining points. The six remaining vectors are \begin{align*} (0,1,0),\ (1,1,0),\ (0,0,1),\ (1,0,1),\ (0,1,1),\ (1,1,1). \end{align*} Their images under $q$ are computed coordinate by coordinate: \begin{align*} q(0,1,0)=(1,0). \end{align*} \begin{align*} q(1,1,0)=(1,0). \end{align*} \begin{align*} q(0,0,1)=(0,1). \end{align*} \begin{align*} q(1,0,1)=(0,1). \end{align*} \begin{align*} q(0,1,1)=(1,1). \end{align*} \begin{align*} q(1,1,1)=(1,1). \end{align*} Thus the contraction has three parallel classes of size $2$, represented by the three nonzero vectors of $\mathbb F_2^2$. Its rank is $2$, because \begin{align*} a(1,0)+b(0,1)=(0,0) \end{align*} implies \begin{align*} (a,b)=(0,0), \end{align*} so $(1,0)$ and $(0,1)$ are independent, and every image lies in the two-dimensional quotient space $\mathbb F_2^2$. Hence contracting a point of the Fano matroid turns the three lines through that point into the three parallel pairs of the quotient rank-$2$ matroid. [/example] The previous examples show that minors detect local configurations. This leads to a major structural philosophy: a minor-closed class should be governed by the minimal matroids that fail to lie in it. [definition: Excluded Minor] Let $\mathcal C$ be a minor-closed class of matroids. A matroid $M$ is an excluded minor for $\mathcal C$ if $M\notin\mathcal C$ and every proper minor of $M$ belongs to $\mathcal C$. [/definition] An excluded minor is a minimal obstruction. The definition is useful because membership in a minor-closed class can be rephrased entirely in terms of whether any of these minimal obstructions occurs as a minor. [quotetheorem:6620] [citeproof:6620] The minor-closed hypothesis is necessary in both directions. For a concrete failure, let $\mathcal C$ be the class containing only $U_{1,1}$. This class is not minor-closed, because deleting the unique element of $U_{1,1}$ gives the empty matroid, which is not in $\mathcal C$. The empty matroid is then an excluded minor under the displayed minimal-obstruction definition, but $U_{1,1}$ itself contains it as a minor while still belonging to $\mathcal C$; the stated characterisation would therefore fail. The proof is also only a minimality argument inside the finite set of minors of $M$; it does not produce an efficient recognition algorithm. The next remark separates this general philosophy from the much harder question of finding finite obstruction lists. [remark: Philosophy Rather Than Classification] The excluded-minor theorem above is a formal minimality statement, not a guarantee that the excluded minors form a finite or manageable list. Deep theorems in matroid theory identify finite excluded-minor lists for special classes, such as binary or regular matroids, and later chapters return to these classification results. [/remark] ## Duality And Minors Duality turns bases into complementary bases, so it should also exchange the two ways of removing an element. The delicate point is that loops and coloops switch under duality: deleting a loop on one side should correspond to contracting a coloop on the other side. If contraction were treated only by the nonloop independent-set definition, this boundary case would be outside the theorem. The next result confirms that the rank-based contraction convention makes deletion in a matroid match contraction in its dual, and contraction in a matroid match deletion in its dual. [quotetheorem:6621] [citeproof:6621] This theorem is one of the main reasons contraction is defined as it is. It ensures that any statement about deletion has a dual statement about contraction, but it also shows why loops and coloops cannot be ignored: they are exchanged by duality. The one-element loop matroid gives the boundary case. If $M=U_{0,1}$, then $M^*=U_{1,1}$ and $e$ is a coloop in $M^*$; the identity $(M\setminus e)^*\cong M^*/e$ requires contraction of that coloop to give the empty matroid. If contraction had been defined only by the nonloop independent-set rule, this side of the identity would not even be covered. The theorem does not say that deletion and contraction are self-dual operations; it says they are interchanged after passing to the dual matroid. It will be used later to dualise minor-closed statements without reproving every deletion argument. [example: Loops And Coloops Under Duality] If $e$ is a loop of $M$, then $e$ belongs to no basis of $M$. A basis of $M^*$ is the complement $E(M)\setminus B$ of a basis $B$ of $M$, so every basis of $M^*$ contains $e$; hence $e$ is a coloop of $M^*$. Since $e$ is a loop in $M$, the loop convention for contraction gives \begin{align*} M/e=M\setminus e. \end{align*} Taking duals of both sides gives \begin{align*} (M/e)^*=(M\setminus e)^*. \end{align*} By *[Duality Interchanges Deletion And Contraction](/theorems/6621)*, \begin{align*} (M\setminus e)^*\cong M^*/e \end{align*} and \begin{align*} (M/e)^*\cong M^*\setminus e. \end{align*} Combining these identities shows that, in this loop case, \begin{align*} M^*/e\cong (M\setminus e)^*=(M/e)^*\cong M^*\setminus e. \end{align*} Thus deletion and contraction agree on the loop side of $M$, while on the dual side the same element is a coloop and deletion is exchanged with contraction. [/example] ## Direct Sums, Connectedness, Separations, And Two-Sums Minors give smaller matroids, but they do not by themselves explain how a large matroid decomposes into pieces. The first decomposition tools are direct sums, connectedness, separations, and two-sums; each measures whether the ground set splits into parts with limited interaction. [definition: Direct Sum] Let $M_1$ and $M_2$ be matroids on disjoint ground sets $E_1$ and $E_2$. Their direct sum is the matroid $M_1\oplus M_2$ on $E_1\cup E_2$ whose independent sets are \begin{align*} \mathcal I(M_1\oplus M_2)=\{I_1\cup I_2: I_1\in\mathcal I(M_1),\ I_2\in\mathcal I(M_2)\}. \end{align*} [/definition] How can we detect the absence of interaction numerically? It is not enough to add the ranks of two arbitrary subsets, since a circuit crossing the two sides would be counted independently on each side but dependent in the whole matroid. The direct sum hypothesis rules out exactly that kind of cross-dependence. The next theorem is needed because rank additivity gives the criterion later used to recognise separations. [quotetheorem:6622] [citeproof:6622] The disjointness of $E_1$ and $E_2$ is essential here, because the proof writes each independent set uniquely as a union of a part from $E_1$ and a part from $E_2$. A specific failure occurs if both summands are copies of $U_{1,1}$ on the same ground element $p$: adding the two ranks would assign rank $2$ to $\{p\}$, which cannot be the rank of a one-element matroid. The theorem is also limited to matroids already presented as a direct sum; it does not assert that every partition with approximately additive ranks comes from such a decomposition. The exact diagnostic is rank additivity across a partition, and that motivates the connectivity function, which measures the failure of such additivity. [definition: Connectivity Function] Let $M$ be a matroid on ground set $E$ with rank function $r$. The connectivity function of $M$ is the function \begin{align*} \lambda_M:2^E\to\mathbb Z \end{align*} defined by \begin{align*} \lambda_M(A)=r(A)+r(E\setminus A)-r(E) \end{align*} for $A\subset E$. [/definition] The value $\lambda_M(A)$ measures how much rank is shared across the partition $(A,E\setminus A)$. A direct sum split is exactly the case where this shared rank is zero. To avoid counting the empty side as a decomposition, the next definition requires both sides of the partition to be nonempty. [definition: Separation] Let $M$ be a matroid on ground set $E$. A partition $(A,E\setminus A)$ with $A\ne\varnothing$ and $E\setminus A\ne\varnothing$ is a separation if \begin{align*} \lambda_M(A)=0. \end{align*} [/definition] This definition is the ordinary $1$-separation convention used here: a separation is a nonempty partition with $\lambda_M(A)=0$. Under this convention a loop or coloop in a matroid with at least two elements gives a separation by taking $A=\{e\}$, while the one-element matroids have no such partition. The complementary notion is needed because minimal-counterexample arguments often begin by showing that no zero-interaction partition can occur. [definition: Connected Matroid] Let $M$ be a matroid. The matroid $M$ is connected if it has no separation. [/definition] Direct sums provide the basic source of separations, while connected matroids are those where no such zero-interaction partition is present. [example: Direct Sum Is Disconnected] Let $M=M_1\oplus M_2$, where $E_1=E(M_1)$ and $E_2=E(M_2)$ are both nonempty. We show that $(E_1,E_2)$ is a separation of $M$. Since the ground sets in a direct sum are disjoint, \begin{align*} E(M)=E_1\cup E_2. \end{align*} Hence \begin{align*} E(M)\setminus E_1=E_2. \end{align*} By the definition of the connectivity function, \begin{align*} \lambda_M(E_1)=r_M(E_1)+r_M(E(M)\setminus E_1)-r_M(E(M)). \end{align*} Substituting the two displayed set identities gives \begin{align*} \lambda_M(E_1)=r_M(E_1)+r_M(E_2)-r_M(E_1\cup E_2). \end{align*} Now apply *Rank In A Direct Sum*. For $E_1$, the intersections with the two summand ground sets are $E_1\cap E_1=E_1$ and $E_1\cap E_2=\varnothing$, so \begin{align*} r_M(E_1)=r_{M_1}(E_1)+r_{M_2}(\varnothing). \end{align*} The empty set has rank $0$, and therefore \begin{align*} r_M(E_1)=r_{M_1}(E_1). \end{align*} Similarly, \begin{align*} r_M(E_2)=r_{M_2}(E_2). \end{align*} Applying the same rank formula to the whole ground set gives \begin{align*} r_M(E_1\cup E_2)=r_{M_1}(E_1)+r_{M_2}(E_2). \end{align*} Therefore \begin{align*} \lambda_M(E_1)=r_{M_1}(E_1)+r_{M_2}(E_2)-\bigl(r_{M_1}(E_1)+r_{M_2}(E_2)\bigr). \end{align*} Expanding the subtraction, \begin{align*} \lambda_M(E_1)=r_{M_1}(E_1)+r_{M_2}(E_2)-r_{M_1}(E_1)-r_{M_2}(E_2). \end{align*} The two rank terms cancel pairwise, so \begin{align*} \lambda_M(E_1)=0. \end{align*} Since both $E_1$ and $E_2$ are nonempty, $(E_1,E_2)$ is a separation. Thus a nontrivial direct sum is not connected. [/example] Direct sums split a matroid completely. The next operation glues two matroids along a single common element and then deletes that common element, producing a controlled interaction rather than a disjoint union. The hypotheses on the common element matter: if the shared element were a loop, circuits through it would not transmit dependence between the two sides, while if it were a coloop, deleting it after gluing would collapse the intended interface. The circuit description below is therefore a theorem-level construction packaged as a definition; under these hypotheses, the displayed circuit family is the circuit set of a matroid. [definition: Two-Sum] Let $M_1$ and $M_2$ be matroids with \begin{align*} E(M_1)\cap E(M_2)=\{p\}. \end{align*} Assume $p$ is neither a loop nor a coloop in either matroid. The two-sum $M_1\oplus_2 M_2$ is the matroid on $(E(M_1)\cup E(M_2))\setminus\{p\}$ whose circuits are the circuits of $M_1\setminus p$, the circuits of $M_2\setminus p$, and all sets of the form \begin{align*} (C_1\cup C_2)\setminus\{p\}, \end{align*} where $C_i$ is a circuit of $M_i$ containing $p$. [/definition] The two-sum is the first nontrivial gluing operation in matroid structure theory. In graphic matroids it corresponds to gluing graphs along an edge and then deleting the glued edge, under the usual hypotheses that prevent degenerate loops or bridges. [remark: Decomposition Tools] Deletion and contraction reduce size, direct sums identify disconnected parts, and two-sums expose a low-order interface between connected pieces. These operations become especially powerful when combined with excluded-minor arguments: a minimal counterexample to a minor-closed statement is often forced to be connected, and sometimes forced to avoid low-order decompositions. [/remark] Minors make it possible to study matroids by removing or contracting elements and by isolating minimal obstructions. With that language in place, we can ask when an abstract matroid actually comes from linear dependence over a field. # 6. Representability over Fields Representability asks when an abstract matroid can be realised by linear dependence among vectors over a field. The chapter assumes the earlier material on independent sets, bases, circuits, rank, deletion, contraction, and minors, together with the linear-algebra facts that column independence is preserved by invertible row operations and non-zero column rescaling. Chapter 1 treated vector matroids as examples, and Chapter 2 showed how their ranks recover dimensions of spans; this chapter reverses the viewpoint and studies which matroids come from matrices, how much of the matrix is remembered by the matroid, and how the answer changes with the field. The main theme is that field arithmetic leaves combinatorial traces: binary matroids, ternary matroids, regular matroids, and orientable matroids form increasingly structured classes with strong minor characterisations. ## Matrix Representations and Projective Equivalence The first problem is to make precise what it means for a matrix to represent a matroid. A matrix has coordinates, scalars, and a chosen order of columns, while a matroid only records which subsets of the ground set are independent. The definition therefore labels columns by the ground set and forgets all data except linear independence. [definition: Matrix Representation] Let $M$ be a matroid on a finite ground set $E$, and let $k$ be a field. A matrix $A$ over $k$ with columns indexed by $E$ is a representation of $M$ over $k$ if, for every $I \subset E$, the set $I$ is independent in $M$ if and only if the columns $(A_e)_{e \in I}$ are linearly independent over $k$. [/definition] The same definition can be phrased as $M = M[A]$, the column matroid of $A$. Loops are represented by zero columns, parallel elements are represented by non-zero scalar multiples of the same column, and the rank of $M$ is the row rank of $A$ after deleting redundant rows. [example: Uniform Matroid From General Position Columns] Let $k$ be an infinite field, choose distinct elements $a_1,\dots,a_n \in k$, and set \begin{align*} v_i=(1,a_i,a_i^2,\dots,a_i^{r-1})^\top \in k^r. \end{align*} We show that the matrix $A=[v_1\ \cdots\ v_n]$ represents $U_{r,n}$. Let $S=\{i_1,\dots,i_m\}$ with $m\le r$. Consider the $m\times m$ minor obtained from the columns indexed by $S$ and the first $m$ rows. Its entry in row $\ell$ and column $p$ is $a_{i_p}^{\ell-1}$, so it is the Vandermonde matrix $V_S$ with determinant \begin{align*} \det(V_S)=\prod_{1\le p<q\le m}(a_{i_q}-a_{i_p}). \end{align*} To justify the formula, view $\det(V_S)$ as a polynomial in $a_{i_1},\dots,a_{i_m}$. Swapping two variables swaps two columns, so the determinant is alternating; hence it is divisible by each factor $a_{i_q}-a_{i_p}$. The determinant has total degree $0+1+\cdots+(m-1)$, the product has the same total degree, and the coefficient of $a_{i_2}a_{i_3}^2\cdots a_{i_m}^{m-1}$ is $1$ on both sides. Therefore the displayed product is the determinant. Since the $a_i$ are distinct, every factor $a_{i_q}-a_{i_p}$ is non-zero, so $\det(V_S)\ne 0$. Thus the columns indexed by $S$ are linearly independent. If $|S|>r$, then the columns indexed by $S$ lie in the $r$-dimensional vector space $k^r$, so they are linearly dependent. Hence the independent column sets of $A$ are exactly the subsets of $\{1,\dots,n\}$ of size at most $r$, which are precisely the independent sets of $U_{r,n}$. [/example] This example shows that representability can depend on the size and characteristic of the field, but it also raises a more basic comparison problem: which changes to the representing matrix are only changes of coordinates? We need an invariance result before we can speak about different matrices representing the same labelled matroid. [quotetheorem:6623] [citeproof:6623] The hypotheses on the operations are essential because the matroid records all column-dependence relations, not merely the row space or the displayed entries. A non-invertible row operation may collapse two distinct non-zero columns to parallel columns or to zero columns; for instance, projecting the two columns $(1,0)^\top$ and $(0,1)^\top$ in $k^2$ onto the first coordinate changes an independent pair into a dependent pair containing a zero column. Multiplying a column by $0$ has the same defect, since it turns a non-loop into a loop. The theorem also does not say that every two matrices representing the same labelled matroid are related by these operations; it only identifies a family of transformations that preserve the represented matroid. The theorem explains why a representation is better viewed projectively: a non-loop element is a point of projective space, and parallel elements are repeated points. To compare two representations without mistaking coordinate choices for new combinatorics, we package exactly these allowed operations into an [equivalence relation](/page/Equivalence%20Relation). [definition: Projective Equivalence] Let $A$ and $B$ be representations over a field $k$ of matroids on the same labelled ground set $E$. They are projectively equivalent if $B = D_1 A D_2$ after possibly deleting or adjoining zero rows, where $D_1$ is an invertible matrix over $k$ and $D_2$ is an invertible diagonal matrix indexed by $E$. [/definition] Projective equivalence is stronger than representing the same matroid: two matrices may define the same independent sets without being obtainable from each other by these operations. The distinction becomes important in representation theory over small fields, where uniqueness of representation is often a structural property rather than part of the definition. [example: Parallel Columns and Projective Points] Let $v\in k^r$ be non-zero, and assume $2\ne 0$ in $k$, so $v$, $2v$, and $-v$ are all non-zero scalar multiples of $v$. Each singleton is independent: if $\alpha v=0$, then $\alpha=0$ because $v\ne 0$; if $\alpha(2v)=0$, then $(2\alpha)v=0$, so $2\alpha=0$ and hence $\alpha=0$; and if $\alpha(-v)=0$, then $(-\alpha)v=0$, so $\alpha=0$. Every pair is dependent, as shown by the non-trivial relations \begin{align*} 2v-(2v)=0. \end{align*} \begin{align*} v+(-v)=0. \end{align*} \begin{align*} (2v)+2(-v)=0. \end{align*} In each displayed relation the coefficients are non-zero because $1\ne 0$ and $2\ne 0$ in $k$. Thus the represented matroid has three parallel non-loop elements: the matroid sees independent singletons and dependent pairs, while projectively all three columns determine the same one-dimensional subspace $kv$. Rescaling the second column by $1/2$ and the third by $-1$ changes the displayed columns to $v,v,v$, so these coordinate choices change the matrix but not the labelled matroid. [/example] ## Binary, Ternary, Regular, and Orientable Matroids The next question is how the choice of field affects the class of representable matroids. Since linear dependence uses addition and scalar multiplication, changing the characteristic can create or destroy dependencies. The most important first classes are named by the fields over which representations exist. [definition: Binary Ternary and Regular Matroids] A matroid $M$ is binary if it is representable over $\mathbb F_2$. It is ternary if it is representable over $\mathbb F_3$. It is regular if it is representable over every field. [/definition] Regularity is a field-independent form of linear representability. Binary and ternary representability are much more restrictive than representability over some large field, because the small fields allow very few possible projective configurations. [example: The Uniform Matroid U24 Is Not Binary] Consider $U_{2,4}$ with ground set $\{1,2,3,4\}$, and suppose for contradiction that it has a representation over $\mathbb F_2$ with columns $c_1,c_2,c_3,c_4$. Since every singleton is independent in $U_{2,4}$, no column is zero: if $c_i=0$, then \begin{align*} 1\cdot c_i=0 \end{align*} would be a non-trivial dependence relation. Since $U_{2,4}$ has rank $2$, the span of all four columns has dimension $2$. Choose an independent pair, say $c_1,c_2$. Over $\mathbb F_2$, each vector in the span of this pair has the form $\alpha c_1+\beta c_2$ with $\alpha,\beta\in\{0,1\}$, so the four possible vectors are \begin{align*} 0c_1+0c_2=0, \end{align*} \begin{align*} 1c_1+0c_2=c_1, \end{align*} \begin{align*} 0c_1+1c_2=c_2, \end{align*} \begin{align*} 1c_1+1c_2=c_1+c_2. \end{align*} The vector $c_1+c_2$ is non-zero, because otherwise $c_1+c_2=0$ would be a non-trivial dependence relation among $c_1$ and $c_2$. Thus the span contains exactly three non-zero vectors: \begin{align*} c_1,\quad c_2,\quad c_1+c_2. \end{align*} All four columns must be non-zero vectors in this three-element set. Hence two columns are equal, say $c_i=c_j$ with $i\ne j$. Then \begin{align*} c_i+c_j=c_i+c_i=(1+1)c_i=0c_i=0 \end{align*} in $\mathbb F_2$, so $\{i,j\}$ is dependent. This contradicts the defining property of $U_{2,4}$ that every pair is independent. Therefore $U_{2,4}$ is not binary. [/example] The obstruction in this example is not accidental. Tutte's theorem says that the absence of this single minor characterises binary matroids. This is the first major sign that representability over a fixed finite field can be controlled by excluded minors. [quotetheorem:6661] [citeproof:6661] The result uses fundamental circuits, which will be developed in the next section. For now, the theorem gives a practical test: binary matroids are exactly those in which the four-point line cannot appear after deleting and contracting elements. The word "minor" is needed here, not just "restriction": the uniform matroid $U_{3,5}$ has no restriction isomorphic to $U_{2,4}$ because every four-element restriction has rank $3$, but contracting any element gives $U_{2,4}$, so it is not binary. This shows why deletion and contraction both belong in the hypothesis. The theorem does not say that a binary matroid has a unique binary representation, nor does it decide representability over fields other than $\mathbb F_2$; it characterises only the existence of at least one binary representation. This still leaves a different question: what if a matroid carries consistent signs of dependencies rather than coordinates over a chosen field? [definition: Orientable Matroid] A matroid $M$ is orientable if its circuits can be assigned signed versions satisfying the signed circuit elimination axioms of oriented matroid theory. [/definition] Representability over an ordered field, such as $\mathbb R$, gives an orientation by recording signs in each linear dependence among a circuit. Orientability is weaker than real representability: it is a combinatorial shadow of signs, not a guarantee that actual real coordinates exist. [example: The Fano and Non-Fano Matroids] Over $\mathbb F_2$, label the seven non-zero vectors of $\mathbb F_2^3$ by \begin{align*} e_1,\ e_2,\ e_3,\ e_1+e_2,\ e_1+e_3,\ e_2+e_3,\ e_1+e_2+e_3. \end{align*} For example, \begin{align*} e_1+e_2+(e_1+e_2)=(1+1)e_1+(1+1)e_2=0e_1+0e_2=0, \end{align*} so $\{e_1,e_2,e_1+e_2\}$ is dependent. Its proper subsets are independent because $e_1$ and $e_2$ are non-zero and not scalar multiples of one another, and $e_1+e_2$ is not equal to $0$, $e_1$, or $e_2$. Hence this triple is a circuit. The same calculation shows that every projective line has the form $\{u,v,u+v\}$ with $u,v$ independent, since \begin{align*} u+v+(u+v)=0. \end{align*} Thus the seven non-zero vectors of $\mathbb F_2^3$ represent the Fano matroid $F_7$. Now suppose the same incidence pattern were represented over a field $k$ of odd characteristic. Choose three non-collinear points and rescale their columns to be $e_1,e_2,e_3$. The point on the line through $e_1$ and $e_2$ may be rescaled to $e_1+e_2$, because a vector in $\operatorname{span}(e_1,e_2)$ not parallel to either basis vector has the form $\alpha e_1+\beta e_2$ with $\alpha,\beta\ne 0$, and column rescaling changes it to $e_1+(\beta/\alpha)e_2$; after rescaling $e_2$ within the chosen coordinates, this point is $e_1+e_2$. Similarly, take the points on the lines through $e_1,e_3$ and through $e_2,e_3$ to be $e_1+e_3$ and $e_2+e_3$. In the Fano incidence pattern these three new points are collinear, so the columns \begin{align*} e_1+e_2,\quad e_1+e_3,\quad e_2+e_3 \end{align*} must be linearly dependent. If \begin{align*} a(e_1+e_2)+b(e_1+e_3)+c(e_2+e_3)=0, \end{align*} then collecting the $e_1,e_2,e_3$ coordinates gives \begin{align*} a+b=0,\qquad a+c=0,\qquad b+c=0. \end{align*} From $a+b=0$ and $a+c=0$ we get $b=-a$ and $c=-a$. Substituting into $b+c=0$ gives \begin{align*} (-a)+(-a)=-2a=0. \end{align*} Since the three points are meant to form a circuit, the dependence has some non-zero coefficient; taking $a\ne 0$ forces $2=0$, contradicting odd characteristic. Therefore $F_7$ is not representable over fields of odd characteristic. Over any field with $2\ne 0$, the displayed seven columns instead represent the non-Fano matroid, obtained by relaxing the circuit-hyperplane \begin{align*} \{e_1+e_2,\ e_1+e_3,\ e_2+e_3\}. \end{align*} Indeed, the preceding coordinate equations show that a dependence among these three columns forces $a=0$ from $-2a=0$, and then $b=0$ and $c=0$, so the triple is independent. The other three-point line dependencies are still witnessed by equations such as \begin{align*} e_1+e_2-(e_1+e_2)=0,\qquad e_1+e_3-(e_1+e_3)=0,\qquad e_2+e_3-(e_2+e_3)=0. \end{align*} In characteristic two, the same relaxed triple cannot become independent because the calculation gives $-2a=0$ for every $a$. Thus the Fano and non-Fano matroids separate binary representability from odd-characteristic representability already in rank $3$. [/example] These examples are the standard warning that a drawing of points and lines does not determine a field-free object unless the incidence axioms are checked in the relevant algebra. The Fano plane is natural over characteristic two, while the non-Fano configuration is natural over odd characteristic. ## Fundamental Circuits and Standard-Form Representations To construct and compare representations, we now ask how much of a matrix is determined once a basis has been chosen. In linear algebra, a basis lets us put a full-rank matrix into identity-plus-extra-columns form. In matroid language, the extra columns are controlled by fundamental circuits. [definition: Fundamental Circuit] Let $M$ be a matroid on $E$, let $B$ be a basis of $M$, and let $e \in E \setminus B$. The fundamental circuit of $e$ with respect to $B$ is the unique circuit contained in $B \cup \{e\}$, denoted $C_B(e)$. [/definition] This definition names the circuit that will become the support of a non-basis column in standard form. The next theorem is needed because the definition depends on existence and uniqueness inside $B \cup \{e\}$, and those facts must be derived from the matroid axioms before the notation $C_B(e)$ can be used reliably. [quotetheorem:6624] [citeproof:6624] The requirement that $B$ be a basis is doing two jobs. If $B$ is only independent, existence may fail: in $U_{2,3}$, an independent singleton $\{b\}$ together with another element $e$ is still independent, so there is no circuit in $\{b,e\}$. If $B$ is only spanning and not independent, uniqueness can fail because circuits already inside $B$ may coexist with circuits using $e$; in $U_{2,4}$, a three-element spanning set contains a circuit, and adjoining a fourth element creates further three-element circuits. The theorem therefore does not describe all circuits of $M$, only the unique circuit created by adding one outside element to a chosen basis. The theorem turns a basis into a coordinate system for circuits. In a matrix representation, after row operations and column rescaling, the basis columns can be made into an identity matrix; the next definition names this normal form so that the circuit supports can be read directly from the matrix. [definition: Standard Form Representation] Let $M$ be a rank-$r$ matroid represented over a field $k$, and let $B=\{b_1,\dots,b_r\}$ be a basis. A representation is in standard form with respect to $B$ if the columns labelled by $B$ form the identity matrix $I_r$, so the full matrix has the block form \begin{align*} A = [I_r \mid D], \end{align*} with the remaining columns indexed by $E \setminus B$. [/definition] Once the matrix is in standard form, a non-basis column may still have several nonzero coordinates, and a priori those coordinates could reflect the chosen linear coordinates rather than the matroid circuit. The obstruction to remove is this coordinate dependence: the support must be shown to coincide exactly with the fundamental circuit support. [quotetheorem:6625] [citeproof:6625] The standard-form and basis hypotheses are what make the support readable. Before normalising the basis columns to $I_r$, the same element may have coordinates whose non-zero entries reflect the chosen ambient basis rather than the fundamental circuit in the matroid. If the displayed columns are not a matroid basis, then adding $e$ need not create a unique fundamental circuit, so there is no single support set for the theorem to recover. Over fields larger than $\mathbb F_2$, the support still identifies which basis elements occur in $C_B(e)$, but it does not determine the coefficients: columns $(1,1)^\top$ and $(1,2)^\top$ over $\mathbb F_3$ have the same support and different dependence coefficients. Thus the binary consequence below is special to the two-element field. This support rule is often the fastest way to build a representation from combinatorial data. Over $\mathbb F_2$, the entries are only $0$ and $1$, so the standard-form matrix is determined by the fundamental circuits relative to $B$. [example: Binary Standard Form From Fundamental Circuits] Let $M$ be binary with basis $B=\{b_1,b_2,b_3\}$ and non-basis elements $e,f$. In standard form over $\mathbb F_2$, the basis columns are $A_{b_1}=(1,0,0)^\top$, $A_{b_2}=(0,1,0)^\top$, and $A_{b_3}=(0,0,1)^\top$. By *Support of a Standard Form Column*, the support of the $e$-column is determined by $C_B(e)\setminus\{e\}=\{b_1,b_3\}$. Thus the only non-zero entries of $A_e$ occur in rows $1$ and $3$. Since the only non-zero element of $\mathbb F_2$ is $1$, we get \begin{align*} A_e=(1,0,1)^\top. \end{align*} The associated circuit dependence is visible from \begin{align*} A_{b_1}+A_{b_3}+A_e=(1,0,0)^\top+(0,0,1)^\top+(1,0,1)^\top=(1+1,0,1+1)^\top=(0,0,0)^\top. \end{align*} Similarly, $C_B(f)\setminus\{f\}=\{b_2,b_3\}$, so the $f$-column has non-zero entries exactly in rows $2$ and $3$. Again using that $1$ is the only non-zero scalar in $\mathbb F_2$, \begin{align*} A_f=(0,1,1)^\top. \end{align*} The corresponding dependence is \begin{align*} A_{b_2}+A_{b_3}+A_f=(0,1,0)^\top+(0,0,1)^\top+(0,1,1)^\top=(0,1+1,1+1)^\top=(0,0,0)^\top. \end{align*} Therefore the standard-form representation has columns ordered as $b_1,b_2,b_3,e,f$, with identity columns for $b_1,b_2,b_3$, $e$-column $(1,0,1)^\top$, and $f$-column $(0,1,1)^\top$. This shows why binary standard-form representations are especially rigid: the fundamental circuits determine the non-basis columns because support and coefficient data coincide over $\mathbb F_2$. [/example] ## Regular Matroids and Total Unimodularity The final problem in this chapter is to recognise matroids representable over every field. A matrix over $\mathbb Z$ can be reduced modulo any prime and interpreted over any field, but arbitrary integer matrices may acquire new dependencies after reduction. Total unimodularity is the condition preventing this loss of information. [definition: Totally Unimodular Matrix] A real matrix $A$ is totally unimodular if every square subdeterminant of $A$ is equal to $0$, $1$, or $-1$. [/definition] For a totally unimodular matrix, every nonsingular square submatrix remains nonsingular over every field after interpreting $1$ and $-1$ in that field. Thus the column-dependence matroid is independent of the field. [quotetheorem:6626] [citeproof:6626] The theorem proves that total unimodularity is a sufficient certificate for regularity, and the determinant condition cannot be weakened to arbitrary integer entries. For example, the one-column matrix $[2]$ represents a non-loop over $\mathbb Q$, but over $\mathbb F_2$ the same column becomes zero and represents a loop; the dependence pattern has changed because the determinant $2$ vanishes after reduction. Total unimodularity prevents this by allowing only determinants that remain non-zero in every field when they are non-zero over $\mathbb Q$. At this point it is still only a sufficient condition, not the definition of regularity, because regularity was defined as representability over every field; the converse theorem below is the result that identifies the two viewpoints. The natural recognition question is whether every regular matroid admits such an integral certificate, or whether some regular matroids require different matrices over different fields. This is a stronger question than the preceding sufficient direction, because regularity is defined by a family of representations, one for each field, while total unimodularity asks for a single integer matrix whose minors control all characteristics at once. Answering this question is also what connects regular matroids to the matrices used in network flows and linear programming. Incidence matrices and network matrices are useful precisely because their determinant behaviour is stable under changes of field; the theorem below says that regular matroids are exactly the matroids with that kind of field-independent integral model. [quotetheorem:6627] [citeproof:6627] For this course, the theorem is mainly used as a recognition principle: regular matroids are precisely those whose dependence relations can be encoded integrally without field-specific accidents. The regularity hypothesis is necessary for such a certificate, because a binary matroid such as the Fano matroid cannot have a totally unimodular representation: if it did, the preceding theorem would make it representable over every field, including fields of odd characteristic. The theorem also does not imply uniqueness of representation; a regular matroid may have many projectively inequivalent matrices, and total unimodularity asserts the existence of at least one well-behaved integral representative. Graphic matroids are the basic family where this certificate appears directly from a familiar matrix. [example: Graphic Matroids Are Regular] Let $G=(V,E)$ be an oriented graph, and let $N$ be its signed vertex-edge incidence matrix: a non-loop edge $e:u\to v$ has entry $-1$ in row $u$, entry $1$ in row $v$, and $0$ in all other rows; a loop has the zero column. Delete one row from each connected component and call the resulting matrix $B$. First, $B$ represents the graphic matroid $M(G)$. If $F\subseteq E$ contains a cycle $e_1,\dots,e_m$, orient the cycle cyclically and choose $\epsilon_j=1$ if the orientation of $e_j$ agrees with the cycle direction and $\epsilon_j=-1$ otherwise. At each vertex of the cycle, one incident cycle edge contributes $1$ and the other contributes $-1$, so \begin{align*} \epsilon_1B_{e_1}+\cdots+\epsilon_mB_{e_m}=0. \end{align*} Thus any edge set containing a cycle is column-dependent. Conversely, suppose $F$ is a forest and \begin{align*} \sum_{e\in F} a_eB_e=0. \end{align*} In any non-empty forest component, choose a leaf vertex whose row was not deleted; this is possible because the deleted row is at most one vertex of that tree. If $e$ is the unique edge of $F$ incident with that leaf, then the leaf row of the displayed relation is either $a_e=0$ or $-a_e=0$, so $a_e=0$. Removing that edge and repeating gives $a_e=0$ for every $e\in F$. Hence the columns indexed by $F$ are independent exactly when $F$ is acyclic, which is the definition of the independent sets of $M(G)$. It remains to check total unimodularity. Let $Q$ be any square submatrix of $B$. If some column of $Q$ is zero, then $\det(Q)=0$. If some column of $Q$ has exactly one non-zero entry, that entry is $1$ or $-1$; expanding along that column gives \begin{align*} \det(Q)=\pm \det(Q'), \end{align*} where $Q'$ is a smaller square submatrix of $B$. Repeating this reduction either reaches a $1\times 1$ matrix with determinant $0$, $1$, or $-1$, or reaches the remaining case in which every column has exactly two non-zero entries. In that remaining case each column contains one $1$ and one $-1$, so the sum of all rows of $Q$ is the zero row: \begin{align*} \text{row}_1(Q)+\cdots+\text{row}_m(Q)=0. \end{align*} The rows are linearly dependent, and therefore $\det(Q)=0$. Thus every square subdeterminant of $B$ is $0$, $1$, or $-1$, so $B$ is totally unimodular. By *Totally Unimodular Matrices Represent Regular Matroids*, the column matroid of $B$, namely $M(G)$, is regular. [/example] Graphic matroids therefore provide the guiding family of regular matroids: their representations come from incidence matrices, the same matrices that appear in network-flow constraints. This gives a concrete reason for caring about regularity beyond field classification. When a combinatorial optimisation problem has a totally unimodular constraint matrix, linear programming relaxations have integral vertices, so the matroidal determinant condition is tied to integer solutions appearing without extra rounding. Later structure theorems explain that regular matroids are built from graphic matroids, cographic matroids, and one exceptional ten-element matroid by controlled sums. The present chapter only needs the representability viewpoint, but this hierarchy is a useful warning: representability classes are not arranged by a single geometric picture. They mix projective geometry over small fields, signed dependence over ordered fields, and integral matrix behaviour from total unimodularity. [remark: Hierarchy of Representation Classes] Every regular matroid is binary and ternary, and every matroid representable over $\mathbb R$ is orientable. The converses fail: $F_7$ is binary but not regular, the non-Fano matroid is representable over odd-characteristic fields but not binary, and there exist orientable matroids not representable over $\mathbb R$. These failures make representability a field-sensitive invariant rather than a single yes-or-no property. [/remark] Representability reveals that linear models of matroids depend sharply on the chosen field. We now return to graphs, the original motivating examples, and study graphic and cographic matroids as especially robust representable classes. # 7. Graphic and Cographic Matroids Graphic matroids are where the course returns to its first motivating example and asks how much graph theory survives after only independence data is kept. The prerequisites are the matroid notions developed earlier in the course, especially circuits, cocircuits, rank, deletion, contraction, duality, and representability, together with the basic graph-theoretic language of cycles, forests, cuts, connected components, and planar embeddings. A graph supplies cycles, cuts, deletion, and contraction, while a matroid supplies circuits, cocircuits, deletion, contraction, and duality. This chapter compares these languages: first by identifying graph operations with matroid operations, then by asking when two graphs give the same matroid, and finally by using duality to detect planarity. ## Cycles, Forests, and Cuts as Matroid Data The first question is: if a matroid is built from a graph, which graph-theoretic features become intrinsic matroid features? For cycle matroids, the answer begins with forests and cycles. Throughout this section, graphs may have loops and parallel edges, since these correspond to loops and parallel elements in the associated matroid. [definition: Cycle Matroid] Let $G$ be a finite graph with edge set $E(G)$. The cycle matroid $M(G)$ is the matroid on ground set $E(G)$ whose independent sets are the edge sets of forests in $G$. [/definition] Thus the circuits of $M(G)$ are precisely the edge sets of graph cycles, with a loop edge forming a one-element circuit and a pair of parallel edges forming a two-element circuit. This is the first example where the matroid remembers dependence but not the full placement of vertices. [example: Cycle Matroid of a Triangle and a Square] Let the edges of $C_3$ be $e_1,e_2,e_3$. A subset with $0$, $1$, or $2$ edges is a forest: with two edges the subgraph is a path on three vertices, and with fewer edges it has no cycle. The full set $\{e_1,e_2,e_3\}$ is the triangle cycle, so it is dependent, and no proper subset is dependent. Hence the independent sets of $M(C_3)$ are exactly the subsets of size at most $2$, which is the definition of the uniform matroid $U_{2,3}$: \begin{align*} M(C_3)\cong U_{2,3}. \end{align*} Similarly, let the edges of $C_4$ be $f_1,f_2,f_3,f_4$ in cyclic order. Any three of these edges form a path on four vertices, so every subset of size at most $3$ is a forest. The full set $\{f_1,f_2,f_3,f_4\}$ is the square cycle, so it is dependent, and it is the only dependent set minimal under inclusion. Thus the independent sets of $M(C_4)$ are exactly the subsets of size at most $3$, giving \begin{align*} M(C_4)\cong U_{3,4}. \end{align*} The same reasoning works for a single cycle of length $n$: every proper subset breaks the cycle and is a forest, while the full $n$-edge set is the unique cycle. Therefore the independent sets are exactly the subsets of size at most $n-1$, so the cycle matroid is $U_{n-1,n}$. In the boundary cases, a graph loop is already a one-edge cycle, so its single edge is a one-element circuit of the matroid; two parallel edges form a two-edge cycle, so the pair is a two-element circuit, which is exactly the matroidal parallel-pair phenomenon. [/example] The example shows that cycle matroids package graph cycles as matroid circuits. To use this package in proofs, we also need a numerical invariant: given a chosen set of edges, how many of them can be kept before a cycle is forced? Write $c(A)$ for the number of connected components of the spanning subgraph $(V(G),A)$. [quotetheorem:6628] [citeproof:6628] The hypothesis that the graph is finite matters because rank is being computed by the size of a maximal independent edge set. For example, in the one-way infinite ray with vertices $v_1,v_2,\dots$ and edges $v_i v_{i+1}$, the whole edge set is a forest and has countably infinite size, while the expression $|V|-c(E)$ is not a finite rank value in the usual matroid sense. One could introduce cardinal-valued rank conventions for infinite matroids, but the finite-rank formula used here is a statement about finite graphs. For the next step, the important consequence of the formula is that it detects when deleting edges increases the number of connected components. That is the rank shadow of a cut, but it is still too coarse for duality. If deleting a set of edges disconnects the graph, then any larger deleted set still disconnects it, while a cocircuit is minimal with respect to the corresponding rank drop. The graph-side notion must therefore isolate those cuts whose every edge is needed to produce the separation; that minimal cut is called a bond. [definition: Bond] Let $G$ be a graph. A bond of $G$ is a nonempty cut $\delta(X)=\{uv \in E(G): u \in X,\ v \notin X\}$ that is minimal under inclusion among nonempty cuts. [/definition] A bond is the graph-theoretic version of a cocircuit in the cycle matroid: removing its edges separates the graph, and removing any smaller chosen part does not. Since the course now wants a matroid whose circuits are cuts rather than cycles, we take the dual of the cycle matroid. [definition: Cographic Matroid] Let $G$ be a finite graph. The cographic matroid of $G$ is the dual matroid \begin{align*} M^*(G):=M(G)^*. \end{align*} A matroid is cographic if it is isomorphic to $M^*(G)$ for some graph $G$. [/definition] This definition will be useful only if the dual has a graph-theoretic interpretation that can be recognised without mentioning bases. The promised interpretation is that the minimal cutsets of $G$ are exactly the circuits of the cographic matroid. [quotetheorem:6629] [citeproof:6629] The word "finite" again keeps the rank-counting argument literal, and the use of bonds rather than cuts is essential. In a four-vertex path with edges $e_1,e_2,e_3$, the set $\{e_1,e_2\}$ disconnects the graph, but it is not a bond because the smaller set $\{e_2\}$ already disconnects it. The theorem does not say that every disconnecting set is a circuit of $M^*(G)$; it singles out the minimal disconnecting sets. This distinction is what makes cographic matroids useful in planar duality, where dual cycles correspond to minimal primal separations rather than to arbitrary deleted edge sets. [example: Cographic Matroid from a Square] Let $G=C_4$ have vertices $v_1,v_2,v_3,v_4$ in cyclic order and edges $e_1=v_1v_2$, $e_2=v_2v_3$, $e_3=v_3v_4$, and $e_4=v_4v_1$. The cuts isolating the four vertices are \begin{align*} \delta(\{v_1\})=\{e_1,e_4\},\quad \delta(\{v_2\})=\{e_1,e_2\},\quad \delta(\{v_3\})=\{e_2,e_3\},\quad \delta(\{v_4\})=\{e_3,e_4\}. \end{align*} The cuts separating opposite pairs of vertices are \begin{align*} \delta(\{v_1,v_2\})=\{e_2,e_4\},\quad \delta(\{v_2,v_3\})=\{e_1,e_3\}. \end{align*} Together these are the six two-element subsets of $E(C_4)$. Removing one edge from $C_4$ leaves a path on four vertices, so no one-edge set is a cut. Removing two adjacent edges isolates the common endpoint, and removing two opposite edges leaves two disjoint single-edge components, so every two-edge subset disconnects $G$. Hence these six sets are exactly the bonds of $C_4$. By *Cocircuits of a Cycle Matroid Are Bonds*, the circuits of $M^*(C_4)$ are exactly these six two-element sets. This also matches the uniform-matroid computation. Since \begin{align*} M(C_4)\cong U_{3,4}, \end{align*} the bases of $M(C_4)$ are the four three-element subsets of $E(C_4)$. The bases of the dual matroid are the complements of these bases, so the bases of $M^*(C_4)$ are the four one-element subsets of $E(C_4)$. Therefore \begin{align*} M^*(C_4)\cong U_{1,4}. \end{align*} In $U_{1,4}$ the independent sets are exactly the sets of size at most $1$, so the minimal dependent sets are exactly the two-element subsets. Thus the circuit computation in the dual matroid is exactly the same as the bond computation in the square. [/example] The square example also suggests that deletion and contraction should have graph-level meanings, because removing or identifying edges changes both cycles and cuts. To pass from examples to minor arguments, we need the exact compatibility between graph operations and matroid operations. [illustration:deletion-contraction-edge] The square example gave a local computation of a cographic matroid, but minor arguments require a uniform rule: when a graph edge is removed or identified, the cycle matroid should undergo the corresponding matroid operation on the same labelled ground set. The next theorem supplies that rule and explains why graphic matroids remain graphic after taking minors. [quotetheorem:6630] [citeproof:6630] The non-loop hypothesis in the contraction argument is needed because a loop cannot be part of a forest; contracting it by the usual matroid convention is therefore the same as deleting it, while topological graph contraction of a loop is not a useful forest operation. The theorem does not say that every matroid minor of a graphic matroid remembers a unique graph minor, since different graph minors can still have the same cycle matroid. It does say that graphic matroids form a minor-closed class, so any forbidden-minor statement for graphs can be translated into a matroidal obstruction once the cycle matroid construction is fixed. ## Whitney 2-Isomorphism and the Information Remembered by a Cycle Matroid The next question is more subtle: if $M(G)$ is known, can the graph $G$ be reconstructed? The answer is no in full generality, because the matroid sees which edge sets contain cycles but does not remember the incidence relation among vertices beyond what cycles force. Whitney's theorem describes the precise ambiguity for connected graphs. [definition: Whitney Flip] Let $G$ be a labelled graph with edge set $E$, and suppose $G=G_1\cup G_2$ where $G_1$ and $G_2$ are edge-disjoint labelled subgraphs with $V(G_1)\cap V(G_2)=\{a,b\}$. A Whitney flip is the operation from such labelled graphs to labelled graphs with the same edge set $E$ obtained by detaching $G_2$ from $a,b$ and reattaching it with the roles of $a$ and $b$ interchanged, leaving all edge labels unchanged. [/definition] A Whitney flip changes the graph but preserves its cycle matroid. Cycles entirely inside one piece are unaffected, and cycles using both pieces still correspond to paths between the same two attachment vertices, only with the attachment labels reversed on one side. [example: Nonisomorphic Graphs with the Same Cycle Matroid] Let $G$ have vertices $a,u,v,b,x$ and labelled edges \begin{align*} p=au,\qquad q=uv,\qquad r=vb,\qquad s=ax,\qquad t=xb. \end{align*} These edges occur in the cyclic order \begin{align*} a \xrightarrow{p} u \xrightarrow{q} v \xrightarrow{r} b \xrightarrow{t} x \xrightarrow{s} a, \end{align*} so $\{p,q,r,s,t\}$ is a graph cycle. If one edge is removed from this five-edge cycle, the remaining graph is a path on the same five vertices; every proper subset of a path is a forest. Hence the only circuit of $M(G)$ is \begin{align*} \{p,q,r,s,t\}. \end{align*} Form $H$ by keeping \begin{align*} p=au,\qquad q=uv,\qquad r=vb \end{align*} and replacing the second path by \begin{align*} s=bx,\qquad t=xa. \end{align*} Then the same labelled edge set occurs in the cyclic order \begin{align*} a \xrightarrow{p} u \xrightarrow{q} v \xrightarrow{r} b \xrightarrow{s} x \xrightarrow{t} a, \end{align*} so the only circuit of $M(H)$ is again $\{p,q,r,s,t\}$. The labelled graphs are nevertheless not isomorphic. Any edge-label-preserving isomorphism must send the endpoint common to $p$ and $q$ to the endpoint common to $p$ and $q$, so it fixes $u$; it must send the endpoint common to $q$ and $r$ to the endpoint common to $q$ and $r$, so it fixes $v$; then the other endpoint of $p$ is fixed as $a$, and the other endpoint of $r$ is fixed as $b$. But in $G$ the edge $s$ is incident with $a$, while in $H$ the edge $s$ is incident with $b$. This contradicts preservation of the label $s$. Thus $G$ and $H$ are nonisomorphic labelled