A square system of equations can be deceptive. Three equations in three unknowns may look determined, but if the third equation is obtained by adding the first two, then only two independent constraints are present. Rank is the invariant that separates apparent size from effective linear information. [example: Redundant Equations] Over $\mathbb{R}$, consider \begin{align*} x_1+x_2=2. \end{align*} \begin{align*} x_2+x_3=3. \end{align*} \begin{align*} x_1+2x_2+x_3=5. \end{align*} Adding the left-hand sides of the first two equations gives \begin{align*} (x_1+x_2)+(x_2+x_3)=x_1+2x_2+x_3, \end{align*} and adding the right-hand sides gives \begin{align*} 2+3=5. \end{align*} Thus the third equation imposes no new condition beyond the first two. To solve, set $x_2=t$ with $t \in \mathbb{R}$. From $x_1+x_2=2$ we get \begin{align*} x_1+t=2, \end{align*} so $x_1=2-t$. From $x_2+x_3=3$ we get \begin{align*} t+x_3=3, \end{align*} so $x_3=3-t$. Hence every solution has the form \begin{align*} (x_1,x_2,x_3)=(2-t,t,3-t). \end{align*} Conversely, substituting this expression into the three equations gives \begin{align*} (2-t)+t=2, \end{align*} \begin{align*} t+(3-t)=3, \end{align*} and \begin{align*} (2-t)+2t+(3-t)=5. \end{align*} So the solution set is the affine line $\{(2-t,t,3-t):t\in\mathbb{R}\}$. The coefficient columns are $(1,0,1)$, $(1,1,2)$, and $(0,1,1)$. They satisfy \begin{align*} (1,1,2)=(1,0,1)+(0,1,1), \end{align*} so the column space is contained in the span of $(1,0,1)$ and $(0,1,1)$. If \begin{align*} \alpha(1,0,1)+\beta(0,1,1)=(0,0,0), \end{align*} then comparing coordinates gives $\alpha=0$, $\beta=0$, and $\alpha+\beta=0$. Thus these two columns are independent, the column space has dimension $2$, and the coefficient matrix has rank $2$. [/example] The example reveals the central problem of the subject: count independence, not presentation. Rank answers the same question for matrices, linear maps, systems of equations, finitely generated modules, and tensor decompositions. Each setting has its own technical language, but the guiding idea is stable: rank measures how many independent simple directions are needed to describe the object. ## Definition A matrix can be read as a list of output vectors. If a right-hand side $b$ is to be produced by a system $Ax=b$, then $b$ must be built as a linear combination of the columns. The primary numerical question is therefore: how many independent output directions do the columns actually generate? [definition: Matrix Rank] Let $k$ be a field, and let $A \in k^{m \times n}$ have columns $a_1, \dots, a_n \in k^m$. The rank of $A$ is \begin{align*} \operatorname{rank}(A) := \dim_k \operatorname{span}_k\{a_1, \dots, a_n\}. \end{align*} [/definition] This definition measures the output content of the matrix without being distracted by how many columns were written down. To use rank in systems, though, we need to refer not only to its dimension but to the actual set of attainable right-hand sides. Naming that set lets us ask whether a given vector $b$ can occur as $Ax$. [definition: Column Space] Let $k$ be a field, and let $A \in k^{m \times n}$ have columns $a_1, \dots, a_n \in k^m$. The column space of $A$ is the subspace \begin{align*} \operatorname{Col}(A) := \operatorname{span}_k\{a_1, \dots, a_n\} \subset k^m. \end{align*} [/definition] With this notation, the rank is $\operatorname{rank}(A)=\dim_k \operatorname{Col}(A)$. Column space records outputs, but row reduction works by changing equations. To connect computation with geometry, we also need the subspace generated by the equations themselves. [definition: Row Space] Let $k$ be a field, and let $A \in k^{m \times n}$ have rows $r_1, \dots, r_m \in k^n$. The row