graphs with the same labelled cycle matroid. [/example] The failure here is not accidental. A cycle matroid records which labelled edge sets close up to form cycles, but it does not record the order in which two-terminal pieces are attached across a two-vertex separation. Whitney's theorem says that this is the only source of non-uniqueness once the graphs are connected and the edge labels are fixed. To state it without blurring the hypotheses, call two connected labelled graphs 2-isomorphic if they are related by a finite sequence of edge-label-preserving graph isomorphisms, Whitney flips along two-vertex separations, and the inverse operations of splitting or identifying a cut vertex in a way that does not change the labelled cycles. [quotetheorem:6631] [citeproof:6631] Connectedness is needed because isolated vertices and separate components can be added or rearranged without changing any edge cycle set; the labelled edge-set hypothesis is needed because otherwise even the comparison of circuits has no fixed ground set. For instance, a triangle with edges labelled $a,b,c$ and a second triangle with labels $x,y,z$ may have isomorphic unlabelled cycle matroids, but there is no statement that the circuit $\{a,b,c\}$ is the same subset as $\{x,y,z\}$ until a bijection between the edge sets has been fixed. The theorem does not claim that $G$ itself is determined by $M(G)$ as an unlabelled graph, nor does it forbid non-isomorphic representatives in the presence of cut vertices or 2-separations. Its value is that it identifies exactly where the lost information lives. The next theorem records the important case where this ambiguity disappears: if the graph has enough vertex-connectivity, there are no cut vertices or two-vertex separations on which Whitney moves can act. [quotetheorem:6632] [citeproof:6632] The assumptions exclude exactly the moves in Whitney's theorem: a graph with fewer than four vertices can satisfy degenerate connectivity conventions, and a graph that is only 2-connected can still admit a Whitney flip. The theorem does not say that every graphic matroid has a unique graph representation; it says uniqueness appears after graph connectivity is strong enough to prevent all genuine 2-isomorphism moves. This boundary is important because planar duality will later use Whitney's theorem to control possible graph representatives of a dual matroid. [example: Why 3-Connectedness Matters] Let $G$ be a labelled theta graph with common end vertices $a,b$ and three internally disjoint labelled paths \begin{align*} P:a=p_0,p_1,\dots,p_m=b,\qquad Q:a=q_0,q_1,\dots,q_n=b,\qquad R:a=r_0,r_1,\dots,r_\ell=b, \end{align*} where the three lengths are not all equal. Since every internal vertex has degree $2$ and the only vertices of degree $3$ are $a$ and $b$, any graph cycle must enter one of the three paths at one end, leave it at the other end, and return along a second path. Hence the cycles of $G$ are exactly \begin{align*} P\cup Q,\qquad P\cup R,\qquad Q\cup R, \end{align*} with the labels inherited from the three paths. Now perform a Whitney flip on the $R$-side across the separating pair $\{a,b\}$: keep $P$ and $Q$ attached as before, but reattach the labelled path $R$ with its endpoints interchanged, so that \begin{align*} R:b=r_0,r_1,\dots,r_\ell=a. \end{align*} Call the resulting labelled graph $H$. The labelled edge sets of the three terminal paths have not changed. In $H$, the same argument shows that every cycle is the union of two of the three $a$-$b$ paths, so the labelled cycles are again \begin{align*} P\cup Q,\qquad P\cup R,\qquad Q\cup R. \end{align*} Thus $G$ and $H$ have the same labelled cycle matroid. The graph is still only $2$-connected: deleting one vertex leaves the remaining graph connected, but deleting both $a$ and $b$ separates the internal vertices of the three paths into different components. Therefore it is not $3$-connected. This is exactly the ambiguity ruled out by Whitney uniqueness: the cycle matroid remembers the three pairwise path-unions that form cycles, but it does not remember which end of a two-terminal piece was attached to $a$ rather than to $b$. [/example] Whitney's theorem is often the point at which graphic matroids stop being examples and start being structural objects. It tells us that matroid theory can identify a graph up to a controlled family of graph transformations. [illustration:whitney-flip-two-separation] ## Planarity Through Matroid Duality The next problem is to recognise planar graphs without drawing every possible embedding. Cycles alone cannot detect planarity: $K_5$ and many planar graphs both have rich cycle spaces, and the obstruction is how cycles interact with cuts in a surface. Plane graph duality exchanges cycles and cuts, while matroid duality exchanges circuits and cocircuits. The compatibility between these two dualities gives a matroidal planarity criterion. [illustration:plane-dual-cycles-bonds] [definition: Plane Dual Graph] Let $G$ be a finite connected graph embedded in the sphere. The plane dual $G^\dagger$ has one vertex for each face of the embedding and one edge $e^\dagger$ crossing each edge $e\in E(G)$, joining the two face-vertices adjacent to $e$. [/definition] The dual graph depends on the embedding, so a priori it is not clear that it has a matroidal meaning independent of the drawing choices. The reason it does is that every dual cycle crosses a primal cut, and every primal bond should appear as a dual cycle. The next theorem records this exact dictionary and is the bridge from topological planar duality to matroid duality. [quotetheorem:6633] [citeproof:6633] Connectedness is used so that the spherical dual has the standard face-vertex construction without separating the discussion component by component. If $G$ is the disjoint union of two plane triangles, the complement in the sphere has three faces rather than the four faces obtained by treating the two triangles separately and adding their duals as independent pieces; the shared outside face couples the components. The theorem therefore needs the connected hypothesis for this clean form. It does not say that the abstract dual graph is independent of embedding as a graph; different embeddings can produce non-isomorphic dual graphs. It says their cycle matroids represent the same matroid dual, which is the topological content of cycle-cut duality and is the input for the planarity criterion. [example: Duality for a Planar Square] Embed $C_4$ in the sphere, with the inside face and outside face as the two dual vertices. Each edge of the square borders both faces, so the dual graph $(C_4)^\dagger$ has two vertices joined by four parallel edges $e_1^\dagger,e_2^\dagger,e_3^\dagger,e_4^\dagger$. In this four-edge parallel multigraph, the empty set is a forest and every one-edge set is a forest. If $i\ne j$, then $e_i^\dagger$ and $e_j^\dagger$ have the same two endpoints, so $\{e_i^\dagger,e_j^\dagger\}$ is a two-edge cycle and is dependent. Hence the independent sets are exactly the subsets of size at most $1$, so \begin{align*} M((C_4)^\dagger)\cong U_{1,4}. \end{align*} This agrees with the matroid dual computation: since \begin{align*} M(C_4)\cong U_{3,4}, \end{align*} the bases of $M(C_4)$ are the four three-element subsets of $\{e_1,e_2,e_3,e_4\}$. The bases of $M(C_4)^*$ are their complements, namely the four one-element subsets. Therefore $M(C_4)^*$ has exactly the one-element bases, so \begin{align*} M(C_4)^*\cong U_{1,4}. \end{align*} Thus the plane dual graph realizes the dual matroid in this example, as predicted by *Cycle Matroid of the Plane Dual*. [/example] The preceding theorem proves that planar graphs have graphic dual matroids. The harder reverse direction asks whether a graph can be embedded in the sphere whenever its matroid dual happens to come from some graph. The reverse direction uses a graph-theoretic duality theorem of Whitney. The theorem is quoted here as an external input: its proof belongs to topological graph theory, and the matroid argument uses it only after the cycle-bond correspondence has been established. [quotetheorem:6634] The plane-dual theorem gave one direction of the story: an embedding produces a graph whose cycle matroid is the dual matroid. Whitney's abstract duality theorem supplies the missing reverse implication by saying that the cycle-bond dictionary is not merely a formal analogy; it is exactly the certificate from which a sphere embedding can be recovered. Both hypotheses matter. Without connectedness, the outside face of an embedding can tie different components together and there is no single clean face-vertex dual construction for the whole graph. Without both directions of the cycle-bond correspondence, a graph may have cycles matching the other graph's bonds on one side while still failing to encode the reverse separation data needed to recover a cellular sphere embedding. This is the point where the chapter can replace a geometric question by a matroid question. To prove that a graph is planar, it is enough to find some graph whose cycles behave like the original graph's bonds and whose bonds behave like the original graph's cycles. Matroid duality is designed to produce exactly that exchange: it turns circuits into cocircuits, the bond theorem identifies cocircuits of a cycle matroid with graph bonds, and Whitney's abstract duality theorem converts the resulting cycle-bond dictionary into a sphere embedding. The planarity criterion packages this translation into a single test. [quotetheorem:6635] [citeproof:6635] Connectedness prevents separate components from requiring separate sphere embeddings and dual vertices; the disconnected case is recovered by applying the result componentwise. The theorem does not say that a particular graph representing $M(G)^*$ is the unique plane dual of $G$, because Whitney flips can change graph representatives without changing the matroid. It transforms planarity into the graphicness of the dual matroid, which is why nonplanar graph obstructions such as $K_5$ and $K_{3,3}$ become detectable as failures of cographic matroids to be graphic. [example: Kuratowski Obstructions in Matroid Language] The graphs $K_5$ and $K_{3,3}$ are finite and connected, and both are nonplanar. If $M(K_5)^*$ were graphic, then *Planarity Criterion by Graphic Duals* would imply that $K_5$ is planar, contradicting the nonplanarity of $K_5$. Therefore \begin{align*} M(K_5)^* \text{ is not graphic.} \end{align*} The same argument gives \begin{align*} M(K_{3,3})^* \text{ is not graphic.} \end{align*} Their cycle matroids are graphic by construction, since $M(K_5)$ is the cycle matroid of the graph $K_5$ and $M(K_{3,3})$ is the cycle matroid of the graph $K_{3,3}$. However, $M(K_5)$ is not cographic: if $M(K_5)$ were cographic, then there would be a graph $H$ with \begin{align*} M(K_5)\cong M(H)^*. \end{align*} Taking duals and using $(N^*)^*\cong N$ gives \begin{align*} M(K_5)^*\cong M(H), \end{align*} so $M(K_5)^*$ would be graphic, contradicting the conclusion above. The same duality argument shows that $M(K_{3,3})$ is not cographic. Thus $M(K_5)$ and $M(K_{3,3})$ are graphic but not cographic. These examples separate the two properties: a cycle matroid is always graphic, but it has a cographic description exactly in the planar cases covered by the planarity criterion. [/example] For disconnected graphs, the statement is handled componentwise: a graph is planar exactly when each connected component is planar, and the cycle matroid is the direct sum of the component cycle matroids. The connected case contains the essential content. ## Regularity of Graphic Matroids The final question in this chapter connects graphic matroids to representability over fields. Chapter 6 introduced representability as a field-dependent property; graphic matroids are better behaved because their standard incidence matrices work over every field after a small orientation choice. [definition: Regular Matroid] A matroid $M$ is regular if it is representable over every field. [/definition] Regularity is strong: it says the same independence structure can be realised linearly in all characteristics. To prove [graphic matroids are regular](/theorems/6636), we need a matrix whose column dependence records cycles and whose construction does not depend on a special field. [definition: Oriented Incidence Matrix] Let $G$ be a finite graph with an arbitrary orientation of each non-loop edge. The oriented incidence matrix $B_G$ has rows indexed by vertices and columns indexed by edges, with column $e=uv$ containing one $1$, one $-1$, and zeros elsewhere according to the chosen orientation; a loop column is zero. [/definition] Changing the orientation of an edge multiplies its column by $-1$, so it does not change the represented matroid over any field of characteristic different from $2$; in characteristic $2$, the signs coincide and the usual unsigned incidence matrix gives the same dependence relations. The remaining issue is to verify that column dependence is controlled by cycles rather than by an artefact of the chosen orientation. [quotetheorem:6636] [citeproof:6636] Finiteness keeps the representation matrix finite, and the zero loop columns are necessary because a graph loop is already a one-element circuit in the cycle matroid. The finiteness condition is not cosmetic: for the one-way infinite ray, the incidence matrix has infinitely many rows and columns, and the finite column-dependence matroid sees every finite edge set as independent, while the graph has infinite paths whose behaviour is not captured by the finite-matrix argument above. Infinite graphic matroids require a separate set of definitions distinguishing finite cycles from topological circles. The theorem does not say that every regular matroid is graphic. For a concrete counterexample, $M(K_5)^*$ is regular because duals of regular matroids are regular, but it is not graphic: if it were graphic, the planarity criterion would force $K_5$ to be planar. The incidence-matrix proof points toward the stronger total-unimodularity viewpoint: after deleting one row from each connected component, the oriented incidence matrix is totally unimodular, which explains why the same independence data survives over all fields and connects graphic matroids with network-flow integrality in combinatorial optimisation. [example: Incidence Representation of a Triangle] Orient the edges as $e_{12}:1\to2$, $e_{23}:2\to3$, and $e_{31}:3\to1$. With the convention that a column has $-1$ at the tail and $1$ at the head, the three incidence columns in $k^3$ are \begin{align*} c_{12}=(-1,1,0),\quad c_{23}=(0,-1,1),\quad c_{31}=(1,0,-1). \end{align*} Their sum is \begin{align*} c_{12}+c_{23}+c_{31}=(-1+0+1,\;1-1+0,\;0+1-1)=(0,0,0). \end{align*} The coefficients in this relation are all $1$, so not all coefficients are zero in any field. Hence the full three-edge set is dependent. Now check the two-edge subsets. If $a c_{12}+b c_{23}=0$, then \begin{align*} a(-1,1,0)+b(0,-1,1)=(-a,\;a-b,\;b)=(0,0,0). \end{align*} The first coordinate gives $a=0$, and the third coordinate gives $b=0$. If $a c_{12}+c c_{31}=0$, then \begin{align*} a(-1,1,0)+c(1,0,-1)=(-a+c,\;a,\;-c)=(0,0,0). \end{align*} The second coordinate gives $a=0$, and the third coordinate gives $c=0$. If $b c_{23}+c c_{31}=0$, then \begin{align*} b(0,-1,1)+c(1,0,-1)=(c,\;-b,\;b-c)=(0,0,0). \end{align*} The first coordinate gives $c=0$, and the second coordinate gives $b=0$. Thus every one-edge and two-edge subset is independent, while the full three-edge set is dependent. The represented matroid has independent sets exactly the subsets of size at most $2$, so it is $U_{2,3}$, the cycle matroid of the triangle. [/example] Regularity places graphic matroids in the intersection of many representability classes. In the next part of the course, this becomes a source of structure theorems: graphic and cographic matroids are fundamental regular matroids, but regular matroids also include objects that are neither graphic nor cographic. Graphic and cographic matroids show how much graph theory is retained by the matroid abstraction. The next chapter moves from single independence systems to situations where several matroid constraints interact, leading to intersection and union theorems. # 8. Matroid Intersection and Union The course has developed matroids as an abstraction of independence, rank, closure, bases, and duality. This chapter uses that machinery to study optimization problems in which more than one independence constraint is present. The main prerequisites are the rank axioms from Chapter 2, restriction and direct sums from Chapter 5, partition and graphic matroids from Chapters 1 and 7, and the circuit-exchange viewpoint behind augmentation. This chapter moves from the structure of a single matroid to the interaction of several matroids on related ground sets. The guiding question is whether the exchange axiom is strong enough to make simultaneous independence behave as well as ordinary independence. Matroid intersection gives a min-max theorem for common independent sets of two matroids, while matroid union turns sums of independent sets into another matroid and gives compact criteria for packing. These results are also the first place where the course's structural language becomes an algorithmic language: a failed search for a larger feasible set produces a subset certificate, and the same certificate is the dual object in the associated rank polyhedron. ## Common Independent Sets of Two Matroids Suppose two matroids encode two different constraints on the same finite set $E$. A set may be legal for each constraint separately, but the optimization problem asks for a set satisfying both constraints at once. The first question is therefore: how large can a set be if it is independent in both matroids? [definition: Common Independent Set] Let $M_1=(E,\mathcal I_1)$ and $M_2=(E,\mathcal I_2)$ be matroids on the same finite ground set $E$. A common independent set of $M_1$ and $M_2$ is a subset $I \subset E$ such that $I \in \mathcal I_1 \cap \mathcal I_2$. [/definition] The family $\mathcal I_1 \cap \mathcal I_2$ is hereditary, but it need not satisfy the matroid exchange axiom. This is why the problem is richer than applying greedy optimization to a new matroid. The next example is the standard model: matching in a bipartite graph can be read as simultaneous independence for two partition constraints. [example: Bipartite Matching as Matroid Intersection] Let $G=(U\sqcup V,E)$ be a finite bipartite graph. For each $u\in U$, write \begin{align*} \delta(u)=\{e\in E:e\text{ is incident with }u\}, \end{align*} and define $M_U$ to be the partition matroid with blocks $\delta(u)$ and capacity $1$ on each block. Thus $F\subset E$ is independent in $M_U$ exactly when \begin{align*} |F\cap \delta(u)|\le 1 \end{align*} for every $u\in U$. Similarly, using the blocks \begin{align*} \delta(v)=\{e\in E:e\text{ is incident with }v\} \end{align*} for $v\in V$, the matroid $M_V$ has independent sets satisfying \begin{align*} |F\cap \delta(v)|\le 1 \end{align*} for every $v\in V$. We show that the common independent sets are exactly the matchings. If $F$ is common independent, then no two edges of $F$ share a vertex in $U$, because each $u\in U$ is incident with at most one edge of $F$. Also no two edges of $F$ share a vertex in $V$, because each $v\in V$ is incident with at most one edge of $F$. Since every edge of a bipartite graph has one endpoint in $U$ and one endpoint in $V$, no two edges of $F$ share any endpoint, so $F$ is a matching. Conversely, if $F$ is a matching, then each vertex of $U$ and each vertex of $V$ is incident with at most one edge of $F$, so $F$ is independent in both $M_U$ and $M_V$. Hence \begin{align*} F\in \mathcal I(M_U)\cap \mathcal I(M_V)\quad\Longleftrightarrow\quad F\text{ is a matching in }G. \end{align*} The corresponding rank functions make the two collision constraints explicit. For every $A\subset E$, \begin{align*} r_U(A)=\sum_{u\in U}\min\{1,|A\cap\delta(u)|\}=|\{u\in U:A\cap\delta(u)\ne\varnothing\}|. \end{align*} Similarly, \begin{align*} r_V(A)=\sum_{v\in V}\min\{1,|A\cap\delta(v)|\}=|\{v\in V:A\cap\delta(v)\ne\varnothing\}|. \end{align*} Thus maximizing the size of a matching is exactly the problem of maximizing $|F|$ over sets $F$ that are independent in both partition matroids. [/example] This example also explains why intersection cannot be reduced to a single local exchange rule. Augmenting a matching may require alternating paths, not merely exchanging one element for another. Matroid intersection generalizes the same augmenting-path principle, with circuits replacing collisions at vertices. [definition: Maximum Common Independent Set Problem] For matroids $M_1=(E,\mathcal I_1)$ and $M_2=(E,\mathcal I_2)$, the maximum common independent set problem is to compute \begin{align*} \max\{|I| : I \in \mathcal I_1 \cap \mathcal I_2\}. \end{align*} [/definition] The quantity above has a dual certificate. If $A \subset E$, then a common independent set $I$ splits as $(I \cap A) \cup (I \setminus A)$. The first part has size at most $r_1(A)$, and the second part has size at most $r_2(E \setminus A)$. Hence every subset $A$ gives the upper bound \begin{align*} |I| \le r_1(A)+r_2(E\setminus A). \end{align*} The theorem of Edmonds says that the best such upper bound is always attained. ## Edmonds' Matroid Intersection Theorem The main problem now is to prove that the elementary upper bounds above are complete. This is a min-max result: the maximum size of a feasible object equals the minimum size of a certificate obstructing larger feasible objects. [quotetheorem:5814] [citeproof:5814] An augmenting-path analogy helps explain the theorem. The exchange digraph packages the fundamental circuits that appear when a new element is added to an independent set. The absence of augmenting paths is converted into a subset $A$ that witnesses optimality. The hypotheses matter in two separate ways. The two matroids must live on the same finite ground set, because the expression $E\setminus A$ compares the two constraints element-by-element; if the constraints are placed on unrelated sets, there is no common subset $I$ to optimize. The matroid exchange axiom is also essential: the proof relies on fundamental circuits and on the ability to turn a failed augmentation into a spanning statement. A concrete hereditary failure occurs on $E=\{a,b,c\}$ with \begin{align*} \mathcal H_1&=\{\varnothing,\{a\},\{b\},\{c\},\{a,b\}\},& \mathcal H_2&=2^E. \end{align*} The common feasible family is $\mathcal H_1$, whose maximal feasible sets $\{a,b\}$ and $\{c\}$ have different sizes and violate augmentation: from $\{c\}$ no element of $\{a,b\}\setminus\{c\}$ can be added. Thus replacing the first matroid by an arbitrary hereditary constraint destroys the exchange mechanism behind the subset certificate. The theorem also does not say that common independent sets form a matroid, nor that an optimal common independent set is unique; it gives a maximum size and a subset certificate for optimality. [example: Konig's Theorem from Matroid Intersection] For the bipartite graph $G=(U\sqcup V,E)$, let $M_U$ and $M_V$ be the two partition matroids from the matching example. Their common independent sets are exactly the matchings, so *Matroid Intersection Theorem* gives \begin{align*} \nu(G)=\max\{|F|:F\text{ is a matching in }G\}=\min_{A\subset E}\bigl(r_U(A)+r_V(E\setminus A)\bigr). \end{align*} For each $A\subset E$, the two partition ranks are \begin{align*} r_U(A)=|\{u\in U:A\cap\delta(u)\ne\varnothing\}|. \end{align*} \begin{align*} r_V(E\setminus A)=|\{v\in V:(E\setminus A)\cap\delta(v)\ne\varnothing\}|. \end{align*} Fix $A\subset E$ and define \begin{align*} X_A=\{u\in U:A\cap\delta(u)\ne\varnothing\}. \end{align*} \begin{align*} Y_A=\{v\in V:(E\setminus A)\cap\delta(v)\ne\varnothing\}. \end{align*} Since $X_A\subset U$ and $Y_A\subset V$, the two sets are disjoint, and therefore \begin{align*} |X_A\cup Y_A|=|X_A|+|Y_A|=r_U(A)+r_V(E\setminus A). \end{align*} The set $X_A\cup Y_A$ is a vertex cover. Indeed, if $e=uv\in A$, then $u\in X_A$, so $e$ is covered by its $U$-endpoint. If $e=uv\in E\setminus A$, then $v\in Y_A$, so $e$ is covered by its $V$-endpoint. Thus every edge is covered. Choosing $A$ to minimize $r_U(A)+r_V(E\setminus A)$ therefore gives a vertex cover of size $\nu(G)$. Conversely, let $C=X\cup Y$ be any vertex cover, with $X\subset U$ and $Y\subset V$. Put \begin{align*} A_X=\{uv\in E:u\in X\}. \end{align*} Every edge of $A_X$ is incident with a vertex of $X$, so \begin{align*} r_U(A_X)\le |X|. \end{align*} If $uv\in E\setminus A_X$, then $u\notin X$. Since $C$ is a vertex cover, the edge $uv$ must have $v\in Y$, and hence every edge of $E\setminus A_X$ is incident with a vertex of $Y$. Therefore \begin{align*} r_V(E\setminus A_X)\le |Y|. \end{align*} It follows that \begin{align*} \min_{A\subset E}\bigl(r_U(A)+r_V(E\setminus A)\bigr)\le r_U(A_X)+r_V(E\setminus A_X)\le |X|+|Y|=|C|. \end{align*} Since this holds for every vertex cover $C$, every vertex cover has size at least $\nu(G)$, while the minimizing-set construction above produces a vertex cover of size $\nu(G)$. Hence the maximum size of a matching equals the minimum size of a vertex cover, which is Konig's theorem in matroid-intersection form. [/example] The theorem is useful beyond graph matching because it separates the two constraints. A linear constraint and a coloring constraint, for instance, can be placed in different matroids and then optimized together. [example: Colorful Linear Independence] Let $V$ be a finite-dimensional vector space over a field $k$, and let $E=E_1\sqcup\cdots\sqcup E_m$ be a finite colored set of vectors in $V$. Let $M_1$ be the vector matroid on $E$, so for every $A\subset E$, \begin{align*} r_1(A)=\dim \operatorname{span}(A). \end{align*} Let $M_2$ be the partition matroid with blocks $E_1,\dots,E_m$ and capacity $1$ on each block. Thus \begin{align*} F\in \mathcal I(M_2)\Longleftrightarrow |F\cap E_j|\le 1 \text{ for every } j. \end{align*} Therefore a common independent set is exactly a linearly independent set of vectors with no repeated color. By *Matroid Intersection Theorem*, the largest size of such a set is \begin{align*} \min_{A\subset E}\bigl(r_1(A)+r_2(E\setminus A)\bigr)=\min_{A\subset E}\bigl(\dim \operatorname{span}(A)+r_2(E\setminus A)\bigr). \end{align*} For the partition matroid $M_2$, the rank of $B\subset E$ is the number of color classes represented in $B$, since one may choose one vector from each represented class and none from an unrepresented class: \begin{align*} r_2(B)=|\{j:B\cap E_j\ne\varnothing\}|. \end{align*} Hence \begin{align*} r_2(E\setminus A)=|\{j:(E\setminus A)\cap E_j\ne\varnothing\}|. \end{align*} We now translate the condition for a colorful independent transversal of size $m$. Since $M_2$ permits at most one vector from each of the $m$ color classes, every common independent set has size at most $m$. Thus such a transversal exists exactly when \begin{align*} \min_{A\subset E}\bigl(\dim \operatorname{span}(A)+r_2(E\setminus A)\bigr)\ge m. \end{align*} For a fixed $A\subset E$, set \begin{align*} J(A)=\{j:E_j\subset A\}. \end{align*} These are precisely the color classes not represented in $E\setminus A$, so \begin{align*} r_2(E\setminus A)=m-|J(A)|. \end{align*} Also $\bigcup_{j\in J(A)}E_j\subset A$, and monotonicity of span gives \begin{align*} \dim \operatorname{span}(A)\ge \dim \operatorname{span}\left(\bigcup_{j\in J(A)}E_j\right). \end{align*} If for every $J\subset\{1,\dots,m\}$ one has \begin{align*} \dim \operatorname{span}\left(\bigcup_{j\in J}E_j\right)\ge |J|, \end{align*} then for every $A\subset E$, \begin{align*} \dim \operatorname{span}(A)+r_2(E\setminus A)\ge |J(A)|+\bigl(m-|J(A)|\bigr)=m. \end{align*} Therefore the minimum in the intersection formula is at least $m$, and because no common independent set can have size larger than $m$, a colorful independent transversal of size $m$ exists. Conversely, suppose $x_1,\dots,x_m$ is a colorful independent transversal with $x_j\in E_j$ for each $j$. For any $J\subset\{1,\dots,m\}$, the set $\{x_j:j\in J\}$ is linearly independent and is contained in $\bigcup_{j\in J}E_j$. Therefore \begin{align*} \dim \operatorname{span}\left(\bigcup_{j\in J}E_j\right)\ge |\{x_j:j\in J\}|=|J|. \end{align*} Thus a colorful independent transversal of size $m$ exists exactly when every selected collection of color classes collectively spans dimension at least the number of selected colors. [/example] The previous example leads to a classical theorem of Rado. Instead of fixing a coloring of an existing vector configuration, start with prescribed sets from which representatives must be chosen and ask whether the representatives can be independent in a matroid. [quotetheorem:6637] [citeproof:6637] [Rado's theorem](/theorems/6637) contains [Hall's marriage theorem](/theorems/2018) as the special case where $M$ is the free matroid on $E$. With a non-free matroid, the same Hall-type condition is strengthened by rank: the union of any group of lists must have enough matroidal dimension, not only enough elements. The rank inequalities are not cosmetic; they are exactly the obstruction. For instance, in a rank-one matroid on $E=\{a,b\}$ with $A_1=A_2=\{a,b\}$, there are enough distinct elements for Hall's condition, but $r(A_1\cup A_2)=1<2$, so two chosen representatives cannot be independent. The hypotheses are also part of the statement, not background decoration. Finiteness lets the copied-ground-set intersection theorem apply and prevents compactness issues that would require an infinite matroid formulation. If one tries to extend the statement without an infinite-matroid axiom, take $E=\mathbb N$ and lists $A_i=\{i,i+1,i+2,\dots\}$ for $i\in\mathbb N$. Every finite subfamily has distinct representatives in the free matroid, but an infinite proof cannot be obtained by minimizing over finitely many copied-ground-set subsets; extra compactness or infinite-matroid assumptions are needed to turn the finite witnesses into a global choice. Thus finiteness is not a cosmetic convenience of the proof. The ambient matroid structure is what turns the obstruction into a rank inequality. If it is replaced by an arbitrary hereditary family, the Rado condition need not be meaningful or sufficient. For example, let $E=\{a,b,c\}$ and \begin{align*} \mathcal H=\{\varnothing,\{a\},\{b\},\{c\},\{a,b\}\}. \end{align*} This is hereditary but not a matroid. With lists $A_1=\{a,b\}$ and $A_2=\{c\}$, each individual list contains a feasible singleton and the two lists have enough distinct elements. Nevertheless there is no feasible two-element representative set, because neither $\{a,c\}$ nor $\{b,c\}$ belongs to $\mathcal H$. The missing exchange axiom is exactly what prevents the local choices from being augmented into a global representative system. The theorem gives existence, not uniqueness, and it does not choose a preferred representative system when several independent transversals exist. These limitations point forward to union: instead of choosing one element from each list under a single independence constraint, union asks when every element can be assigned to one of several independent layers. [example: Hall's Theorem as the Free-Matroid Case] Let $M$ be the free matroid on a finite set $E$. By definition of the free matroid, every subset of $E$ is independent, so the rank of any $B\subset E$ is \begin{align*} r(B)=\max\{|I|:I\subseteq B,\ I\text{ is independent in }M\}=|B|. \end{align*} Applying *Rado's Theorem* to subsets $A_1,\dots,A_m\subset E$, the required condition is that for every $J\subset\{1,\dots,m\}$, \begin{align*} r\left(\bigcup_{i\in J}A_i\right)\ge |J|. \end{align*} Since $M$ is free, substituting $B=\bigcup_{i\in J}A_i$ into $r(B)=|B|$ gives \begin{align*} r\left(\bigcup_{i\in J}A_i\right) =\left|\bigcup_{i\in J}A_i\right|. \end{align*} Thus Rado's condition is exactly \begin{align*} \left|\bigcup_{i\in J}A_i\right|\ge |J| \end{align*} for every $J\subset\{1,\dots,m\}$. In the free matroid there is no independence restriction beyond distinctness of the chosen representatives, because every set of distinct elements is independent. Therefore *Rado's Theorem* specializes precisely to Hall's theorem for distinct representatives, and the general matroid version replaces the cardinality of a union by its matroid rank. [/example] ## The Union of Matroids Intersection combines constraints by requiring a set to satisfy all of them. Union asks the complementary question: when can a set be decomposed into several pieces, each satisfying one constraint? The answer is unexpectedly stable: the possible unions of independent sets again form a matroid. [definition: Matroid Union] Let $M_1=(E,\mathcal I_1),\dots,M_k=(E,\mathcal I_k)$ be matroids on the same finite ground set $E$. The union family is \begin{align*} \mathcal I_1\vee\cdots\vee\mathcal I_k =\{I_1\cup\cdots\cup I_k : I_j\in\mathcal I_j \text{ for } 1\le j\le k\}. \end{align*} [/definition] This definition says what decompositions are allowed, but it does not by itself say that we have a matroid. The theorem below supplies both the matroid structure and the rank formula. Its rank formula should be compared with the intersection formula: the minimization now occurs inside a set $X$, and the penalty $|X\setminus A|$ pays for elements not handled by the ranks of the constituent matroids. [quotetheorem:5815] [citeproof:5815] The union theorem turns decomposition questions into rank inequalities, but the rank formula is still written as a minimization over all subsets. The shared ground set is part of the content. If $M_1$ is the free matroid on $E_1=\{a\}$ and $M_2$ is the free matroid on $E_2=\{a,b\}$ where the two symbols $a$ are not identified by a specified common ambient set, then the expression $\{a\}\cup\{a\}$ is ambiguous: it could mean one element after identification or two labelled elements before identification. These choices give different ranks on the would-be union. Placing both matroids on the same ground set fixes which elements are the same and makes the decomposition $I=I_1\cup\cdots\cup I_k$ a statement about one subset of one set. The matroid hypotheses are also used in the copied-ground-set intersection argument; an arbitrary hereditary family can fail augmentation after union. For example, on $E=\{a,b,c\}$ let \begin{align*} \mathcal H_1&=\{\varnothing,\{a\},\{b\},\{c\},\{a,b\}\},& \mathcal H_2&=\{\varnothing\}. \end{align*} Both families are hereditary, and their union family is just $\mathcal H_1$. The sets $\{a,b\}$ and $\{c\}$ violate augmentation, since neither $\{a,c\}$ nor $\{b,c\}$ belongs to $\mathcal H_1$. The theorem should not be read as giving a unique decomposition of an independent set in the union: in the union of two free matroids on the same set, the same subset can be split between the two copies in many different ways. The first useful specialization asks whether the entire ground set can be decomposed into $k$ independent pieces in one fixed matroid. This converts the formula into a clean necessary and sufficient inequality. [quotetheorem:6638] [citeproof:6638] The covering criterion is sharp because the subset inequalities test exactly the local density of the matroid. If $M$ is the rank-one matroid on a three-element set and $k=2$, then the inequality fails at $A=E$ since $|E|=3>2r(E)=2$; correspondingly, two independent sets cover at most two elements. Finiteness is used when the union rank formula turns all covering obstructions into a finite minimum. Without a finite-ground-set theorem or an infinite-matroid replacement, the statement no longer follows from the preceding argument; for instance, an infinite family of finite obstructions can be checked on every finite restriction while a global cover requires an additional compactness principle. Thus the finite hypothesis belongs to the theorem being used, not just to notation. The exchange axiom is also necessary. On $E=\{a,b,c,d\}$ let \begin{align*} \mathcal H=\{\varnothing\} \cup\{\{x\}:x\in E\} \cup\{\{a,b\},\{a,c\},\{b,c\}\}. \end{align*} This hereditary family is not a matroid: the feasible sets $\{a,b\}$ and $\{d\}$ violate augmentation. If $\rho(A)=\max\{|I|:I\subseteq A,\ I\in\mathcal H\}$, then $|A|\le 2\rho(A)$ holds for every $A\subset E$: pairs involving $d$ have $\rho=1$, every three-element subset contains one of $\{a,b\},\{a,c\},\{b,c\}$, and $\rho(E)=2$. Nevertheless $E$ is not a union of two feasible sets. Any two-set cover would have to cover $d$ together with at least one of $a,b,c$, but no two-element set containing $d$ is feasible, and the remaining alternative would require a three-element feasible set. The obstruction is not a density inequality; it is the absence of augmentation. The matroid exchange axiom is what makes density inequalities sufficient for a cover. The theorem also only asserts the existence of a cover by independent sets, not a canonical cover and not a cover by bases. For graphic matroids, independent sets are forests. The criterion therefore measures when the edge set of a graph can be covered by a bounded number of forests, which is the matroidal form of arboricity. [example: Covering a Graph by Forests] Let $G=(V,E)$ be a finite graph, and let $M(G)$ be its graphic matroid, whose independent sets are exactly the forests in $G$. For $A\subset E$, write $V(A)$ for the vertices incident with at least one edge of $A$, and let $c(A)$ be the number of connected components of the subgraph $(V(A),A)$. If those components have vertex sets $W_1,\dots,W_{c(A)}$, then a maximal forest in the component on $W_i$ has $|W_i|-1$ edges. Therefore the graphic-matroid rank of $A$ is \begin{align*} r(A)=\sum_{i=1}^{c(A)}(|W_i|-1). \end{align*} Expanding the sum gives \begin{align*} r(A)=\sum_{i=1}^{c(A)}|W_i|-\sum_{i=1}^{c(A)}1. \end{align*} Since the component vertex sets partition $V(A)$, this becomes \begin{align*} r(A)=|V(A)|-c(A). \end{align*} If $A$ spans all of $V$ and isolated vertices are counted as components, then $V(A)=V$, so the same formula reads \begin{align*} r(A)=|V|-c(A). \end{align*} Applying *Independent Set Covering Criterion* to the graphic matroid $M(G)$, the edge set $E$ is a union of $k$ independent sets of $M(G)$ if and only if every $A\subset E$ satisfies \begin{align*} |A|\le k r(A). \end{align*} Because independent sets in $M(G)$ are forests, this says that $G$ can be decomposed into $k$ forests if and only if every edge set $A\subset E$ satisfies \begin{align*} |A|\le k\bigl(|V(A)|-c(A)\bigr). \end{align*} Thus the obstruction is exactly a subgraph whose number of edges is larger than $k$ times its graphic-matroid rank; equivalently, dense subgraphs are the only obstructions to decomposing $G$ into $k$ forests. [/example] Covering by forests is a covering problem; spanning tree packing is the corresponding packing problem for bases. To obtain disjoint bases, we ask for the union of $k$ independent sets in the same matroid to have the largest possible total size $k r(E)$. The union rank formula turns that demand into the following subset inequalities. [quotetheorem:6639] [citeproof:6639] The base packing theorem is stronger than merely covering many elements: each packed independent set must reach full rank. The inequality detects when there is not enough room outside a subset to complete every base. For example, in the uniform matroid $U_{2,3}$, two disjoint bases would require four distinct elements, but $E$ has only three; the condition already fails at $A=\varnothing$ because $|E|=3<2r(E)=4$. The finite-ground-set hypothesis again comes from the union theorem used in the proof: the minimum over $A\subset E$ is a finite combinatorial certificate, while an infinite base-packing statement needs a separate infinite-matroid framework. The exchange axiom is needed here as well. In the hereditary family \begin{align*} \mathcal H=\{\varnothing,\{a\},\{b\},\{c\},\{a,b\}\} \end{align*} on $E=\{a,b,c\}$, the maximal feasible sets do not all have the same size, so there is no well-defined collection of bases with a common rank. If one declares the largest feasible sets to play the role of bases, then $\{a,b\}$ is the only such set and no rank inequality can capture the failure of $\{c\}$ to augment toward it. Matroid exchange is the condition that makes "base" a stable object and lets packing be tested by ranks of subsets. The theorem gives a criterion for existence of disjoint bases, but it does not count the number of packings or distinguish between different possible base decompositions. In graph language, disjoint bases are edge-disjoint spanning trees. The theorem supplies a compact rank condition, and the classical partition form is obtained by evaluating the rank inequality on edge cuts determined by vertex partitions. [example: Spanning Tree Packing in Graphic Matroids] Let $G=(V,E)$ be a connected finite graph, and let $M(G)$ be its graphic matroid. The bases of $M(G)$ are the spanning trees of $G$: independent sets are forests, and a maximal forest in a connected graph has $|V|-1$ edges and spans all vertices. Thus $k$ pairwise disjoint bases of $M(G)$ are exactly $k$ edge-disjoint spanning trees. By *[Matroid Base Packing Theorem](/theorems/6639)*, such a packing exists if and only if every $A\subset E$ satisfies \begin{align*} |E\setminus A|\ge k\bigl(r(E)-r(A)\bigr). \end{align*} Since $G$ is connected, the full graphic rank is \begin{align*} r(E)=|V|-1. \end{align*} Now let $\mathcal P=\{P_1,\dots,P_t\}$ be a partition of $V$, and let $A_{\mathcal P}$ be the set of edges whose two endpoints lie in the same part of $\mathcal P$. Then $E\setminus A_{\mathcal P}$ is exactly the set of edges crossing between distinct parts of the partition. Write the connected components of the spanning subgraph $(V,A_{\mathcal P})$ as $W_1,\dots,W_c$, including isolated vertices as one-vertex components. In the component with vertex set $W_i$, a maximal forest has $|W_i|-1$ edges, so the graphic rank is \begin{align*} r(A_{\mathcal P})=\sum_{i=1}^{c}(|W_i|-1). \end{align*} Expanding the sum gives \begin{align*} r(A_{\mathcal P})=\sum_{i=1}^{c}|W_i|-\sum_{i=1}^{c}1. \end{align*} The component vertex sets $W_1,\dots,W_c$ partition $V$, and $\sum_{i=1}^{c}1=c$, so \begin{align*} r(A_{\mathcal P})=|V|-c. \end{align*} Therefore \begin{align*} r(E)-r(A_{\mathcal P})=(|V|-1)-(|V|-c)=c-1. \end{align*} Each part $P_j$ contains at least one component of $(V,A_{\mathcal P})$, because all edges of $A_{\mathcal P}$ stay inside parts of $\mathcal P$. Hence $c\ge t=|\mathcal P|$, and so \begin{align*} r(E)-r(A_{\mathcal P})=c-1\ge |\mathcal P|-1. \end{align*} Substituting $A=A_{\mathcal P}$ into the base-packing inequality gives \begin{align*} |E\setminus A_{\mathcal P}|\ge k\bigl(r(E)-r(A_{\mathcal P})\bigr). \end{align*} Using the previous identity and inequality, this becomes \begin{align*} |E\setminus A_{\mathcal P}|\ge k(c-1)\ge k(|\mathcal P|-1). \end{align*} Thus the matroid packing criterion implies that every vertex partition $\mathcal P$ has at least $k(|\mathcal P|-1)$ crossing edges. The converse direction, namely that these partition inequalities imply all rank inequalities \begin{align*} |E\setminus A|\ge k\bigl(r(E)-r(A)\bigr), \end{align*} is the full graphic specialization known as the *Nash-Williams-Tutte Spanning Tree Packing Theorem*. This example uses the forward implication to show how the matroid rank obstruction becomes the familiar crossing-edge obstruction for packing spanning trees. [/example] The chapter's two operations fit together conceptually. Intersection handles simultaneous feasibility and leads to augmenting-path algorithms and Hall-type rank conditions. Union handles decomposability and leads to covering and packing criteria. Both theorems show that matroid rank functions provide the right language for certificates: every failure of feasibility is witnessed by a subset inequality. Matroid intersection and union turn rank inequalities into certificates for optimization and feasibility. We now specialize these ideas to matroids built from matchings and linkages, where Hall-type conditions and augmenting paths become structural constructions. # 9. Transversal, Gammoid, and Matching Matroids Transversal matroids are the matroids that record when a set of objects can be assigned to distinct admissible labels. They connect the exchange axiom with Hall-type matching conditions, so this chapter turns a theorem about bipartite matchings into a matroid construction. Building on the matching and intersection viewpoint from Chapter 8, the same idea reappears in directed networks: replacing matched edges by vertex-disjoint directed paths gives gammoids, another important family of matroids. ## Systems of Distinct Representatives Suppose a family of subsets $A_1, \dots, A_m$ of a set $E$ is interpreted as a list of allowed choices: representative $i$ may choose an element from $A_i$, and no two representatives may choose the same element. Which subsets of $E$ can appear as the chosen elements? The answer gives a matroid because augmenting a partial assignment is governed by alternating-path arguments from matching theory. [definition: Partial Transversal] Let $E$ be a finite set and let $A_1, \dots, A_m \subseteq E$. A subset $X \subseteq E$ is a partial transversal of $(A_i)_{i=1}^m$ if there exists an injective map $\varphi: X \to \{1, \dots, m\}$ such that $x \in A_{\varphi(x)}$ for every $x \in X$. [/definition] Equivalently, $X$ can be matched into the index set $\{1, \dots, m\}$ in the bipartite graph with left vertex set $E$, right vertex set $\{1, \dots, m\}$, and edge $x i$ whenever $x \in A_i$. This matching picture turns the local representative condition into a whole independence system, so the next definition packages all partial transversals from one presentation. [definition: Transversal Matroid] Let $E$ be a finite set and let $A_1, \dots, A_m \subseteq E$. The transversal matroid presented by $(A_i)_{i=1}^m$ is the set system $M[A_1, \dots, A_m]$ on ground set $E$ whose independent sets are the partial transversals of $(A_i)_{i=1}^m$. [/definition] The word "presented" matters: different families of sets can define isomorphic, or even identical, transversal matroids. The presentation is extra data, while the matroid remembers only which subsets of $E$ can be assigned distinct representatives. The first question is therefore whether this construction always satisfies the exchange axiom rather than merely producing an arbitrary hereditary set system. [quotetheorem:6640] [citeproof:6640] The finiteness hypotheses keep the matching problem inside the usual finite augmenting-path theorem; for infinite families, the same statement needs extra compactness assumptions and is not a formal consequence of this proof. The theorem does not say that an arbitrary element of $J\setminus I$ can be added: with $A_1=\{a,b\}$ and $A_2=\{b\}$, the set $I=\{b\}$ cannot be augmented by $a$ if $b$ is already forced to the only useful copy in a chosen matching, but a rerouting may make augmentation possible. This is why the next step is to replace explicit rerouting by Hall inequalities, which detect the obstruction without choosing a particular matching. [example: Job Assignment Matroid] Let $E$ be a set of applicants, and let $A_i \subseteq E$ be the applicants qualified for job $i$, for $1 \le i \le m$. By the definition of $M[A_1,\dots,A_m]$, a subset $X\subseteq E$ is independent exactly when there is an injective assignment $\varphi:X\to \{1,\dots,m\}$ such that $x\in A_{\varphi(x)}$ for every applicant $x\in X$. For the presentation \begin{align*} A_1=\{a,b\},\qquad A_2=\{b,c\},\qquad A_3=\{c,d\}, \end{align*} the set $\{a,c,d\}$ is independent. Indeed, define $\varphi(a)=1$, $\varphi(c)=2$, and $\varphi(d)=3$. Then \begin{align*} a\in A_1,\qquad c\in A_2,\qquad d\in A_3, \end{align*} and the values $1,2,3$ are distinct, so $\varphi$ is an injective qualified-job assignment. In contrast, $\{a,b,c,d\}$ is dependent because an injective map from a four-element set to $\{1,2,3\}$ would require four distinct jobs, but only three jobs are available. Now change the presentation to \begin{align*} A_1=A_2=\{a,b\},\qquad A_3=\{c,d\}. \end{align*} The set $\{a,b,c\}$ is independent, witnessed by $\varphi(a)=1$, $\varphi(b)=2$, and $\varphi(c)=3$, since \begin{align*} a\in A_1,\qquad b\in A_2,\qquad c\in A_3. \end{align*} However, $\{a,b,c,d\}$ is dependent already because the subset $\{c,d\}$ has only one available job. Explicitly, \begin{align*} A_1\cap\{c,d\}=\varnothing, \end{align*} \begin{align*} A_2\cap\{c,d\}=\varnothing, \end{align*} and \begin{align*} A_3\cap\{c,d\}=\{c,d\}. \end{align*} Therefore \begin{align*} \{i\in\{1,2,3\}:A_i\cap\{c,d\}\ne\varnothing\}=\{3\}. \end{align*} Thus the two applicants $c$ and $d$ collectively have only one admissible job, so no injective assignment can include both of them. This is the local obstruction that Hall-type conditions detect. [/example] Examples of this kind also show that transversal matroids generalise partition matroids. If the sets $A_i$ are pairwise disjoint blocks, then a partial transversal chooses at most one element from each block. [example: Partition Matroids as Transversal Matroids] Let $E=E_1 \cup \cdots \cup E_m$ be a disjoint union, and present a transversal matroid by setting $A_i=E_i$ for $1\le i\le m$. We show that its independent sets are exactly the subsets $X\subseteq E$ satisfying $|X\cap E_i|\le 1$ for every $i$. First suppose $X$ is independent in $M[A_1,\dots,A_m]$. Then there is an injective map $\varphi:X\to\{1,\dots,m\}$ such that $x\in A_{\varphi(x)}$ for every $x\in X$. Since $A_i=E_i$, this means $x\in E_{\varphi(x)}$. If $x\in X\cap E_i$, then $x\in E_i$ and $x\in E_{\varphi(x)}$. Because the blocks $E_1,\dots,E_m$ are disjoint, this forces $\varphi(x)=i$. Hence every element of $X\cap E_i$ is sent to the single index $i$. Since $\varphi$ is injective, at most one element can be sent to $i$, so \begin{align*} |X\cap E_i|\le 1. \end{align*} Conversely, suppose $|X\cap E_i|\le 1$ for every $i$. For each $x\in X$, there is a unique index $i$ with $x\in E_i$, because the sets $E_i$ form a disjoint union of $E$. Define $\varphi(x)=i$ for this unique index. Then $x\in E_i=A_i=A_{\varphi(x)}$. If $\varphi(x)=\varphi(y)=i$, then $x,y\in X\cap E_i$; since $|X\cap E_i|\le 1$, we get $x=y$. Thus $\varphi$ is injective, so $X$ is a partial transversal. Therefore the independent sets are exactly those meeting each block in at most one element. More generally, to