space of $A$ is the subspace \begin{align*} \operatorname{Row}(A) := \operatorname{span}_k\{r_1, \dots, r_m\} \subset k^n. \end{align*} [/definition] A problem remains: the same [linear map](/page/Linear%20Map) can be represented by many matrices after changing bases. To avoid making rank depend on a coordinate choice, we need an intrinsic definition using the image of the map itself. [definition: Rank of a Linear Map] Let $k$ be a field, let $V$ and $W$ be finite-dimensional $k$-vector spaces, and let $T: V \to W$ be a $k$-linear map. The rank of $T$ is \begin{align*} \operatorname{rank}(T) := \dim_k \operatorname{im}(T). \end{align*} [/definition] This leaves a common decision problem: sometimes the exact rank matters less than whether it is as large as the dimensions allow. That question motivates the language of full rank. [definition: Full Rank] Let $k$ be a field, and let $A \in k^{m \times n}$. The matrix $A$ has full rank if \begin{align*} \operatorname{rank}(A)=\min\{m,n\}. \end{align*} [/definition] Full rank turns independence into a yes-or-no test, but rectangular matrices force us to separate two different ways of being maximal. When a map has more output coordinates than input coordinates, it cannot reach every output vector; the useful question is instead whether it loses any input information. This is the situation measured by full column rank: every column must contribute a new independent direction, so distinct input vectors remain distinguishable after applying $A$. [definition: Full Column Rank] Let $k$ be a field, and let $A \in k^{m \times n}$. The matrix $A$ has full column rank if \begin{align*} \operatorname{rank}(A)=n. \end{align*} [/definition] Full column rank says that no input direction is lost. When there are at least as many columns as rows, the opposite question becomes more important: can the columns generate the whole target space, so that every possible right-hand side is reachable? [definition: Full Row Rank] Let $k$ be a field, and let $A \in k^{m \times n}$. The matrix $A$ has full row rank if \begin{align*} \operatorname{rank}(A)=m. \end{align*} [/definition] In a rectangular matrix, full column rank separates having no lost input directions from full row rank, which separates reaching every output direction. [example: A Rank-Two Matrix Without Matrix Notation] Let $A: \mathbb{Q}^4 \to \mathbb{Q}^3$ be the linear map whose columns are \begin{align*} a_1=(1,0,1),\quad a_2=(2,1,3),\quad a_3=(0,1,1),\quad a_4=(1,3,4). \end{align*} We compute the span of the columns. First, \begin{align*} 2a_1+a_3=2(1,0,1)+(0,1,1)=(2,0,2)+(0,1,1)=(2,1,3)=a_2. \end{align*} Also, \begin{align*} a_1+3a_3=(1,0,1)+3(0,1,1)=(1,0,1)+(0,3,3)=(1,3,4)=a_4. \end{align*} Thus each of $a_1,a_2,a_3,a_4$ lies in $\operatorname{span}_{\mathbb{Q}}\{a_1,a_3\}$, so \begin{align*} \operatorname{span}_{\mathbb{Q}}\{a_1,a_2,a_3,a_4\}=\operatorname{span}_{\mathbb{Q}}\{a_1,a_3\}. \end{align*} It remains to check that $a_1$ and $a_3$ are independent. Suppose \begin{align*} \alpha a_1+\beta a_3=(0,0,0) \end{align*} with $\alpha,\beta \in \mathbb{Q}$. Substituting $a_1=(1,0,1)$ and $a_3=(0,1,1)$ gives \begin{align*} \alpha(1,0,1)+\beta(0,1,1)=(\alpha,\beta,\alpha+\beta)=(0,0,0). \end{align*} Comparing the first coordinate gives $\alpha=0$, and comparing the second coordinate gives $\beta=0$. Hence $a_1$ and $a_3$ form an independent spanning set for the column space. Therefore the column space has dimension $2$, so $\operatorname{rank}(A)=2$. [/example] ## Rank and Solving Linear Systems ### Consistency The first major use of rank is to decide when equations can be solved. Adding the right-hand side as one more column tests whether the requested output was already reachable