allow up to $r_i$ elements from $E_i$, replace the single set $E_i$ by $r_i$ identical copies \begin{align*} A_{i,1}=E_i,\quad A_{i,2}=E_i,\quad \dots,\quad A_{i,r_i}=E_i. \end{align*} An injective assignment can send elements of $X\cap E_i$ only to the $r_i$ copies indexed by $i$, so it exists exactly when \begin{align*} |X\cap E_i|\le r_i \end{align*} for every $i$. This is precisely the partition matroid with block capacities $r_i$. [/example] ## Hall's Theorem as a Rank Condition The next question is when a specified set $X \subseteq E$ is independent in a transversal matroid without explicitly finding a matching. Hall's theorem converts this into inequalities over all subfamilies. In matroid language, these inequalities become the independence criterion for the presentation. [quotetheorem:2579] [citeproof:2579] The finiteness of the family is used in the induction and in the finite matching interpretation; infinite Hall theorems require a separate statement. Hall's theorem does not construct a preferred system of representatives or count all systems of representatives: it only decides existence. Its obstruction is local rather than global, as $B_1=B_2=\{a\}$ shows: the union of all sets has size $1<2$, and the subfamily $\{1,2\}$ already witnesses failure. For transversal matroids, Hall's theorem is applied after restricting the presentation to a chosen subset $X$. The right side of the bipartite graph is the family of indices, and the left side is the set that we are trying to match. [quotetheorem:6641] [citeproof:6641] The condition depends on checking every subset $Y\subseteq X$, not just $X$ itself. For example, three elements may all be adjacent only to two indices even when the whole presentation has many unused indices elsewhere; that three-element subset is dependent because it has neighbourhood size $2$. The criterion does not identify which matching should be used, nor does it give a rank formula for arbitrary subsets without optimisation. Its role here is to turn independence into a family of explicit inequalities, which is the form needed for Rado's matroidal generalisation. [example: Matchable Subsets in a Bipartite Graph] Let $G$ be a bipartite graph with left vertex set $E$ and right vertex set $R=\{1,\dots,m\}$. For each right vertex $i\in R$, set \begin{align*} A_i=\{x\in E:x\text{ is adjacent to }i\}. \end{align*} Then a subset $X\subseteq E$ is independent in $M[A_1,\dots,A_m]$ exactly when there is an injective map $\varphi:X\to R$ with $x\in A_{\varphi(x)}$ for every $x\in X$. Since $x\in A_{\varphi(x)}$ means precisely that $x$ is adjacent to $\varphi(x)$ in $G$, such a map is the same thing as a matching of the vertices of $X$ into $R$. Suppose $u,v,w\in E$ are three distinct left vertices whose neighbourhood is contained in two right vertices, say \begin{align*} N_G(\{u,v,w\})\subseteq \{p,q\}\subseteq R. \end{align*} Let $Y=\{u,v,w\}$. The indices available to elements of $Y$ are exactly the right neighbours of $Y$: \begin{align*} \{i\in R:A_i\cap Y\ne\varnothing\}=\{i\in R:\text{ some }x\in Y\text{ is adjacent to }i\}=N_G(Y). \end{align*} Because $N_G(Y)\subseteq\{p,q\}$, we get \begin{align*} \left|\{i\in R:A_i\cap Y\ne\varnothing\}\right|=|N_G(Y)|\le |\{p,q\}|=2<3=|Y|. \end{align*} Thus the Hall inequality fails for the subset $Y$, so $Y$ is not independent by the Hall criterion for transversal independence. Since every subset of an independent set is independent, any set of left vertices containing $u,v,w$ must also be dependent. [/example] Hall's theorem decides whether a whole set is matchable. Rado's theorem is the version that simultaneously accounts for a matroid constraint on the chosen representatives, and it is often the most useful form for applications. [quotetheorem:6642] [citeproof:6642] The rank hypothesis is essential because cardinality alone cannot see dependence inside $N$. For instance, in a vector matroid on two parallel nonzero vectors $u,v$, the family $B_1=B_2=\{u,v\}$ has enough distinct elements for Hall's theorem, but $r_N(B_1\cup B_2)=1<2$, so two linearly independent representatives do not exist. Rado's theorem does not specify a canonical choice of representatives, and it does not replace the need to know the rank function of $N$. It recovers Hall's theorem when $N$ is the free matroid on $E$, and it explains why transversal matroids are natural: they are exactly the independence structures obtained by asking for representatives subject only to being distinct. [example: Representatives with a Linear Constraint] Let $E$ be a finite set of vectors in a vector space over a field $k$, and let $N$ be the vector matroid on $E$. For subsets $B_1,\dots,B_k\subseteq E$, *Rado Theorem* says that there are distinct representatives $b_i\in B_i$ such that $\{b_1,\dots,b_k\}$ is independent in $N$ exactly when \begin{align*} r_N\left(\bigcup_{i\in S}B_i\right)\ge |S| \end{align*} for every $S\subseteq\{1,\dots,k\}$. In a vector matroid, the rank of a set of vectors is the dimension of its linear span, so this condition becomes \begin{align*} \dim_k\operatorname{span}\left(\bigcup_{i\in S}B_i\right)\ge |S| \end{align*} for every $S\subseteq\{1,\dots,k\}$. This is stronger than the cardinality inequality \begin{align*} \left|\bigcup_{i\in S}B_i\right|\ge |S|, \end{align*} because vectors can be distinct while still spanning too small a subspace. For example, if $u$ and $v$ are distinct nonzero parallel vectors, then \begin{align*} \operatorname{span}\{u,v\}=\operatorname{span}\{u\}, \end{align*} so \begin{align*} \dim_k\operatorname{span}\{u,v\}=1 \end{align*} even though \begin{align*} |\{u,v\}|=2. \end{align*} Thus the representative problem is not only to choose different vectors from the sets $B_i$, but to choose them so that no chosen vector lies in the span of the others. [/example] ## Gammoids from Directed Linkages Matching is a special case of routing disjoint paths: an edge in a bipartite graph is a path of length one from an element to a position. Gammoids replace admissible assignments by directed paths ending in a fixed set of sinks. The exchange axiom now comes from rerouting vertex-disjoint paths rather than alternating matched edges. [definition: Linkable Set] Let $D=(V,A)$ be a finite directed graph and let $T \subseteq V$. A subset $X \subseteq V$ is linkable to $T$ if there exist pairwise vertex-disjoint directed paths $(P_x)_{x \in X}$ such that $P_x$ starts at $x$ and ends at a vertex of $T$, and the terminal vertices of the paths are distinct. [/definition] A path of length zero is allowed when $x \in T$, so elements already in the target set can be linked to themselves. Vertex-disjointness includes the endpoints, which is why the terminals are automatically distinct once the paths are pairwise vertex-disjoint. Having named the linkable sets, we now collect them into the matroid candidate associated to a directed network. [definition: Gammoid] Let $D=(V,A)$ be a finite directed graph and let $T \subseteq V$. The gammoid represented by $(D,T)$ is the set system on $V$ whose independent sets are the subsets linkable to $T$. [/definition] The term gammoid is used for matroids obtained from directed linkages, so the definition is only justified once linkable sets are known to satisfy exchange. Some sources reserve "strict gammoid" for the case where the ground set is all of $V$; others allow a specified ground subset $E \subseteq V$ and take independent sets $X \subseteq E$ linkable to $T$ in $D$. [quotetheorem:6643] [citeproof:6643] Vertex-disjointness is the crucial hypothesis: if paths were only edge-disjoint, two chosen sources could collide at an internal vertex and then separate again, which would describe a different independence system. The theorem does not say that any individually linkable vertices can be linked simultaneously; two sources may both have all directed routes passing through the same bottleneck vertex. This bottleneck phenomenon is the directed analogue of a Hall obstruction, and it is exactly what the following examples make visible. [example: Linkage Matroid from a Directed Network] Let $D$ have vertices $s_1,s_2,a,b,t_1,t_2$ and directed edges $s_1a$, $s_2a$, $s_2b$, $at_1$, $bt_2$. Take $T=\{t_1,t_2\}$ and ground set $E=\{s_1,s_2,a,b\}$. The directed paths from $s_1$ to $T$ are forced to use \begin{align*} s_1\to a\to t_1, \end{align*} while $b$ has the path \begin{align*} b\to t_2. \end{align*} These two paths have vertex sets $\{s_1,a,t_1\}$ and $\{b,t_2\}$, which are disjoint, and their terminal vertices $t_1$ and $t_2$ are distinct. Hence $\{s_1,b\}$ is linkable. The set $\{s_1,s_2\}$ is also linkable. Use \begin{align*} P_{s_1}: s_1\to a\to t_1,\qquad P_{s_2}: s_2\to b\to t_2. \end{align*} Their vertex sets are \begin{align*} V(P_{s_1})=\{s_1,a,t_1\},\qquad V(P_{s_2})=\{s_2,b,t_2\}, \end{align*} and these sets are disjoint, so the two paths form a valid linkage to $T$. However, $\{s_1,s_2,a\}$ is not linkable. Any linkage of three vertices to $T=\{t_1,t_2\}$ would require three pairwise distinct terminal vertices in $T$, because vertex-disjoint paths cannot share an endpoint. But \begin{align*} |T|=|\{t_1,t_2\}|=2<3=|\{s_1,s_2,a\}|, \end{align*} so three distinct terminals are not available. There is also a visible bottleneck: the only path from $s_1$ to $T$ is $s_1\to a\to t_1$, and every path from $a$ to $T$ begins with $a\to t_1$, so paths starting at $s_1$ and $a$ would share the vertex $a$ and the terminal $t_1$. Thus the independent sets record vertex-capacity constraints in the directed network, not merely reachability to the sink set. [/example] Transversal matroids appear inside the gammoid framework by orienting a bipartite graph from elements to indices and taking the indices as sinks. A matching becomes a collection of vertex-disjoint directed paths of length one. [example: Transversal Matroids as Gammoids] Given a presentation $A_1,\dots,A_m\subseteq E$, take new vertices $t_1,\dots,t_m$ and form a directed graph $D$ with vertex set $E\cup\{t_1,\dots,t_m\}$ and with a directed edge \begin{align*} x\to t_i \end{align*} exactly when $x\in A_i$. Let $T=\{t_1,\dots,t_m\}$. We show that, for every $X\subseteq E$, being linkable to $T$ in $D$ is the same as being a partial transversal of $(A_i)_{i=1}^m$. Suppose first that $X$ is a partial transversal. Then there is an injective map $\varphi:X\to\{1,\dots,m\}$ such that $x\in A_{\varphi(x)}$ for every $x\in X$. By the definition of the edges of $D$, each membership $x\in A_{\varphi(x)}$ gives an edge \begin{align*} x\to t_{\varphi(x)}. \end{align*} For each $x\in X$, let $P_x$ be the one-edge directed path $x\to t_{\varphi(x)}$. If $x\ne y$, then the starting vertices $x$ and $y$ are distinct, and injectivity gives \begin{align*} \varphi(x)\ne \varphi(y), \end{align*} so \begin{align*} t_{\varphi(x)}\ne t_{\varphi(y)}. \end{align*} Thus the paths $P_x$ and $P_y$ have no common vertex. Hence the family $(P_x)_{x\in X}$ is a linkage of $X$ to $T$. Conversely, suppose $X$ is linkable to $T$. Then there are pairwise vertex-disjoint directed paths $(P_x)_{x\in X}$, where $P_x$ starts at $x$ and ends in $T$. Since the only directed edges leaving vertices of $E$ are edges of the form $x\to t_i$, and the vertices $t_i$ have no outgoing edges in this construction, each path starting at $x\in E$ and ending in $T$ must have the form \begin{align*} P_x:\ x\to t_i \end{align*} for some $i\in\{1,\dots,m\}$. Define $\varphi(x)=i$ when $P_x$ ends at $t_i$. The edge $x\to t_i$ exists only when $x\in A_i$, so \begin{align*} x\in A_{\varphi(x)} \end{align*} for every $x\in X$. If $x\ne y$, then the paths $P_x$ and $P_y$ are vertex-disjoint, so they cannot have the same terminal vertex; hence \begin{align*} t_{\varphi(x)}\ne t_{\varphi(y)} \end{align*} and therefore $\varphi(x)\ne\varphi(y)$. Thus $\varphi$ is injective, so $X$ is a partial transversal. Therefore the independent subsets of $E$ in this gammoid are exactly the partial transversals of $(A_i)_{i=1}^m$, which is precisely the transversal matroid $M[A_1,\dots,A_m]$. [/example] ## Duality Between Transversal Matroids and Strict Gammoids The final structural point of the chapter is a duality theorem. Transversal matroids are defined by matchings into sets, while strict gammoids are defined by linkages to sinks. The theorem says that these two constructions are dual to each other. [definition: Strict Gammoid] A matroid $M$ is a strict gammoid if there exists a finite directed graph $D=(V,A)$ and a subset $T\subseteq V$ such that the ground set of $M$ is $V$ and the independent sets of $M$ are exactly the subsets of $V$ linkable to $T$. [/definition] The adjective "strict" removes the option of restricting the ground set to only some of the vertices. This distinction is useful because the duality statement is cleanest in the strict case. [quotetheorem:6644] This theorem should be read as a structural bridge rather than as a new elementary construction in the present course. It says that the matching model and the directed-linkage model are not competing sources of examples: after duality, each supplies the other. The strictness hypothesis keeps the ground set aligned with the directed graph, which is why the statement is cleaner for strict gammoids than for arbitrary restrictions of gammoid presentations. This duality is important because the class of transversal matroids is not closed under duality by definition, but it becomes duality-stable after adjoining strict gammoids. It also helps explain why matching and directed routing are two faces of the same matroidal phenomenon. Transversal and gammoid constructions show how matching and routing naturally produce matroids, and how duality links the two. The next chapter shifts from constructing matroids to measuring them, using flats, minors, and rank to build polynomial invariants. # 10. Characteristic Polynomials and the Tutte Polynomial This chapter turns the structural language of flats and minors into numerical invariants. The guiding question is how much of a matroid can be detected by counting flats, colorings, bases, and spanning sets. The prerequisites are the rank function, closure, and flats from Chapter 2, deletion and contraction and direct sums from Chapter 5, duality from Chapter 4, and the graphic matroid construction from Chapter 7. We first attach a Möbius function to the lattice of flats, then use it to define the characteristic polynomial, and finally place that polynomial inside the two-variable Tutte polynomial. ## Möbius Inversion on the Lattice of Flats The previous chapters developed flats as closed sets, but now we ask what information is contained in the whole inclusion poset of flats. Ordinary inclusion-exclusion over all subsets is usually too fine: many subsets have the same closure and therefore impose the same rank or spanning condition. Since flats form a finite lattice, they support a Möbius function, which is the combinatorial analogue of inclusion-exclusion after subsets with the same closure have been identified. [definition: Lattice of Flats] Let $M$ be a matroid on a finite ground set $E$. The lattice of flats $L(M)$ is the set of flats of $M$, partially ordered by inclusion. [/definition] The meet of two flats is their intersection, while their join is the closure of their union. Thus $L(M)$ is not merely a poset: it records how spans combine in the matroid. [example: Flats of a Uniform Matroid] Let $M=U_{r,n}$ on ground set $E$. In a uniform matroid, a set $A\subseteq E$ has rank \begin{align*} r_M(A)=\min(|A|,r). \end{align*} If $|A|<r$ and $e\in E\setminus A$, then \begin{align*} r_M(A\cup\{e\})=\min(|A|+1,r)=|A|+1>|A|=r_M(A), \end{align*} so no element outside $A$ lies in $\operatorname{cl}(A)$. Hence $\operatorname{cl}(A)=A$, and every subset of size $<r$ is a flat. If $|A|\geq r$, then \begin{align*} r_M(A)=\min(|A|,r)=r. \end{align*} For every $e\in E\setminus A$, \begin{align*} r_M(A\cup\{e\})=\min(|A|+1,r)=r=r_M(A), \end{align*} so every element of $E\setminus A$ lies in $\operatorname{cl}(A)$. Thus $\operatorname{cl}(A)=E$. Therefore such an $A$ is a flat only when $A=E$. Consequently, \begin{align*} L(U_{r,n})=\{A\subseteq E:|A|<r\}\cup\{E\}. \end{align*} So the flat lattice agrees with the Boolean lattice below rank $r$, and all subsets of size at least $r$ collapse to the single top flat $E$. [/example] This example shows that $L(M)$ can differ from the full Boolean lattice precisely at the top ranks. To use the lattice for counting, we need coefficients that undo cumulative summation over intervals. These coefficients are the Möbius function, and the inversion theorem below is the formal version of inclusion-exclusion that will later convert Boolean sums into sums over flats. [definition: Möbius Function of a Finite Poset] Let $P$ be a finite poset. Its Möbius function is the function $\mu:P\times P\to \mathbb Z$ determined by $\mu(x,x)=1$, by $\mu(x,y)=0$ if $x\nleq y$, and by \begin{align*} \sum_{x\leq z\leq y}\mu(x,z)=0 \end{align*} whenever $x<y$. [/definition] For a loopless matroid, the bottom flat is $\varnothing$, and we usually write $\mu_F=\mu(\varnothing,F)$. If loops are present, the bottom flat is $\operatorname{cl}(\varnothing)$, the set of loops, so any formula using $\varnothing$ as the bottom flat must be read under a loopless convention. These integers are alternating inclusion-exclusion weights attached to flats. The next theorem is needed because the characteristic polynomial will be built from exactly these interval-inversion coefficients, not from arbitrary signs. [quotetheorem:6645] [citeproof:6645] The finiteness hypothesis is doing real work: all interval sums are finite, so the incidence-algebra inverse exists without convergence issues. For example, in the poset $(\mathbb N,\leq)$ the interval $[0,n]$ is finite, but the lower set below no top element is finite; trying to invert a relation such as $g(n)=\sum_{m\leq n}f(m)$ after passing to a formal top would require an infinite sum of coefficients with no value in an arbitrary abelian group. In matroid language, this is why the later characteristic-polynomial formula is stated for finite ground sets: an infinite-rank free matroid would have infinitely many flats below a formal top, so the coefficient of that top could not be recovered by a finite alternating sum. There is also a matroid-specific reason not to weaken the role of the flat lattice to rank data alone. In $U_{2,4}$, the four singleton flats and the top flat all have the ranks forced by the uniform matroid, but the top Möbius coefficient is determined by how all lower flats sit below the same top element; replacing $L(M)$ by only the sequence of Whitney numbers would lose the interval relation that recovers that coefficient. The theorem also does not say that every useful invariant is determined by the lattice of flats alone; it says that whenever a quantity is obtained by summing over lower flats, there is a canonical way to recover the unsummed contribution. This is the mechanism behind the characteristic polynomial, where geometric containment data becomes alternating enumerative data. ## The Characteristic Polynomial The central counting problem is to assign a polynomial to a matroid that behaves like the chromatic polynomial of a graph and like the complement-counting polynomial of a hyperplane arrangement. Counting flats by rank alone loses the alternating cancellations that make inclusion-exclusion work, while summing over all subsets counts many different sets with the same closure as though they imposed different geometric conditions. The lattice of flats supplies the right finite poset, and the rank function supplies the powers. [definition: Characteristic Polynomial] Let $M$ be a finite matroid on $E$ with rank function $r:2^E\to \mathbb Z_{\geq 0}$. The characteristic polynomial of $M$ is the polynomial $\chi_M\in \mathbb Z[q]$, where $\mathbb Z[q]$ denotes the ring of polynomials in the variable $q$ with integer coefficients, defined as follows. If $M$ has a loop, then \begin{align*} \chi_M(q)=0. \end{align*} If $M$ is loopless, with lattice of flats $L(M)$, then \begin{align*} \chi_M(q)=\sum_{F\in L(M)}\mu(\varnothing,F)q^{r(E)-r(F)}. \end{align*} [/definition] The exponent measures the codimension of the flat. In representable examples this mirrors the dimension of the corresponding intersection of hyperplanes. [example: Characteristic Polynomial of a Uniform Matroid] Assume $1\leq r\leq n$, and let $M=U_{r,n}$ on ground set $E$. From the description of flats of a uniform matroid, \begin{align*} L(U_{r,n})=\{A\subseteq E:|A|<r\}\cup\{E\}. \end{align*} If $A\subseteq E$ has $|A|=k<r$, then every subset of $A$ is also a flat, so the interval $[\varnothing,A]$ is the Boolean lattice of subsets of $A$. We show that $\mu(\varnothing,A)=(-1)^k$. For $k=0$, this is $\mu(\varnothing,\varnothing)=1=(-1)^0$. For $k>0$, assume the formula is known for all proper subsets of $A$. The Möbius recursion gives \begin{align*} 0=\sum_{B\subseteq A}\mu(\varnothing,B). \end{align*} Separating the top term and grouping the proper subsets by size gives \begin{align*} 0=\mu(\varnothing,A)+\sum_{j=0}^{k-1}\binom{k}{j}(-1)^j. \end{align*} Hence \begin{align*} \mu(\varnothing,A)=-\sum_{j=0}^{k-1}\binom{k}{j}(-1)^j. \end{align*} Since \begin{align*} 0=(1-1)^k=\sum_{j=0}^{k}\binom{k}{j}(-1)^j, \end{align*} we have \begin{align*} \sum_{j=0}^{k-1}\binom{k}{j}(-1)^j=-(-1)^k. \end{align*} Therefore \begin{align*} \mu(\varnothing,A)=(-1)^k=(-1)^{|A|}. \end{align*} It remains to compute the Möbius coefficient of the top flat $E$. Since $\varnothing<E$, the recursion on $[\varnothing,E]$ gives \begin{align*} 0=\sum_{F\leq E}\mu(\varnothing,F). \end{align*} The flats below $E$ are the subsets of $E$ of size less than $r$, together with $E$ itself, so \begin{align*} 0=\mu(\varnothing,E)+\sum_{k=0}^{r-1}\binom{n}{k}(-1)^k. \end{align*} Thus \begin{align*} \mu(\varnothing,E)=-\sum_{k=0}^{r-1}(-1)^k\binom{n}{k}. \end{align*} Using Pascal's identity, for $k\geq 0$ we write $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$, with $\binom{n-1}{-1}=0$. Then \begin{align*} \sum_{k=0}^{r-1}(-1)^k\binom{n}{k}=\sum_{k=0}^{r-1}(-1)^k\binom{n-1}{k}+\sum_{k=1}^{r-1}(-1)^k\binom{n-1}{k-1}. \end{align*} Reindexing the second sum by $j=k-1$ gives \begin{align*} \sum_{k=1}^{r-1}(-1)^k\binom{n-1}{k-1}=-\sum_{j=0}^{r-2}(-1)^j\binom{n-1}{j}. \end{align*} Therefore all terms cancel except the final $k=r-1$ term in the first sum: \begin{align*} \sum_{k=0}^{r-1}(-1)^k\binom{n}{k}=(-1)^{r-1}\binom{n-1}{r-1}. \end{align*} Hence \begin{align*} \mu(\varnothing,E)=(-1)^r\binom{n-1}{r-1}. \end{align*} Now use the definition of the characteristic polynomial. Since $r_M(E)=r$, a proper flat $A$ with $|A|=k<r$ has rank $k$ and contributes $(-1)^k q^{r-k}$. There are $\binom{n}{k}$ such flats of size $k$. The top flat $E$ contributes \begin{align*} \mu(\varnothing,E)q^{r-r}=(-1)^r\binom{n-1}{r-1}. \end{align*} Combining these contributions gives \begin{align*} \chi_{U_{r,n}}(q)=\sum_{k=0}^{r-1}(-1)^k\binom{n}{k}q^{r-k}+(-1)^r\binom{n-1}{r-1}. \end{align*} Thus the uniform matroid agrees with the Boolean lattice through rank $r-1$, while all subsets of size at least $r$ are represented by the single top flat $E$. The assumption $r\geq 1$ matters: if $n>0$, then $U_{0,n}$ has every element as a loop, so its characteristic polynomial is $0$ by definition. [/example] The uniform computation shows that characteristic polynomials package many Möbius coefficients into one object. For graphic matroids, the same package should recover a more familiar polynomial: the chromatic polynomial. The following result identifies the exact normalization and explains why each connected component contributes one free color choice outside the matroidal data. [quotetheorem:6646] [citeproof:6646] The hypothesis that $M(G)$ is the graphic matroid matters because the rank formula $r(A)=|V|-c(V,A)$ is what translates monochromatic edge constraints into matroid rank. A concrete failure occurs for the non-graphic matroid $U_{2,4}$: its characteristic polynomial is $q^2-4q+3$, so the expression $q\chi_{U_{2,4}}(q)=q(q-1)(q-3)$ would take the value $-2$ at $q=2$, while a graph-coloring count can never be negative. The result also does not claim that the graph itself is determined by $\chi_{M(G)}(q)$; non-isomorphic graphs can have the same graphic matroid, and different matroids can share a characteristic polynomial. For a connected graph, this says $P_G(q)=q\chi_{M(G)}(q)$, so the matroid characteristic polynomial is the chromatic polynomial with the unavoidable choice of a color on each connected component removed. [example: Triangle Graph] Let $G=C_3$, with edge set $E$ of size $3$. In the graphic matroid $M(G)$, a set of edges is independent exactly when it contains no cycle. The only cycle in $C_3$ uses all three edges, so the independent sets are precisely the subsets of $E$ of size at most $2$. Hence $M(G)=U_{2,3}$. Using the uniform-matroid characteristic-polynomial formula with $r=2$ and $n=3$, we get \begin{align*} \chi_{M(G)}(q)=\sum_{k=0}^{1}(-1)^k\binom{3}{k}q^{2-k}+(-1)^2\binom{2}{1}. \end{align*} The two terms in the sum are \begin{align*} (-1)^0\binom{3}{0}q^2=q^2. \end{align*} and \begin{align*} (-1)^1\binom{3}{1}q=-3q. \end{align*} The top-flat contribution is \begin{align*} (-1)^2\binom{2}{1}=2. \end{align*} Therefore \begin{align*} \chi_{M(G)}(q)=q^2-3q+2. \end{align*} Also, \begin{align*} (q-1)(q-2)=q^2-2q-q+2=q^2-3q+2. \end{align*} Thus \begin{align*} \chi_{M(G)}(q)=(q-1)(q-2). \end{align*} Since $C_3$ is connected, $c(G)=1$. By *Chromatic Polynomial from a Graphic Matroid*, \begin{align*} P_G(q)=q^{c(G)}\chi_{M(G)}(q)=q\chi_{M(G)}(q). \end{align*} Substituting the computed characteristic polynomial gives \begin{align*} P_G(q)=q(q^2-3q+2). \end{align*} Using the factorization above, \begin{align*} P_G(q)=q(q-1)(q-2). \end{align*} This matches the elementary coloring count: choose the first vertex in $q$ ways, the second in $q-1$ ways, and the third in $q-2$ ways because it is adjacent to both earlier vertices. [/example] The same definition also reflects the hyperplane-arrangement motivation. If a matroid is represented by normals to hyperplanes over a finite field, then the characteristic polynomial counts points in the complement for sufficiently compatible field sizes. [remark: Arrangement Motivation] For a central arrangement $\mathcal A=\{H_e:e\in E\}$ of hyperplanes, the intersection lattice is ordered by reverse inclusion: larger flats in the matroid correspond to smaller intersections $\bigcap_{e\in F}H_e$. The lattice of flats of the associated matroid records exactly which subfamilies have the same intersection, so the characteristic polynomial is the Möbius-weighted sum over these intersections rather than over all subsets of hyperplanes. Over finite fields, the finite-field method applies inclusion-exclusion to the complement $\mathbb F_q^d\setminus\bigcup_{e\in E}H_e$. A point survives precisely when it avoids every hyperplane, and the number of points lying in an intersection indexed by a flat $F$ is $q^{d-r(F)}$ after the normals have rank $r(F)$. Grouping the inclusion-exclusion sum by intersections gives $\chi_M(q)$, for field sizes where the represented intersection lattice is the intended one. [/remark] This motivation explains why deletion and contraction should enter the theory: removing a hyperplane and restricting to it are the arrangement analogues of deleting and contracting an element of the matroid. ## The Tutte Polynomial and Deletion-Contraction The characteristic polynomial is powerful, but it records only one slice of the deletion-contraction behaviour of a matroid. For example, evaluating the Tutte polynomial at $(1,1)$ will count bases, while the characteristic polynomial alone cannot recover the number of bases of a general matroid. The Tutte polynomial is designed to keep both spanning failure and independence failure at once, producing the universal two-variable invariant governed by deletion and contraction, with special rules for loops and coloops. [definition: Tutte Polynomial] Let $M$ be a finite matroid on $E$ with rank function $r:2^E\to \mathbb Z_{\geq 0}$. The Tutte polynomial of $M$ is the polynomial $T_M\in \mathbb Z[x,y]$, where $\mathbb Z[x,y]$ denotes the ring of polynomials in the variables $x$ and $y$ with integer coefficients, defined by \begin{align*} T_M(x,y)=\sum_{A\subset E}(x-1)^{r(E)-r(A)}(y-1)^{|A|-r(A)}. \end{align*} [/definition] The two exponents separate two defects. The first measures how far $A$ is from spanning, and the second measures how far $A$ is from being independent. [example: Tutte Polynomial of a Boolean Matroid] Let $M$ be the rank $n$ matroid on ground set $E$ in which every element is a coloop. Then $|E|=n$, every subset $A\subset E$ is independent, and hence \begin{align*} r(A)=|A|. \end{align*} Also $r(E)=n$. Substituting these values into the definition of the Tutte polynomial gives \begin{align*} T_M(x,y)=\sum_{A\subset E}(x-1)^{n-|A|}(y-1)^{|A|-|A|}. \end{align*} Since $|A|-|A|=0$ and $(y-1)^0=1$, this becomes \begin{align*} T_M(x,y)=\sum_{A\subset E}(x-1)^{n-|A|}. \end{align*} There are $\binom{n}{k}$ subsets of $E$ of size $k$, so grouping by $k=|A|$ gives \begin{align*} T_M(x,y)=\sum_{k=0}^{n}\binom{n}{k}(x-1)^{n-k}. \end{align*} By the [binomial theorem](/theorems/750), \begin{align*} x^n=((x-1)+1)^n=\sum_{k=0}^{n}\binom{n}{k}(x-1)^{n-k}. \end{align*} Therefore \begin{align*} T_M(x,y)=x^n. \end{align*} This shows that, under the Tutte polynomial convention used here, each coloop contributes one factor of $x$. [/example] The Boolean matroid computation isolates the contribution of coloops, so it raises the next structural question: how does one peel off a single element from an arbitrary matroid without losing control of the polynomial? The answer must distinguish loops, coloops, and ordinary elements, because those are exactly the cases in which deletion and contraction affect rank differently. The following theorem gives the recurrence that turns the subset definition into a usable computational tool. [quotetheorem:6647] [citeproof:6647] The three cases cannot be merged without changing the polynomial: loops never increase rank and coloops always increase rank when added to a spanning-independent core. The one-element loop matroid has $T_M(x,y)=y$, whereas treating the loop as an ordinary element would give $T_{M\setminus e}+T_{M/e}=1+1=2$. Dually, the one-element coloop matroid has $T_M(x,y)=x$, while the ordinary recurrence would again give $2$. The theorem is a recurrence, not merely a computational shortcut, because it characterises how the two rank defects in the subset expansion behave under the two minor operations. This is why graph coloring, spanning trees, and arrangement complements can all appear as different evaluations of the same rank expansion. [example: Spanning Trees from the Tutte Polynomial] Let $G$ be a connected graph with edge set $E$, and let $M(G)$ be its graphic matroid. In a graphic matroid, independent sets are forests, so a basis is an independent edge set of rank $r(E)$; because $G$ is connected, this means a maximal forest on all vertices, equivalently a spanning tree. Evaluate the subset expansion at $(x,y)=(1,1)$: \begin{align*} T_{M(G)}(1,1)=\sum_{A\subseteq E}(1-1)^{r(E)-r(A)}(1-1)^{|A|-r(A)}. \end{align*} Since $1-1=0$, this is \begin{align*} T_{M(G)}(1,1)=\sum_{A\subseteq E}0^{r(E)-r(A)}0^{|A|-r(A)}. \end{align*} The exponents are nonnegative because $r(A)\leq r(E)$ and $r(A)\leq |A|$. The first factor equals $1$ exactly when $r(E)-r(A)=0$, that is, when $A$ spans $M(G)$; otherwise it equals $0$. The second factor equals $1$ exactly when $|A|-r(A)=0$, that is, when $A$ is independent; otherwise it equals $0$. Thus a summand contributes $1$ exactly for subsets $A$ that are both spanning and independent, and contributes $0$ for all other subsets. Therefore \begin{align*} T_{M(G)}(1,1)=\#\{A\subseteq E:A\text{ is spanning and independent in }M(G)\}. \end{align*} The spanning independent sets are exactly the bases, so \begin{align*} T_{M(G)}(1,1)=\#\{\text{bases of }M(G)\}. \end{align*} Since the bases of $M(G)$ are the spanning trees of $G$, we get \begin{align*} T_{M(G)}(1,1)=\#\{\text{spanning trees of }G\}. \end{align*} For the cycle graph $C_n$, there are $n$ edges. Removing one edge destroys the unique cycle and leaves a connected graph on the same vertex set with $n-1$ edges, hence a spanning tree. Conversely, every spanning tree of $C_n$ has $n-1$ edges, so it is obtained from the full $n$-edge cycle by omitting exactly one edge. There are $n$ choices for the omitted edge, and consequently \begin{align*} T_{M(C_n)}(1,1)=n. \end{align*} This evaluation shows that the point $(1,1)$ of the Tutte polynomial counts bases, and in the graphic case those bases are precisely spanning trees. [/example] The spanning-tree specialization illustrates how a single point of the Tutte plane can encode a classical invariant. The characteristic polynomial should also occupy a specific place in this plane, since both invariants are built from rank data. The next theorem gives the exact substitution and supplies a bridge from the flat-lattice definition to deletion-contraction. [quotetheorem:6648] [citeproof:6648] The loop case explains why the convention in the definition of $\chi_M(q)$ is not cosmetic: for a single loop, the Tutte polynomial is $y$, so the specialization gives $0$, matching the definition of $\chi_M(q)$ but not any formula that pretends $\varnothing$ is the bottom flat. The rank factor is also essential. For the one-element coloop matroid, $T_M(1-q,0)=1-q$, and the sign $(-1)^{r(E)}$ converts this to $q-1$, the correct characteristic polynomial. The substitution $y=0$ also has content: it cancels subsets with positive nullity and leaves precisely the alternating closure data encoded by the Möbius function. Thus the characteristic polynomial is not an isolated construction; it is the $y=0$ boundary of the Tutte polynomial, after a sign-normalized change of variables. This specialization loses information: many distinct Tutte polynomials have the same value along the line $(1-q,0)$, so $\chi_M(q)$ cannot recover the full Tutte polynomial. [example: Chromatic Polynomial as a Tutte Specialization] Let $G$ be a graph with $c(G)$ connected components, edge set $E$, and graphic matroid $M(G)$. By *Chromatic Polynomial from a Graphic Matroid*, \begin{align*} P_G(q)=q^{c(G)}\chi_{M(G)}(q). \end{align*} By *Characteristic Polynomial as a Tutte Specialization*, applied to $M(G)$, \begin{align*} \chi_{M(G)}(q)=(-1)^{r(E)}T_{M(G)}(1-q,0). \end{align*} Substituting the second identity into the first gives \begin{align*} P_G(q)=q^{c(G)}\left((-1)^{r(E)}T_{M(G)}(1-q,0)\right). \end{align*} Since multiplication of polynomial factors is associative, this is \begin{align*} P_G(q)=q^{c(G)}(-1)^{r(E)}T_{M(G)}(1-q,0). \end{align*} Thus the chromatic polynomial is obtained from the Tutte polynomial of the graphic matroid by setting $x=1-q$ and $y=0$, multiplying by the sign $(-1)^{r(E)}$, and then multiplying by $q^{c(G)}$. The deletion-contraction recurrence for graph coloring is therefore the specialization of *Deletion-Contraction for the Tutte Polynomial* under this substitution. When $G$ is connected, $c(G)=1$, so the prefactor is $q^{c(G)}=q$, which records the free choice of one global color on the single connected component. [/example] ## The Beta Invariant The final problem in this chapter is to extract connectedness information from a polynomial with many coefficients. Neither the degree nor the value at a single generic point gives a reliable connectedness test, because direct sums multiply Tutte polynomials and can hide their decomposition across many terms. The beta invariant isolates the first connected contribution: it vanishes on disconnected matroids, equals $1$ for circuits, and counts several natural objects in special classes. [definition: Beta Invariant] Let $M$ be a finite matroid of positive rank. The beta invariant of $M$ is the integer $\beta(M)\in \mathbb Z$ defined by \begin{align*} \beta(M)=(-1)^{r(E)-1}\chi_M'(1). \end{align*} [/definition] Through the Tutte specialization, this derivative is also encoded in the coefficient of $x$ in $T_M(x,y)$ when $M$ has no loops or coloops. The derivative form is often the most direct for computations from flats. [quotetheorem:6649] [citeproof:6649] These properties make $\beta(M)$ a compact test for whether the Tutte polynomial contains genuinely connected information, but each hypothesis is doing work. Positive rank excludes the rank-zero matroids, whose characteristic polynomial is either $1$ in the empty case or $0$ in the presence of loops, so the derivative formula would not measure connected structure. The duality clause also needs the positive-rank hypothesis on the dual matroid: a positive-rank Boolean matroid has rank-zero dual, so the beta invariant of the dual is not defined under the present convention. The requirement that the connected matroid have at least two elements excludes the one-element coloop, for which $\chi_M(q)=q-1$ and $\beta(M)=1$ although the connectedness statement is not the phenomenon being detected. The disconnected example also matters: for $M=U_{1,2}\oplus U_{1,2}$, the Tutte polynomial is $(x+y)^2$, and the degree-one beta coefficient vanishes even though each summand is connected. The theorem does not make $\beta(M)$ a complete invariant; many connected matroids have the same beta invariant. Its role is to detect the first nonzero connected coefficient, not to reconstruct the matroid. [example: Beta Invariant of a Circuit] Let $M=U_{n-1,n}$ with $n\geq 2$, so $M$ is the matroid of a single $n$-element circuit and $r(E)=n-1$. Using the uniform-matroid characteristic-polynomial formula with $r=n-1$ gives \begin{align*} \chi_M(q)=\sum_{k=0}^{n-2}(-1)^k\binom{n}{k}q^{n-1-k}+(-1)^{n-1}\binom{n-1}{n-2}. \end{align*} Since $\binom{n-1}{n-2}=n-1$, this becomes \begin{align*} \chi_M(q)=\sum_{k=0}^{n-2}(-1)^k\binom{n}{k}q^{n-1-k}+(-1)^{n-1}(n-1). \end{align*} We rewrite this polynomial in a form whose derivative at $q=1$ is visible. By the binomial theorem, \begin{align*} (q-1)^n=\sum_{k=0}^{n}\binom{n}{k}(-1)^kq^{n-k}. \end{align*} Adding $(-1)^n(q-1)$ gives \begin{align*} (q-1)^n+(-1)^n(q-1)=\sum_{k=0}^{n}\binom{n}{k}(-1)^kq^{n-k}+(-1)^nq-(-1)^n. \end{align*} The $k=n$ term in the sum is $(-1)^n$, so it cancels with $-(-1)^n$. The $k=n-1$ term is $n(-1)^{n-1}q$, and \begin{align*} n(-1)^{n-1}q+(-1)^nq=(-1)^{n-1}(n-1)q. \end{align*} Therefore \begin{align*} (q-1)^n+(-1)^n(q-1)=\sum_{k=0}^{n-2}(-1)^k\binom{n}{k}q^{n-k}+(-1)^{n-1}(n-1)q. \end{align*} Dividing by $q$ gives \begin{align*} \frac{(q-1)^n+(-1)^n(q-1)}{q}=\sum_{k=0}^{n-2}(-1)^k\binom{n}{k}q^{n-1-k}+(-1)^{n-1}(n-1). \end{align*} Hence \begin{align*} \chi_M(q)=\frac{(q-1)^n+(-1)^n(q-1)}{q}. \end{align*} For example, when $n=3$, \begin{align*} \chi_M(q)=\frac{(q-1)^3-(q-1)}{q}. \end{align*} Expanding the numerator gives \begin{align*} (q-1)^3-(q-1)=q^3-3q^2+3q-1-q+1. \end{align*} Combining like terms gives \begin{align*} (q-1)^3-(q-1)=q^3-3q^2+2q. \end{align*} Thus \begin{align*} \chi_M(q)=q^2-3q+2. \end{align*} To compute the beta invariant, put $t=q-1$. Then $q=t+1$ and \begin{align*} \chi_M(q)=\frac{t^n+(-1)^nt}{t+1}. \end{align*} Since $dt/dq=1$, differentiating with respect to $q$ is the same as differentiating with respect to $t$. The quotient rule gives \begin{align*} \chi_M'(q)=\frac{(nt^{n-1}+(-1)^n)(t+1)-(t^n+(-1)^nt)}{(t+1)^2}. \end{align*} At $q=1$ we have $t=0$. Because $n\geq 2$, $0^{n-1}=0$, so \begin{align*} \chi_M'(1)=\frac{(-1)^n\cdot 1-0}{1^2}. \end{align*} Therefore \begin{align*} \chi_M'(1)=(-1)^n. \end{align*} Using the definition of the beta invariant and $r(E)=n-1$, \begin{align*} \beta(M)=(-1)^{r(E)-1}\chi_M'(1). \end{align*} Substituting $r(E)=n-1$ and $\chi_M'(1)=(-1)^n$ gives \begin{align*} \beta(M)=(-1)^{n-2}(-1)^n. \end{align*} Since $(-1)^{n-2}(-1)^n=(-1)^{2n-2}=1$, we get \begin{align*} \beta(M)=1. \end{align*} Thus an $n$-element circuit has beta invariant $1$, reflecting that all of its dependence is concentrated in one minimal circuit. [/example] The chapter therefore ends with a hierarchy of invariants. The lattice of flats gives Möbius weights; those weights define $\chi_M(q)$; the Tutte polynomial contains $\chi_M(q)$ as a specialization; and the beta invariant extracts a connectedness-sensitive coefficient from this structure. The Tutte and characteristic polynomials compress structural information into numerical invariants while still remembering deletion, contraction, flats, and connectivity. We now move from counting invariants to geometric shadows, where matroids appear inside tropical linear spaces and valuated exchange. # 11. Matroids, Tropical Geometry, and Valuated Matroids This chapter brings matroids into contact with tropical geometry. The guiding theme is that ordinary linear algebra over a valued field has a shadow in piecewise-linear geometry, and the shadow remembers matroidal data. We use circuits, flats, and bases from Chapters 1 and 2, polyhedral language from Chapter 3, and representability language from Chapter 6; Plucker coordinates enter here as the new linear-algebraic input. We move from the Bergman fan of an unvalued matroid, through tropical linear spaces attached to realizations, to valuated matroids and their polyhedral subdivisions of hypersimplices. ## Tropicalization and Linear Dependence What remains of a linear subspace when addition is replaced by taking minima? Tropical geometry answers this by recording leading valuations, so the first task is to identify the matroidal condition that survives tropicalization. A naive attempt would tropicalize each linear equation by taking a single minimum, but that would miss the algebraic fact that a sum can vanish only through cancellation among at least two lowest-order terms. [definition: Tropical Semiring Convention] The tropical semiring is $(\mathbb R \cup \{\infty\}, \oplus, \odot)$, where \begin{align*} \oplus &: (\mathbb R \cup \{\infty\})^2 \to \mathbb R \cup \{\infty\}, \qquad a \oplus b = \min(a,b). \end{align*} The tropical multiplication is \begin{align*} \odot &: (\mathbb R \cup \{\infty\})^2 \to \mathbb R \cup \{\infty\}, \qquad a \odot b = a + b. \end{align*} [/definition] The minimum convention is natural for valuations: the valuation of a sum is at least the minimum valuation of the summands, with equality unless cancellation occurs among terms of minimal valuation. Tropical linear equations therefore do not ask for an ordinary sum to vanish; they ask for the minimum among several affine-linear expressions to be attained at least twice. The next definition isolates the basic tropical equation whose intersections will model linear spaces. [definition: Tropical Hyperplane] For $a=(a_1,\dots,a_n) \in (\mathbb R \cup \{\infty\})^n$ with at least two finite coordinates, the tropical hyperplane determined by $a$ is the subset \begin{align*} H_a \subset \mathbb R^n \end{align*} consisting of all $w \in \mathbb R^n$ such that the minimum of $a_i+w_i$ over all $i$ with $a_i \ne \infty$ is attained for at least two indices. [/definition] This condition is the tropical replacement for a linear equation. If a classical relation $\sum_i c_i x_i=0$ is evaluated at a point over a valued field, cancellation can occur only when at least two terms have the lowest valuation. The smallest example is the tropical line, where a single circuit supplies the whole condition. [example: Tropical Line From Uniform Rank Two Matroid] For the uniform matroid $U_{2,3}$, the only circuit is $C=\{1,2,3\}$. Thus the Bergman circuit condition says that a class of $w=(w_1,w_2,w_3)$ belongs to the tropical linear space exactly when \begin{align*} \min\{w_1,w_2,w_3\} \end{align*} is attained for at least two of the three indices. To see the shape in the quotient by $\mathbb R(1,1,1)$, let \begin{align*} m=\min\{w_1,w_2,w_3\}. \end{align*} Replacing $w$ by \begin{align*} w-m(1,1,1)=(w_1-m,w_2-m,w_3-m) \end{align*} does not change its class in $\mathbb R^3/\mathbb R(1,1,1)$, and the new coordinates are nonnegative with minimum $0$. Since the minimum must occur at least twice, the normalized vector has one of the forms \begin{align*} (a,0,0),\qquad (0,a,0),\qquad (0,0,a) \end{align*} with $a\ge 0$. For $a=0$ these three forms give the same vertex, and for $a>0$ they give three rays in directions represented by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ modulo the diagonal. Thus the tropical line is exactly three rays meeting at one vertex. [/example] The example already shows the central feature of the chapter. A tropical linear space is controlled by which subsets are minimally dependent, and those subsets are the circuits of a matroid. ## The Bergman Fan of a Matroid Given an abstract matroid with no coordinates, can we still construct the tropical object that would have arisen from a realization? The Bergman fan answers this question by imposing the circuit minimum condition directly from the matroid. This is stronger than intersecting arbitrary tropical hyperplanes attached to an arbitrary presentation: different presentations can introduce redundant equations, while the circuits are the minimal dependences that cannot be omitted without changing the tropical condition. [definition: Bergman Fan] Let $M$ be a matroid on finite ground set $E$. The Bergman fan $\widetilde{\mathcal B}(M)$ is the set of all $w \in \mathbb R^E$ such that, for every circuit $C$ of $M$, the minimum of $\{w_e : e \in C\}$ is attained at least twice. [/definition] The diagonal line $\mathbb R(1,\dots,1)$ is always contained in the lineality space, since adding the same constant to every coordinate does not change where minima are attained. Many authors work with the quotient $\mathcal B(M)=\widetilde{\mathcal B}(M)/\mathbb R(1,\dots,1)$. A graphic matroid makes the condition concrete because its circuits are graph cycles. [example: Graphic Bergman Fan] Let $M(G)$ be the graphic matroid of a connected graph $G$. Its ground set is $E(G)$, and its circuits are exactly the edge sets of simple cycles in $G$. Therefore the definition of the Bergman fan says that $w\in\mathbb R^{E(G)}$ lies in $\widetilde{\mathcal B}(M(G))$ exactly when, for every simple cycle $C\subseteq E(G)$, the minimum \begin{align*} \min\{w_e:e\in C\} \end{align*} is attained by at least two edges of $C$. For a triangle with edges $e_{12},e_{23},e_{31}$, there is only one simple cycle: \begin{align*} C=\{e_{12},e_{23},e_{31}\}. \end{align*} Thus the Bergman condition is precisely that \begin{align*} \min\{w_{12},w_{23},w_{31}\} \end{align*} is attained at least twice. This is the same repeated-minimum condition that defines the tropical line for $U_{2,3}$. For a square with vertices $1,2,3,4$ and diagonal $13$, write the five edge weights as \begin{align*} w_{12},\quad w_{23},\quad w_{34},\quad w_{41},\quad w_{13}. \end{align*} The simple cycles are the triangle using the diagonal and the path $1-2-3$, \begin{align*} C_1=\{12,23,13\}, \end{align*} the triangle using the diagonal and the path $1-4-3$, \begin{align*} C_2=\{13,34,41\}, \end{align*} and the four-cycle around the square, \begin{align*} C_3=\{12,23,34,41\}. \end{align*} Hence $w$ belongs to $\widetilde{\mathcal B}(M(G))$ exactly when each of the three minima \begin{align*} \min\{w_{12},w_{23},w_{13}\}, \end{align*} \begin{align*} \min\{w_{13},w_{34},w_{41}\}, \end{align*} and \begin{align*} \min\{w_{12},w_{23},w_{34},w_{41}\} \end{align*} is attained at least twice. Thus the graphic Bergman fan records exactly the edge-weight assignments in which no cycle has a unique lightest edge. [/example] The circuit definition is compact, but the fan structure is most transparent when expressed in terms of flats. Chains of flats give cones, and this description shows which data of the matroid the Bergman fan remembers. We therefore introduce the cones that will form the combinatorial pieces of the fan. [definition: Flag Cone] Let $M$ be a matroid on $E$, and let \begin{align*} \varnothing = F_0 \subsetneq F_1 \subsetneq \cdots \subsetneq F_k = E \end{align*} be a chain of flats. The associated flag cone is the cone in $\mathbb R^E$ consisting of all vectors constant on each difference $F_i \setminus F_{i-1}$ and weakly decreasing along the chain. [/definition] Flag cones organize the Bergman fan into a polyhedral fan, but the circuit definition and the flat-chain definition look like different objects. The key obstruction is to prove that the repeated-minimum condition on every circuit is equivalent to lying in cones determined only by the lattice of flats. [quotetheorem:6650] [citeproof:6650] This theorem turns a local circuit condition into a global combinatorial fan. The loopless hypothesis is essential: if $e$ is a loop, the circuit $\{e\}$ forces a unique minimum on a singleton, so the finite-coordinate Bergman fan is empty even though the lattice of flats can still be discussed. The statement is also limited to the unvalued setting; two realizations with the same flats but different coefficient valuations can have different tropical linear spaces, which is why valuated matroids enter later. The forward use of the theorem is that it gives a computable test for Bergman membership: sort the coordinates of $w$ from large to small, form the strict superlevel sets, and check that each set is a flat. [remark: Loops And Coloops In The Bergman Fan] If $M$ has a loop $e$, then $\{e\}$ is a circuit, and the minimum on that circuit is attained only once. Thus the Bergman fan is empty for a matroid with loops under the finite-coordinate convention. Coloops impose no circuit condition, so they contribute extra lineality directions. [/remark] ## Tropical Linear Spaces From Realizations How does the Bergman fan arise from an actual linear subspace over a valued field? The connection passes through valuations of points and Plucker coordinates. [definition: Valued Field For Tropicalization] A valued field is a field $K$ equipped with a map $\nu:K \to \mathbb R \cup \{\infty\}$ such that $\nu(0)=\infty$, $\nu(xy)=\nu(x)+\nu(y)$, and $\nu(x+y)\ge \min(\nu(x),\nu(y))$ for all $x,y \in K$. [/definition] The valued-field axioms are the mechanism that turns cancellation in ordinary algebra into repeated minima in tropical algebra. Puiseux series provide the standard source of examples: the valuation records the lowest exponent of the parameter. To connect this mechanism with a specific linear subspace, we now collect the valuation vectors of its