from the coefficient columns. [definition: Augmented Matrix] Let $k$ be a field, let $A \in k^{m \times n}$, and let $b \in k^m$. The augmented matrix of the system $Ax=b$ is the matrix $(A \mid b) \in k^{m \times (n+1)}$ obtained by adjoining $b$ as the final column of $A$. [/definition] The problem is now reduced to recognizing when the final column creates a new independent direction. This is exactly the obstruction to solving the system: if the requested output is already in the column space, the system is consistent, and otherwise it is not. This motivates the rank criterion for consistency. [quotetheorem:7990] The theorem converts a question about equations into a comparison of two dimensions. The next example shows the failure mode: the right-hand side asks for a direction outside the span of the coefficient columns. [example: An Inconsistent System] Let $A:\mathbb{R}^2\to\mathbb{R}^2$ have both columns equal to $(1,2)$, and let $b=(1,3)$. For $x=(x_1,x_2)\in\mathbb{R}^2$, the output is \begin{align*} Ax=x_1(1,2)+x_2(1,2)=(x_1+x_2)(1,2). \end{align*} Thus every vector in $\operatorname{Col}(A)$ is a scalar multiple of $(1,2)$, and $(1,2)$ itself occurs by taking $x_1=1$ and $x_2=0$. Hence \begin{align*} \operatorname{Col}(A)=\operatorname{span}_{\mathbb{R}}\{(1,2)\}. \end{align*} Since $(1,2)\neq(0,0)$, this span has dimension $1$, so $\operatorname{rank}(A)=1$. Now compare $b=(1,3)$ with the column direction $(1,2)$. If $b=\lambda(1,2)$ for some $\lambda\in\mathbb{R}$, then comparing first coordinates gives $\lambda=1$, while comparing second coordinates gives $3=2\lambda$. Substituting $\lambda=1$ into the second equation gives $3=2$, a contradiction. Therefore $b\notin\operatorname{Col}(A)$. The augmented matrix $(A\mid b)$ has columns $(1,2),(1,2),(1,3)$. Its column space is \begin{align*} \operatorname{span}_{\mathbb{R}}\{(1,2),(1,3)\}. \end{align*} To compute its dimension, suppose \begin{align*} \alpha(1,2)+\beta(1,3)=(0,0). \end{align*} Comparing coordinates gives $\alpha+\beta=0$ and $2\alpha+3\beta=0$. From $\alpha+\beta=0$ we get $\alpha=-\beta$, and substituting into the second equation gives \begin{align*} 2(-\beta)+3\beta=\beta=0. \end{align*} Thus $\beta=0$ and then $\alpha=0$, so $(1,2)$ and $(1,3)$ are independent. Hence $\operatorname{rank}(A\mid b)=2$. Solving $Ax=b$ would require \begin{align*} (x_1+x_2)(1,2)=(1,3), \end{align*} which would put $b$ in $\operatorname{Col}(A)$. Since $b\notin\operatorname{Col}(A)$, the system $Ax=b$ has no solution. [/example] ### Degrees of Freedom Consistency is only the first question. Once a solution exists, rank also tells how many parameters remain free. To state that precisely, we need the part of the domain that the map sends to zero. [definition: Kernel] Let $k$ be a field, let $V$ and $W$ be $k$-vector spaces, and let $T: V \to W$ be a $k$-linear map. The kernel of $T$ is \begin{align*} \ker(T) := \{v \in V : T(v)=0\}. \end{align*} [/definition] The kernel is a subspace, but the next question is quantitative: how many independent directions are invisible? This motivates assigning a dimension to the kernel. [definition: Nullity] Let $k$ be a field, let $V$ and $W$ be finite-dimensional $k$-vector spaces, and let $T: V \to W$ be a $k$-linear map. The nullity of $T$ is \begin{align*} \operatorname{nullity}(T) := \dim_k \ker(T). \end{align*} [/definition] Rank measures what survives and nullity measures what disappears, but we still need an accounting theorem saying that no domain dimension is missing. That need leads to rank-nullity. [quotetheorem:385] This theorem explains why a consistent system with $n$ unknowns and rank $r$ has $n-r$ independent parameters. The next example gives a map where both