nonzero-coordinate points; that collection is the tropical linear space attached to the realization. [definition: Tropical Linear Space Of A Realized Matroid] Let $L \subset K^E$ be a linear subspace over a valued field $K$. The tropicalization $\operatorname{Trop}(L)$ is the set of all valuation vectors $\nu(x) \in \mathbb R^E$ arising from points $x \in L$ with all coordinates nonzero, together with its closure in $\mathbb R^E$. [/definition] The closure is needed because limits of valuation vectors often correspond to degenerations rather than individual points with all coordinates nonzero. For comparison theorems, we also use the standard extension convention: if a lift exists only after replacing $K$ by a valued extension $K'$, its valuation vector is included before taking closure. This convention prevents the tropical set from depending on accidental restrictions of the original value group or on the failure of a required algebraic lift to lie in $K$ itself. The following theorem identifies the defining equations of this tropicalization when the realization is defined over a subfield whose nonzero elements all have valuation $0$. Without that restriction, the same underlying matroid can produce shifted circuit equations, so the ordinary Bergman fan loses the coefficient-valuation data. [definition: Scalar Extension Convention] If $K'/K$ is a valued [field extension](/page/Field%20Extension) and $L\subset K^E$ is a $K$-linear subspace, then $L\otimes_K K'$ denotes the scalar extension of $L$ from $K$ to $K'$. Concretely, it is the $K'$-linear subspace of $(K')^E$ obtained by allowing the same linear combinations as in $L$, but with coefficients in the larger field $K'$. This is the ambient space used when a tropical point has to be lifted after extending the valued field. [/definition] With this convention in place, the next theorem compares two ways of extracting a tropical object from the same realization. One side starts from valuation vectors of points in the linear space, allowing extension fields when a lift is not available over $K$ itself. The other side uses only the ordinary matroid of the coordinate functionals and imposes the Bergman repeated-minimum condition on every circuit. The zero-valuation hypothesis on the base field is what makes these two constructions match without additional coefficient shifts. [quotetheorem:6651] [citeproof:6651] The theorem explains why matroid theorists can study these tropical linear spaces without choosing coordinates, provided the tropicalization is taken with the extension-and-closure convention. The zero-valuation hypothesis cannot be dropped: the columns $(1,0),(0,1),(1,t)$ over Puiseux series still have the uniform rank two matroid, but the relation involving the third column carries a $t$-weight and shifts the corresponding tropical equation. The theorem is also a torus statement followed by closure; points with zero coordinates are represented only as limiting directions, not as direct valuation vectors in $\mathbb R^E$. This limitation is the reason the next section records the valuations of Plucker coordinates rather than only their vanishing pattern. ## Valuated Matroids and Tropical Plucker Relations Ordinary matroids remember which bases exist. What structure records not only the set of bases, but also a weight attached to each basis in a way compatible with exchange? A rank two configuration over Puiseux series may have all six pairs as bases, while one minor has valuation $1$ and the others have valuation $0$; the ordinary matroid sees only six nonzero minors and forgets the order of vanishing that changes the tropical linear space. [definition: Valuated Matroid] Let $E$ be a finite set and let $r \in \mathbb N$. A valuated matroid of rank $r$ on $E$ is a function \begin{align*} p: \binom{E}{r} \longrightarrow \mathbb R \cup \{\infty\} \end{align*} whose finite-valued subsets form the bases of a matroid and which satisfies the tropical Plucker relations: for every $S\subset E$ with $|S|=r-1$ and every $T\subset E$ with $|T|=r+1$, the minimum \begin{align*} \min_{i\in T\setminus S}\bigl(p(S\cup\{i\})+p(T\setminus\{i\})\bigr) \end{align*} is attained at least twice whenever the displayed set of terms is nonempty. [/definition] The definition refers to the tropical Plucker relations, but the raw Plucker equations are not the most usable form in matroid theory. Since the ordinary basis exchange axiom was the organizing principle for unvalued matroids, we want a version that compares two bases by exchanging elements. The next theorem gives exactly this exchange form. [quotetheorem:6652] [citeproof:6652] The inequality says that exchanging $e$ out of $B$ can be balanced by exchanging some $f$ out of $B'$ without increasing the total weight. The nonempty-basis hypothesis is necessary because the function identically $\infty$ satisfies no meaningful exchange geometry and would not define a rank $r$ matroid. The valuation inequality is stronger than ordinary exchange: on the bases of $U_{2,4}$, the assignment $p(12)=p(34)=0$ and $p(13)=p(14)=p(23)=p(24)=10$ breaks the relation for $B=12$, $B'=34$, and $e=1$. The theorem is limited to basis weights; it does not itself describe the tropical linear space, but it gives the exchange mechanism used to build valuated circuits next. [example: Constant Valuation From An Ordinary Matroid] Let $M$ be a rank $r$ matroid on $E$ with set of bases $\mathcal B$. Define $p$ on the $r$-subsets of $E$ by setting $p(B)=0$ when $B\in\mathcal B$ and $p(B)=\infty$ when $B\notin\mathcal B$. The finite-valued $r$-subsets are exactly the bases of $M$, so they form a nonempty matroid basis set. To verify the valuated exchange inequality, take $B,B'\in\mathcal B$ and $e\in B\setminus B'$. By the *symmetric basis exchange theorem*, there is an element $f\in B'\setminus B$ such that $B-e+f\in\mathcal B$ and $B'-f+e\in\mathcal B$. Hence \begin{align*} p(B)+p(B')=0+0=0. \end{align*} For the exchanged bases, the same definition gives \begin{align*} p(B-e+f)+p(B'-f+e)=0+0=0. \end{align*} Therefore \begin{align*} p(B)+p(B')=p(B-e+f)+p(B'-f+e). \end{align*} In particular, \begin{align*} p(B)+p(B')\ge p(B-e+f)+p(B'-f+e). \end{align*} By the basis-exchange form of the tropical Plucker relations, $p$ is a valuated matroid. Thus ordinary matroids sit inside valuated matroids as the special case where every basis has valuation $0$ and every nonbasis has valuation $\infty$. [/example] To build a valuated matroid from linear algebra, the candidate weights must be the valuations of maximal minors. The obstruction is that valuations remember only leading orders, so the construction works only if the Plucker equations still force the tropical exchange inequalities after possible cancellations. [quotetheorem:6653] [citeproof:6653] The theorem supplies the promised bridge from valued linear algebra to valuated matroids. These basis weights are not the final tropical equations by themselves: from them we form fundamental valuated circuits, and those circuits define the tropical hyperplanes whose intersection is the tropical linear space. The matrix hypothesis matters: for rank two on four elements, the assignment $p(12)=p(34)=0$ and $p(13)=p(14)=p(23)=p(24)=10$ makes the tropical Plucker minimum $p(12)+p(34)$ unique, so it cannot arise as valuations of maximal minors. The theorem is not a realizability theorem in the opposite direction; many valuated matroids are not obtained from a matrix over a chosen valued field. A small Puiseux-series matrix shows how a single parameter can change basis weights while leaving the underlying set of bases nearly unchanged. [example: Puiseux Matrix Valuated Matroid] Let $K=\mathbb C\{\{t\}\}$, with $\nu(t)=1$ and $\nu(c)=0$ for every nonzero constant $c\in\mathbb C$. For the columns \begin{align*} a_1=(1,0),\quad a_2=(0,1),\quad a_3=(1,1),\quad a_4=(1,t), \end{align*} the rank two basis weight $p(ij)$ is the valuation of the determinant with columns $a_i$ and $a_j$. Using $\det((x,y),(u,v))=xv-uy$, the six determinants are \begin{align*} \det(a_1,a_2)=1\cdot 1-0\cdot 0=1. \end{align*} \begin{align*} \det(a_1,a_3)=1\cdot 1-1\cdot 0=1. \end{align*} \begin{align*} \det(a_1,a_4)=1\cdot t-1\cdot 0=t. \end{align*} \begin{align*} \det(a_2,a_3)=0\cdot 1-1\cdot 1=-1. \end{align*} \begin{align*} \det(a_2,a_4)=0\cdot t-1\cdot 1=-1. \end{align*} \begin{align*} \det(a_3,a_4)=1\cdot t-1\cdot 1=t-1. \end{align*} Therefore the corresponding valuations are \begin{align*} p(12)=\nu(1)=0,\quad p(13)=\nu(1)=0,\quad p(14)=\nu(t)=1. \end{align*} The remaining three are \begin{align*} p(23)=\nu(-1)=0,\quad p(24)=\nu(-1)=0,\quad p(34)=\nu(t-1)=0, \end{align*} where $\nu(t-1)=0$ because $t-1$ has nonzero constant term $-1$. All six two-element subsets have finite weight, so the underlying matroid is $U_{2,4}$. The only rank two tropical Plucker relation on four elements compares the three sums $p(12)+p(34)$, $p(13)+p(24)$, and $p(14)+p(23)$. Substituting the computed values gives \begin{align*} p(12)+p(34)=0+0=0. \end{align*} \begin{align*} p(13)+p(24)=0+0=0. \end{align*} \begin{align*} p(14)+p(23)=1+0=1. \end{align*} The minimum is $0$, and it is attained by the first two sums. Thus these minors define a valuated matroid, and the single raised weight $p(14)=1$ records valuation data that the underlying uniform matroid does not see. [/example] ## Tropical Linear Spaces of Valuated Matroids Once bases have valuations, the tropical linear space is no longer only a Bergman fan. What replaces the unweighted circuit condition? A valuated circuit records not only the support of a dependence, but also the relative valuations of its coefficients. For non-uniform underlying matroids, circuit supports may have fewer than $r+1$ elements, so the correct construction uses fundamental circuits relative to bases rather than all $(r+1)$-subsets. [definition: Valuated Circuit] Let $p$ be a rank $r$ valuated matroid on $E$, and let $\mathcal B=\{B:p(B)<\infty\}$ be its underlying set of bases. For a basis $B\in\mathcal B$ and an element $e\in E\setminus B$, the fundamental valuated circuit $c^{B,e}\in(\mathbb R\cup\{\infty\})^E$ is defined by \begin{align*} c^{B,e}_e = p(B). \end{align*} For $f\in B$, set \begin{align*} c^{B,e}_f = p(B-f+e). \end{align*} For $g\notin B\cup\{e\}$, set \begin{align*} c^{B,e}_g = \infty. \end{align*} A valuated circuit of $p$ is any tropical scalar translate $c^{B,e}+\lambda(1,\dots,1)$ of such a fundamental valuated circuit, after restricting to its finite support. The finite support is the ordinary fundamental circuit of $e$ with respect to $B$ in the underlying matroid. [/definition] Every ordinary circuit of the underlying matroid appears as the support of a fundamental circuit after choosing a basis that contains the circuit with one element removed. The tropical hyperplane condition below is unchanged by adding the same constant to all finite entries of $c$, so including tropical scalar translates is a representative convention rather than extra geometry. Valuated circuits are tropical analogues of linear dependence vectors, so a point should be excluded exactly when some weighted dependence has a unique lowest term. The definition below turns that obstruction into the repeated-minimum condition with the circuit coefficients included. [definition: Tropical Linear Space Of A Valuated Matroid] Let $p$ be a valuated matroid on $E$. The tropical linear space $\mathcal L(p)$ is the set of all $w \in \mathbb R^E$ such that, for every valuated circuit $c$, the minimum of $c_e+w_e$ over $e\in E$ is attained at least twice. [/definition] When $p$ is the constant valuation of an ordinary matroid, this definition recovers the Bergman fan. The membership test is direct: for every valuated circuit $c$, add the vector $w$ coordinatewise to $c$ and check that the least finite value occurs at least twice. The realizable case should also recover the tropicalization of a linear space over a valued field. The next theorem states this compatibility and explains why valuated matroids are the right combinatorial objects for tropical linear spaces. [quotetheorem:6654] [citeproof:6654] This result is the valuated counterpart of the Bergman fan theorem. The torus and closure conventions are necessary: a point of $L$ with a zero coordinate has valuation $\infty$ in that coordinate and is seen in the finite tropical space only through limits of nonzero-coordinate points. The valued-field-extension clause prevents a second failure, where a tropical point satisfies all initial equations but needs a larger valued field for an actual lift. The theorem is limited to realizable valuated matroids; the next section removes coordinates and studies the polyhedral condition that characterizes all valuated matroids. For a small torus example, the line $x+y+1=0$ in $(K^\times)^2$ has tropicalization consisting of valuation pairs where $\min(w_x,w_y,0)$ is attained at least twice; the limit direction with $w_x\to\infty$ records approach to the coordinate boundary rather than an actual finite point of the torus. For a field-extension example, over a non-algebraically closed valued field a tropical initial equation may have a solution only after adjoining a root with the required valuation, so the lifting statement must allow valued extensions. ## Matroid Subdivisions of Hypersimplices What polyhedral object stores a valuated matroid? The answer is a regular subdivision of the hypersimplex, whose cells are matroid polytopes. A general subdivision of a hypersimplex is not enough: if a cell has vertices corresponding to bases $12$ and $34$ but omits every exchange vertex $13,14,23,24$, then the vertex set violates ordinary basis exchange and cannot represent a matroid. Before discussing subdivisions, we need the ambient polytope whose vertices are indexed by possible bases. [definition: Hypersimplex] For $0\le r\le n$, the hypersimplex $\Delta(r,n)$ is the polytope \begin{align*} \Delta(r,n)=\operatorname{conv}\{\mathbf{1}_B : B\subset \{1,\dots,n\},\ |B|=r\} \subset \mathbb R^n. \end{align*} [/definition] The hypersimplex supplies one vertex for every candidate basis, but a matroid usually permits only some of those candidates. The next object is needed to place a single ordinary matroid inside this ambient polytope: we keep the vertices corresponding to actual bases and take their convex hull. This motivates the following definition, whose edge directions reflect basis exchange. [definition: Matroid Polytope] Let $M$ be a rank $r$ matroid on $\{1,\dots,n\}$ with bases $\mathcal B(M)$. Its matroid polytope is \begin{align*} P(M)=\operatorname{conv}\{\mathbf{1}_B : B\in \mathcal B(M)\} \subset \Delta(r,n). \end{align*} [/definition] Matroid polytopes translate ordinary basis exchange into local polyhedral geometry: their edges have directions of the form $e_i-e_j$. A valuated matroid varies basis weights across the hypersimplex, so a single matroid polytope is no longer enough. We need a subdivision whose cells are matroid polytopes, allowing different ordinary matroids to control different lower faces. [definition: Matroid Subdivision] A matroid subdivision of $\Delta(r,n)$ is a polyhedral subdivision in which every cell is the matroid polytope of some rank $r$ matroid on $\{1,\dots,n\}$. [/definition] A valuated matroid produces a regular subdivision by lifting each vertex $\mathbf{1}_B$ to height $p(B)$ and projecting the lower faces. Infinite basis weights delete the corresponding vertices. To read the subdivision from heights, choose a linear functional $u\in \mathbb R^n$, minimize $p(B)+\sum_{i\in B}u_i$ over all $r$-subsets $B$, and take the convex hull of the minimizers; varying $u$ gives all lower cells. The remaining question is whether the valuated exchange axiom is exactly the condition that all lower cells are matroid polytopes, and the answer is the main polyhedral characterization. The height construction by itself produces a regular subdivision, but the cells need not respect basis exchange. The obstruction to a matroid subdivision is an exposed face whose minimizing bases fail to form the bases of any ordinary matroid. [quotetheorem:6655] This theorem gives the third dictionary entry of the chapter: matroids correspond to matroid polytopes, valuated matroids correspond to regular matroid subdivisions, and tropical linear spaces are dual to the corresponding subdivision data. Regularity is necessary in the statement: the non-Pappus matroid gives a matroid polytope subdivision phenomenon that cannot be produced by a single height function over the Pappus realization constraints, so matroidal cells alone do not specify a valuated matroid. The finite nonempty-basis condition is also necessary: if every $r$-subset has height $\infty$, there is no lower subdivision of $\Delta(r,n)$ with vertices to test exchange on. The matroidal-cell condition is separate from regularity: the failed exchange cell described above would produce a lower face whose vertices do not form the bases of any matroid. The Puiseux example above gives a small subdivision of the rank two hypersimplex and illustrates the height-reading procedure. [example: Subdivision From The Puiseux Matrix] For the Puiseux matrix above, the rank two basis weights on the six vertices of $\Delta(2,4)$ are \begin{align*} p(12)=p(13)=p(23)=p(24)=p(34)=0,\qquad p(14)=1. \end{align*} For $u=(u_1,u_2,u_3,u_4)$, the lower cell exposed by $u$ is found by minimizing $p(ij)+u_i+u_j$ over all two-element subsets $\{i,j\}$. Thus the adjusted heights are $u_1+u_2$ for $12$, $u_1+u_3$ for $13$, $1+u_1+u_4$ for $14$, $u_2+u_3$ for $23$, $u_2+u_4$ for $24$, and $u_3+u_4$ for $34$. Taking $u=(0,0,0,0)$ gives adjusted heights $0$ on $12,13,23,24,34$ and adjusted height $1$ on $14$. Hence the minimizers are \begin{align*} 12,\ 13,\ 23,\ 24,\ 34, \end{align*} so one lower cell is \begin{align*} \operatorname{conv}\{12,13,23,24,34\}. \end{align*} Equivalently, the affine height function $\ell_0(x)=0$ agrees with the lifted heights on those five vertices and satisfies \begin{align*} \ell_0(\mathbf{1}_{14})=0<1=p(14), \end{align*} so this face lies on the lower hull. A lower cell containing the raised vertex is obtained by choosing \begin{align*} u=(0,1,1,0). \end{align*} Then the adjusted height of $12$ is \begin{align*} p(12)+u_1+u_2=0+0+1=1. \end{align*} The adjusted height of $13$ is \begin{align*} p(13)+u_1+u_3=0+0+1=1. \end{align*} The adjusted height of $14$ is \begin{align*} p(14)+u_1+u_4=1+0+0=1. \end{align*} The adjusted height of $23$ is \begin{align*} p(23)+u_2+u_3=0+1+1=2. \end{align*} The adjusted height of $24$ is \begin{align*} p(24)+u_2+u_4=0+1+0=1. \end{align*} The adjusted height of $34$ is \begin{align*} p(34)+u_3+u_4=0+1+0=1. \end{align*} Therefore the minimizers are \begin{align*} 12,\ 13,\ 14,\ 24,\ 34, \end{align*} and the corresponding lower cell is \begin{align*} \operatorname{conv}\{12,13,14,24,34\}. \end{align*} The supporting affine height function is \begin{align*} \ell_1(x)=1-x_2-x_3. \end{align*} It has value $0$ on $\mathbf{1}_{12}$, $\mathbf{1}_{13}$, $\mathbf{1}_{24}$, and $\mathbf{1}_{34}$, value $1$ on $\mathbf{1}_{14}$, and value $-1$ on $\mathbf{1}_{23}$. Thus it agrees with $p$ on $12,13,14,24,34$ and lies strictly below $p(23)=0$, as required for a lower face. The two maximal cells meet along \begin{align*} \operatorname{conv}\{12,13,24,34\}. \end{align*} The first cell has all two-element bases except $14$, which is the rank two matroid in which $1$ and $4$ are parallel. The second cell has all two-element bases except $23$, which is the rank two matroid in which $2$ and $3$ are parallel. Thus the lower subdivision consists of matroid polytopes, and the raised height $p(14)=1$ records exactly the valuation data that subdivides the ordinary octahedron $\Delta(2,4)$. [/example] The chapter completes the tropical thread by showing that matroid theory is not only an abstraction of independence. It is also the combinatorial skeleton of tropical linear geometry: circuits become tropical equations, flats become cones of the Bergman fan, bases become vertices of hypersimplices, and valuated exchange becomes regular matroid subdivision. Tropical geometry presents matroids as the combinatorial framework underlying piecewise-linear linear algebra. The course closes by returning to global structure, using the full collection of matroid tools to study log-concavity, minors, and modern structural theorems. # 12. Log-Concavity and Modern Structure Theory This closing chapter of the matroid theory course uses the machinery developed earlier: independence and rank, bases, flats, duality, minors, graphic and cographic matroids, and the Tutte polynomial. The guiding question is how much global structure is forced by these local axioms. We prove the elementary log-concavity computations visible in familiar examples, record the modern Hodge-theoretic theorems that explain the general pattern, and close by placing regular matroids in the excluded-minor and decomposition viewpoint. ## Counting Sequences Attached to a Matroid Which numerical shadows of a matroid remember its combinatorial structure? Chapters 2 through 5 developed rank, flats, circuits, duality, and minors, while Chapter 10 introduced the Tutte polynomial. We now focus on sequences extracted from these objects, because log-concavity turns out to be a strong signature of matroidal structure. The first sequence counts independent sets by size. It is the $f$-vector of the independence complex, so it packages the most direct enumeration of independence. [definition: Independent Set Sequence] Let $M$ be a matroid of rank $r$ on a finite ground set $E$. For $0 \le k \le r$, let \begin{align*} I_k(M) := |\{A \subset E : A \text{ is independent and } |A| = k\}|. \end{align*} The sequence $(I_0(M), I_1(M), \dots, I_r(M))$ is the independent set sequence of $M$. [/definition] This sequence records all sizes of independent sets, but it forgets which elements occur together. Two matroids can have the same numbers $I_k(M)$ while having quite different collections of bases, so a single-variable count is too coarse for questions about correlation or element-by-element dependence. To retain that information, we need a multivariate polynomial whose monomials are indexed by bases. [definition: Basis Generating Polynomial] Let $M$ be a matroid on $E = \{e_1, \dots, e_n\}$ with set of bases $\mathcal{B}(M)$. The basis generating polynomial of $M$ is the polynomial $B_M \in \mathbb Z[x_1, \dots, x_n]$, where this denotes polynomials in the variables $x_1,\dots,x_n$ with integer coefficients, defined by \begin{align*} B_M(x_1,\dots,x_n) := \sum_{B \in \mathcal{B}(M)} \prod_{e_i \in B} x_i. \end{align*} [/definition] Equivalently, $B_M$ is the element of $\mathbb Z[x_1,\dots,x_n]$ obtained by the same formula. The unweighted specialization $x_1=\cdots=x_n=1$ gives the number of bases, while refined specializations count bases according to how many elements they contain from a chosen subset. Bases, however, still miss how dependence is organised below the top rank: circuits can force cancellations that are invisible in a positive counting polynomial. The next sequence comes from the lattice of flats, whose Mobius function records those cancellations across ranks. [definition: Characteristic Polynomial] Let $M$ be a rank-$r$ matroid with lattice of flats $L(M)$. If $M$ has a loop, the characteristic polynomial is $\chi_M(q):=0$. If $M$ is loopless, let $\mu: L(M) \times L(M) \to \mathbb Z$ be the Mobius function of $L(M)$, and define \begin{align*} \chi_M(q) := \sum_{F \in L(M)} \mu(\varnothing,F) q^{r-r(F)}. \end{align*} [/definition] The loop convention matches the deletion-contraction normalization used for graphic matroids and agrees with the subset-expansion definition of the characteristic polynomial. For any matroid, write \begin{align*} \chi_M(q) = \sum_{k=0}^{r} (-1)^k w_k(M) q^{r-k} \end{align*} and call $w_k(M)$ the Whitney numbers of the first kind; if $M$ has a loop, all these coefficients are zero. For a loopless matroid, $\chi_M(1)=0$, so the reduced characteristic polynomial $\bar{\chi}_M(q)=\chi_M(q)/(q-1)$ is often the cleaner invariant. Before discussing the general theory, it is useful to see the expected shape in the most symmetric family of matroids. [example: Uniform Matroid Independent Sets] For the uniform matroid $U_{r,n}$, a subset is independent exactly when its size is at most $r$. Therefore, for $0 \le k \le r$, the independent sets of size $k$ are precisely the $k$-element subsets of the $n$-element ground set, so \begin{align*} I_k(U_{r,n})=\binom{n}{k}. \end{align*} To check log-concavity, fix $1 \le k \le r-1$. Since $r \le n$, we have $k+1 \le n$, so the relevant binomial coefficients are nonzero. From the factorial formula, \begin{align*} \frac{\binom{n}{k}}{\binom{n}{k-1}}=\frac{\frac{n!}{k!(n-k)!}}{\frac{n!}{(k-1)!(n-k+1)!}}=\frac{(k-1)!(n-k+1)!}{k!(n-k)!}=\frac{n-k+1}{k}. \end{align*} Similarly, \begin{align*} \frac{\binom{n}{k}}{\binom{n}{k+1}}=\frac{\frac{n!}{k!(n-k)!}}{\frac{n!}{(k+1)!(n-k-1)!}}=\frac{(k+1)!(n-k-1)!}{k!(n-k)!}=\frac{k+1}{n-k}. \end{align*} Multiplying these two identities gives \begin{align*} \frac{\binom{n}{k}^2}{\binom{n}{k-1}\binom{n}{k+1}}=\frac{n-k+1}{k}\cdot\frac{k+1}{n-k}=\frac{k+1}{k}\cdot\frac{n-k+1}{n-k}. \end{align*} Both factors are greater than $1$, so \begin{align*} \binom{n}{k}^2 \ge \binom{n}{k-1}\binom{n}{k+1}. \end{align*} Thus the independent set sequence of $U_{r,n}$ is log-concave, and the computation shows why binomial coefficients form the baseline for the normalized inequalities in Mason's conjecture. [/example] Uniform matroids demonstrate the expected shape of the independent set sequence, but they do not yet say what inequality is being tested for a general sequence. The first obstruction is purely numerical: adjacent terms must be constrained strongly enough to rule out isolated spikes. [definition: Log-Concave Sequence] A sequence $(a_0, a_1, \dots, a_m)$ of nonnegative [real numbers](/page/Real%20Numbers) is log-concave if \begin{align*} a_k^2 \ge a_{k-1}a_{k+1} \end{align*} for every $1 \le k \le m-1$. [/definition] Log-concavity says that the ratios $a_k/a_{k-1}$ do not increase where the ratios are defined. This is the right first test for sequences such as characteristic-polynomial coefficients, but independent-set numbers have an additional ambient source of growth. Even the free matroid on $n$ elements has independent set counts $\binom{n}{k}$, before any dependence relations have appeared. Mason's conjecture therefore needs a definition that removes this Boolean contribution and asks whether the remaining normalized sequence is log-concave. [definition: Ultra Log-Concave Sequence] Let $n \in \mathbb N$. A sequence $(a_0, a_1, \dots, a_n)$ of nonnegative real numbers is ultra log-concave of order $n$ if the normalized sequence \begin{align*} \left(\frac{a_k}{\binom{n}{k}}\right)_{k=0}^{n} \end{align*} is log-concave. [/definition] This normalization removes the universal binomial growth already present before the independence constraints are applied. For a rank-$r$ matroid on $n$ elements, one extends the independent set sequence by setting $I_k(M)=0$ for $k>r$. The theorem below is the resulting form of Mason's enumerative prediction for matroids. [quotetheorem:6656] This theorem is one of the major modern results in matroid theory. The ground-set size $n$ is essential in the normalization: it measures the ambient number of possible $k$-subsets before independence is imposed, while the rank $r$ only records the largest possible independent size. The conclusion does not identify the independent sets or determine the matroid; it gives a universal inequality that every exchange system must satisfy. In the sequel, this is the independent-set analogue of the Hodge-theoretic inequalities for characteristic-polynomial coefficients. ## Characteristic Polynomials and Hodge Theory Why should coefficients of a characteristic polynomial satisfy inequalities resembling the [hard Lefschetz theorem](/theorems/3876)? The characteristic polynomial is defined from the lattice of flats, so at first it appears to be a Mobius-inversion invariant rather than a geometric one. The modern insight is that a matroid has a combinatorial Chow ring whose intersection theory behaves like the [cohomology ring](/theorems/2271) of a smooth projective variety. We first record the log-concavity statement in the coefficient convention fixed above. It was conjectured by Rota and Welsh and proved by Adiprasito, Huh, and Katz. [quotetheorem:6657] The theorem applies to every matroid: when loops are present the coefficient sequence is zero by convention, and the loopless case is the substantive Hodge-theoretic statement. That universality is the point of the Chow-ring construction: it builds the needed intersection theory directly from the lattice of flats, so there is no ambient variety in the hypotheses. The matroidal origin of the polynomial is doing real work: an arbitrary alternating-sign polynomial need not have log-concave absolute coefficients, for instance $q^2-q+3$ would give $(1,1,3)$ and violates $1^2\ge 1\cdot 3$. The statement does not give a combinatorial injection proving the inequalities, and it does not say that the coefficients determine the lattice of flats. Instead, it shows that the Mobius cancellations in the characteristic polynomial obey the same kind of positivity constraint that appears in projective geometry. The following explanation records the strategy so that the theorem is not a black box. [explanation: Hodge-Theoretic Strategy] The Chow ring $A^*(M)$ is generated by variables indexed by proper nonempty flats, modulo relations encoding incomparability of flats and linear equivalences among flats containing a given element. Its degree map plays the role of integration over a compact variety. Two distinguished nef classes, usually denoted $\alpha$ and $\beta$, produce intersection numbers \begin{align*} \deg(\alpha^{r-1-k}\beta^k) \end{align*} for $0 \le k \le r-1$, which recover the coefficients of the reduced characteristic polynomial. The hard Lefschetz theorem supplies isomorphisms obtained by multiplying by powers of a Lefschetz class. The Hodge-Riemann relations give a signature constraint on an associated [bilinear form](/page/Bilinear%20Form). Applied to the span of two consecutive monomials in $\alpha$ and $\beta$, this signature constraint becomes the desired log-concavity inequality. [/explanation] This strategy explains why log-concavity appears in a setting with no ambient variety. It also explains why the statement is stronger than a formal property of Mobius inversion: the inequality is produced by an intersection pairing with controlled signature, not by coefficient signs alone. A small characteristic-polynomial computation shows the numerical form of the theorem before we connect it to graphs. [example: Characteristic Polynomial of a Uniform Matroid] For $U_{2,4}$ on ground set $E=\{1,2,3,4\}$, the flats are $\varnothing$, the four singleton sets, and $E$. Their ranks are $r(\varnothing)=0$, $r(\{i\})=1$ for $1\le i\le 4$, and $r(E)=2$. The Mobius function of the flat lattice is determined by $\mu(\varnothing,\varnothing)=1$ and, for each nonempty flat $G$, by the recursion \begin{align*} \sum_{\varnothing \le F \le G}\mu(\varnothing,F)=0. \end{align*} For a singleton flat $\{i\}$ this gives \begin{align*} 0=\mu(\varnothing,\varnothing)+\mu(\varnothing,\{i\})=1+\mu(\varnothing,\{i\}), \end{align*} so $\mu(\varnothing,\{i\})=-1$ for each $i$. For the top flat $E$, the same recursion gives \begin{align*} 0=\mu(\varnothing,\varnothing)+\sum_{i=1}^{4}\mu(\varnothing,\{i\})+\mu(\varnothing,E)=1+4(-1)+\mu(\varnothing,E), \end{align*} and hence $\mu(\varnothing,E)=3$. Substituting these values into the definition of the characteristic polynomial gives \begin{align*} \chi_{U_{2,4}}(q)=\mu(\varnothing,\varnothing)q^{2-r(\varnothing)}+\sum_{i=1}^{4}\mu(\varnothing,\{i\})q^{2-r(\{i\})}+\mu(\varnothing,E)q^{2-r(E)}. \end{align*} Using the ranks listed above, this becomes \begin{align*} \chi_{U_{2,4}}(q)=1\cdot q^2+\sum_{i=1}^{4}(-1)q+3q^0. \end{align*} Since $\sum_{i=1}^{4}(-1)q=-4q$ and $q^0=1$, we obtain \begin{align*} \chi_{U_{2,4}}(q)=q^2-4q+3. \end{align*} Comparing with $\chi_M(q)=w_0q^2-w_1q+w_2$, we get $(w_0,w_1,w_2)=(1,4,3)$. The only log-concavity inequality is \begin{align*} w_1^2=4^2=16 \ge 3 = 1\cdot 3=w_0w_2. \end{align*} This computation shows explicitly how the Mobius values of the flat lattice become the numerical inequality for the characteristic-polynomial coefficients. [/example] The characteristic polynomial also connects back to graphs. To compare the matroid invariant with the older graph invariant, we need the deletion-contraction relationship between the characteristic polynomial of a graphic matroid and the chromatic polynomial of the graph. [quotetheorem:6646] [citeproof:6646] The connectedness hypothesis fixes the normalization. If $G$ has $c$ connected components, then a proper colouring may choose an independent global colour shift on each component, and the corresponding relation becomes $P_G(q)=q^c\chi_{M(G)}(q)$ rather than $q\chi_{M(G)}(q)$. The factor $q$ in the connected case is also not optional: for a tree with $n$ vertices, $P_G(q)=q(q-1)^{n-1}$ while $\chi_{M(G)}(q)=(q-1)^{n-1}$. The theorem does not say that chromatic data determine the graph or even the graphic matroid; nonisomorphic graphs can have the same chromatic polynomial. Its use here is more specific: it transfers log-concavity statements for characteristic-polynomial coefficients into the older language of graph colouring. Basis enumeration supplies another graph invariant, since bases of a graphic matroid are spanning trees. [example: Spanning Trees as Bases] Let $G$ be a connected finite graph with edge set $E(G)=\{e_1,\dots,e_m\}$. In the graphic matroid $M(G)$, independent sets are forests, so a basis is a maximal acyclic set of edges. If such a forest had two connected components, then connectedness of $G$ gives an edge of $G$ joining two different components; adding that edge cannot create a cycle, because a cycle would require an existing path between its endpoints inside the forest. Hence a maximal forest in a connected graph is connected, and therefore it is a spanning tree. Conversely, a spanning tree is acyclic, and adding any edge not already in it creates a cycle along the unique path between the endpoints in the tree, so it is maximal acyclic. Thus the bases of $M(G)$ are exactly the spanning trees of $G$. By the definition of the basis generating polynomial, \begin{align*} B_{M(G)}(x_1,\dots,x_m)=\sum_{B\in \mathcal B(M(G))}\prod_{e_i\in B}x_i. \end{align*} Substituting $x_1=\cdots=x_m=1$ gives \begin{align*} B_{M(G)}(1,\dots,1)=\sum_{B\in \mathcal B(M(G))}\prod_{e_i\in B}1. \end{align*} For each basis $B$, the product $\prod_{e_i\in B}1$ equals $1$, so \begin{align*} B_{M(G)}(1,\dots,1)=\sum_{B\in \mathcal B(M(G))}1=|\mathcal B(M(G))|. \end{align*} Since $\mathcal B(M(G))$ is the set of spanning trees of $G$, this becomes \begin{align*} B_{M(G)}(1,\dots,1)=\tau(G), \end{align*} where $\tau(G)$ denotes the number of spanning trees of $G$. For the cycle graph $C_n$, there are $n$ edges. Deleting one edge from the cycle leaves a connected acyclic graph on the same vertices, hence a spanning tree. Conversely, any spanning tree of $C_n$ has $n-1$ edges, while $C_n$ has $n$ edges, so it is obtained by deleting exactly one edge from the cycle. Therefore the spanning trees of $C_n$ are in bijection with the $n$ choices of a deleted edge, and \begin{align*} \tau(C_n)=n. \end{align*} Thus the specialization $B_{M(G)}(1,\dots,1)$ recovers the classical spanning-tree count. [/example] ## Mason's Conjecture and Independent-Set Log-Concavity The characteristic-polynomial theorem concerns flats and Mobius functions, while Mason's conjecture concerns independent sets. Why should these be related? The modern answer passes through stronger polynomial classes whose coefficients obey Alexandrov-Fenchel-type inequalities. The following theorem records the strongest form needed in this course. It refines ordinary log-concavity by comparing independent set numbers with the binomial model on the full ground set. [quotetheorem:6658] This is the same assertion as the [strong Mason conjecture](/theorems/6656), written in normalized form. The hypothesis that $n$ is the ground-set size is not cosmetic: take $U_{1,n}$, extend its independent-set sequence by zeros, and compare using rank $r=1$; the denominators $\binom{1}{k}$ are not even available for $k>1$, so the normalization cannot encode the zero tail forced by rank. Even if one restricts attention to $0\le k\le r$, rank normalization fails for the uniform matroid $U_{2,4}$: its independent-set sequence is $(1,4,6)$, and dividing by $\binom{2}{k}$ gives $(1,2,6)$, which is not log-concave because $2^2<1\cdot 6$. The matroid hypothesis is also necessary; the arbitrary nonnegative sequence $(1,1,3)$ of order $2$ has normalized sequence $(1,1/2,3)$, violating log-concavity since $(1/2)^2<1\cdot 3$. The theorem still does not describe which subsets are independent; it controls only the profile of how many survive at each size. This profile control is the enumerative input that lets later structural results be compared across minors and decompositions without remembering every independent set. Uniform matroids allow a direct check and show why the ground-set binomial normalization has the right scale. [example: Mason Inequality for Uniform Matroids] For $U_{r,n}$, a subset of the $n$-element ground set is independent exactly when its size is at most $r$. Thus for $0\le k\le r$ we have \begin{align*} I_k(U_{r,n})=\binom{n}{k}, \end{align*} and for $r<k\le n$ we have \begin{align*} I_k(U_{r,n})=0. \end{align*} Mason's normalization uses the ground-set denominator $\binom{n}{k}$, so for $0\le k\le r$, \begin{align*} a_k:=\frac{I_k(U_{r,n})}{\binom{n}{k}}=\frac{\binom{n}{k}}{\binom{n}{k}}=1, \end{align*} while for $r<k\le n$, \begin{align*} a_k:=\frac{I_k(U_{r,n})}{\binom{n}{k}}=\frac{0}{\binom{n}{k}}=0. \end{align*} Therefore the normalized sequence consists of $r+1$ ones followed by $n-r$ zeros. We verify log-concavity by checking every $1\le k\le n-1$. If $k\le r-1$, then $a_{k-1}=a_k=a_{k+1}=1$, so \begin{align*} a_k^2=1^2=1=1\cdot 1=a_{k-1}a_{k+1}. \end{align*} If $k=r$ and $r<n$, then $a_{r-1}=1$, $a_r=1$, and $a_{r+1}=0$, hence \begin{align*} a_r^2=1^2=1\ge 0=1\cdot 0=a_{r-1}a_{r+1}. \end{align*} If $k\ge r+1$, then $a_k=0$ and $a_{k+1}=0$, while $a_{k-1}$ is either $1$ or $0$, so \begin{align*} a_k^2=0^2=0=a_{k-1}\cdot 0=a_{k-1}a_{k+1}. \end{align*} Thus the ground-set-normalized independent set sequence of $U_{r,n}$ is log-concave. This calculation also shows why the normalization must use $n$ rather than $r$: for $U_{2,4}$, rank normalization gives \begin{align*} \frac{1}{\binom{2}{0}}=1, \end{align*} \begin{align*} \frac{4}{\binom{2}{1}}=2, \end{align*} and \begin{align*} \frac{6}{\binom{2}{2}}=6. \end{align*} The resulting sequence $(1,2,6)$ is not log-concave, because \begin{align*} 2^2=4<6=1\cdot 6. \end{align*} [/example] The example is simple because every allowable subset is independent. In a general matroid, dependence removes subsets in a highly constrained way, and the theorem says that the exchange axiom controls those removals well enough to preserve ultra log-concavity. ## Excluded Minors and Regular Matroids Log-concavity gives universal inequalities, but another kind of modern structure theorem classifies matroids by forbidden minors and decompositions. The guiding question is: when a representability condition is inherited by minors, can it be characterized by a finite list of excluded minors, and can the matroids satisfying it be built from elementary pieces? Regular matroids are the central case because they are representable over every field. They include graphic and cographic matroids, and they are rigid enough to admit a deep decomposition theorem. [definition: Regular Matroid] A matroid $M$ is regular if it is representable over every field. [/definition] Regularity is hard to test directly from the definition because it quantifies over all fields. A usable certificate has to be visible in one real matrix while still forcing every square minor to avoid field-dependent denominators and determinants. The determinant obstruction is local: one bad square submatrix can introduce a coefficient that vanishes or becomes noninvertible after changing fields. The following matrix condition rules out exactly that kind of field-dependent failure. [definition: Totally Unimodular Matrix] A real matrix $A$ is totally unimodular if every square submatrix of $A$ has determinant in $\{-1,0,1\}$. [/definition] A matroid is regular precisely when it has a representation by a totally unimodular matrix over $\mathbb R$. Testing all fields directly would be unusable, and testing all real representations would still miss the minor-minimal reasons regularity can fail. Since regularity is minor-closed, the next natural question is which minors are the minimal obstructions. [quotetheorem:6659] The theorem shows that regularity, although defined by representability over all fields, has a finite forbidden-minor characterization. Each excluded minor marks a different failure mode: $U_{2,4}$ is representable over fields with at least three elements but is not binary, so omitting it would admit matroids that already fail regularity over $\mathbb F_2$. The Fano matroid $F_7$ is binary but not representable over fields of characteristic other than $2$, while $F_7^*$ has the same regularity failure after duality; omitting either one would allow a binary matroid that is not regular. For example, a criterion forbidding only $U_{2,4}$ and $F_7$ would wrongly accept $F_7^*$ itself, and a criterion forbidding only $U_{2,4}$ and $F_7^*$ would wrongly accept $F_7$. The theorem does not construct a regular representation or decompose the matroid; it only says what cannot appear as a minor. The next definition names the exceptional primitive matroid needed before Seymour's stronger assembly theorem can be stated. [definition: The Matroid R Ten] The matroid $R_{10}$ is the standard ten-element internally $4$-connected regular matroid which is neither graphic nor cographic. [/definition] Up to isomorphism, $R_{10}$ is the unique internally $4$-connected regular matroid with this exceptional role in Seymour's decomposition theorem. The internal $4$-connectedness qualifier matters: without it, sums involving graphic, cographic, or smaller regular pieces can produce further ten-element regular matroids that are neither graphic nor cographic. The role of $R_{10}$ is to prevent the regular class from being generated by graphic and cographic matroids alone. Excluded minors say which obstructions are absent, but they do not explain how a large regular matroid is assembled. Once this exceptional piece is admitted, the remaining assembly operations are low-order matroid sums. [definition: Low-Order Matroid Sums] A $1$-sum is the direct sum $M_1\oplus M_2$ of matroids on disjoint ground sets. A $2$-sum is the operation $M_1\oplus_2 M_2$ defined earlier by gluing two matroids along one common element $p$ which is neither a loop nor a coloop, and then deleting $p$ from the glued ground set. A $3$-sum is the analogous Seymour gluing operation along a common three-element circuit-cocircuit, usually a triangle or triad, with that common set deleted after the compatible gluing. In the theorem below, "obtained by $1$-, $2$-, and $3$-sums" means obtained by repeatedly applying these permitted low-order gluings under their standard compatibility hypotheses. [/definition] These sums are introduced because forbidden minors describe what regular matroids avoid, but not how regular matroids are built. Graphic and cographic matroids account for the graph-theoretic sources of regularity, while $R_{10}$ is the exceptional primitive piece that cannot be removed from the theory. The point of the decomposition theorem is to say that these pieces, joined only by controlled low-order gluings, already generate the entire regular class. [quotetheorem:6660] The theorem is the matroid analogue of a decomposition theorem in graph minor theory: it turns a global representability condition into a recipe using primitive pieces and controlled gluing. The hypotheses on the pieces are essential. If $U_{2,4}$ is allowed as a primitive piece, then the one-term construction already produces a nonregular matroid; if $F_7$ is allowed, the same failure occurs over fields of characteristic not equal to $2$. The restrictions on the sums are also essential: a permitted $2$-sum or $3$-sum glues along a small common restriction with compatibility conditions, while an uncontrolled identification of elements can create a minor isomorphic to one of the excluded obstructions. The converse is therefore part of the structure theorem, not a formality: the listed operations are exactly the ones known to preserve the regular class when applied to the listed pieces. What the theorem does not provide is a short proof or a canonical decomposition; its value here is that it reduces questions about regular matroids to graphic, cographic, and $R_{10}$ building blocks. The following example indicates how the recipe is used. [example: Building Regular Matroids] Start with three regular building blocks: a graphic matroid $M(G)$, a cographic matroid $M(H)^*$, and the exceptional regular matroid $R_{10}$. The graphic and cographic pieces are regular, and $R_{10}$ is regular by definition of its role in *[Seymour Decomposition Theorem for Regular Matroids](/theorems/6660)*. If $M_1$ and $M_2$ are two regular matroids already constructed, and the hypotheses for a $2$-sum are satisfied, then the converse direction of *Seymour Decomposition Theorem for Regular Matroids* says that \begin{align*} M_1 \oplus_2 M_2 \end{align*} is regular. Similarly, if the hypotheses for a $3$-sum are satisfied, the same theorem gives that \begin{align*} M_1 \oplus_3 M_2 \end{align*} is regular. Thus the construction is inductive. At stage $0$, every available piece is regular. If every matroid constructed up to stage $t$ is regular, then performing one allowed $1$-sum, $2$-sum, or $3$-sum with another regular constructed piece produces a regular matroid at stage $t+1$, by the closure assertion in *Seymour Decomposition Theorem for Regular Matroids*. Therefore every matroid obtained after finitely many such sums is regular. The final matroid may not come with an obvious graph whose graphic matroid it is, or an obvious graph whose cographic matroid it is, because the construction may have glued together graphic, cographic, and $R_{10}$ pieces through several low-order sums; regularity is preserved by the permitted assembly even when the final presentation no longer visibly comes from a single graph. [/example] ## How the Two Themes Fit Together Why do enumerative inequalities and decomposition theorems belong in the same closing chapter? The common difficulty is that the matroid axioms are local, while the conclusions are global: log-concavity constrains entire counting sequences, and decomposition theorems classify whole minor-closed families. The chapter has presented these as two different forms of structure. Log-concavity theorems say that every matroid carries hidden positivity in its counting sequences. Decomposition theorems say that special minor-closed classes, such as regular matroids, have a controlled assembly from primitive components. These viewpoints reinforce the earlier parts of the course. Independence axioms and rank submodularity lead to enumerative sequences; flats and Mobius inversion lead to characteristic polynomials; representability and minors lead to excluded-minor theory. Modern matroid theory shows that these elementary-looking objects sit inside a much larger structure involving Hodge theory, Lorentzian polynomials, and graph-minor-style decompositions. [remark: End of Course Perspective] Matroids began in this course as a common language for vector independence and forests in graphs. By the end, they also serve as a meeting point for algebraic geometry, optimization, and structural graph theory. The power of the abstraction is that the same exchange axiom supports greedy algorithms, duality, minors, polynomial invariants, and deep positivity theorems. [/remark] ## Connections and Further Reading The graph-theoretic side of the course continues naturally in [Cambridge II Graph Theory](/page/Cambridge%20II%20Graph%20Theory), where cycle spaces, cuts, connectivity, and spanning trees can be studied before or after their matroidal abstraction. Graphic and cographic matroids explain why planar duality, bonds, and tree packings are not isolated graph facts but examples of duality and minors in a larger independence theory. The optimization side points toward [Combinatorial Optimisation](/page/Combinatorial%20Optimisation). The greedy theorem is the cleanest matroidal optimization result, while matroid intersection, union, base polytopes, and packing theorems show how linear programming and min-max structure replace greedy choices when one matroid constraint is no longer enough. Representable and valuated matroids connect the subject to algebraic geometry. Readers who want the geometric background can compare the Plucker-coordinate and tropical chapters with [Algebraic Geometry Overview](/page/Algebraic%20Geometry%20Overview) and with the symmetric-function and positivity themes in [Algebraic Combinatorics I: Symmetric Functions](/page/Algebraic%20Combinatorics%20I%3A%20Symmetric%20Functions). This is the setting in which the Hodge-theoretic proofs of log-concavity and the polyhedral theory of tropical linear spaces become part of the same picture. For a next pass through the subject, the standard route is to study Oxley's text for the core theory, Welsh for the older combinatorial viewpoint, Schrijver for polyhedral and optimization methods, and White's volumes for representability and structural matroid theory. These sources also make clear which results in this page are elementary course tools and which are deep theorems included as orientation points. ## References - James Oxley, *Matroid Theory*, Oxford University Press. - Dominic Welsh, *Matroid Theory*, Academic Press. - Alexander Schrijver, *Combinatorial Optimization: Polyhedra and Efficiency*, Springer. - Neil White, editor, *Theory of Matroids* and *Combinatorial Geometries*, Cambridge University Press.