parts can be read without row reduction. [example: Differentiation on Polynomial Spaces] Let $P_3(\mathbb{R})$ be the real [vector space](/page/Vector%20Space) of polynomials of degree at most $3$, and define $D:P_3(\mathbb{R})\to P_2(\mathbb{R})$ by $D(p)=p'$. We compute the image and kernel of $D$ explicitly. If $q\in P_2(\mathbb{R})$, write \begin{align*} q(x)=a+bx+cx^2 \end{align*} with $a,b,c\in\mathbb{R}$. Define \begin{align*} p(x)=ax+\frac{b}{2}x^2+\frac{c}{3}x^3. \end{align*} Then $p\in P_3(\mathbb{R})$, and differentiating term by term gives \begin{align*} D(p)(x)=p'(x)=a+bx+cx^2=q(x). \end{align*} Thus every element of $P_2(\mathbb{R})$ lies in $\operatorname{im}(D)$, so $\operatorname{im}(D)=P_2(\mathbb{R})$. Since $1,x,x^2$ form a basis of $P_2(\mathbb{R})$, we have $\dim_{\mathbb{R}}P_2(\mathbb{R})=3$, and therefore $\operatorname{rank}(D)=3$. Now let $p\in\ker(D)$. Write \begin{align*} p(x)=a+bx+cx^2+dx^3 \end{align*} with $a,b,c,d\in\mathbb{R}$. Then \begin{align*} D(p)(x)=b+2cx+3dx^2. \end{align*} The condition $D(p)=0$ says \begin{align*} b+2cx+3dx^2=0 \end{align*} as a polynomial, so its coefficients are zero: \begin{align*} b=0,\quad 2c=0,\quad 3d=0. \end{align*} Because $2$ and $3$ are nonzero in $\mathbb{R}$, this gives $b=c=d=0$. Hence $p(x)=a$, so $\ker(D)$ is exactly the space of constant polynomials. This space has basis $1$, so $\operatorname{nullity}(D)=1$. Finally, $1,x,x^2,x^3$ form a basis of $P_3(\mathbb{R})$, so $\dim_{\mathbb{R}}P_3(\mathbb{R})=4$. The rank-nullity count in this example is therefore \begin{align*} 4=3+1. \end{align*} Differentiation loses exactly the one-dimensional direction of constant polynomials and reaches every quadratic-or-lower polynomial. [/example] The next natural problem is to recognize when a square linear map has no loss at all. Rank gives a single test that unifies injectivity, surjectivity, and invertibility. [quotetheorem:7991] For rectangular matrices, the same theorem separates into two statements: full column rank gives injectivity, while full row rank gives surjectivity. This distinction is essential in least-squares problems, presentations of modules, and exact sequences. ## Rank Under Operations ### Row Reduction and Normal Form Rank is useful because it survives controlled rewriting. Gaussian elimination changes the equations but not the space of consequences they generate. The permitted rewrites are the elementary row operations. [definition: Elementary Row Operation] Let $k$ be a field. An elementary row operation on $k^{m \times n}$ is a map $E: k^{m \times n} \to k^{m \times n}$ of one of the following forms: For distinct $p,q \in \{1,\dots,m\}$, row interchange sends \begin{align*} A &\mapsto A', \end{align*} where the $p$-th row of $A'$ is the $q$-th row of $A$, the $q$-th row of $A'$ is the $p$-th row of $A$, and all other rows are unchanged. For $p \in \{1,\dots,m\}$ and $\lambda \in k \setminus \{0\}$, row scaling sends \begin{align*} A &\mapsto A', \end{align*} where the $p$-th row of $A'$ is $\lambda$ times the $p$-th row of $A$, and all other rows are unchanged. For distinct $p,q \in \{1,\dots,m\}$ and $\lambda \in k$, row replacement sends \begin{align*} A &\mapsto A', \end{align*} where the $p$-th row of $A'$ is the $p$-th row of $A$ plus $\lambda$ times the $q$-th row of $A$, and all other rows are unchanged. [/definition] Because each elementary row operation can be reversed, a sequence of such operations gives a meaningful [equivalence relation](/page/Equivalence%20Relation). We need this relation to state exactly what row reduction preserves. [definition: Row Equivalence] Let $k$ be a field, and let $A,B \in k^{m \times n}$. The matrices $A$ and $B$ are row equivalent if $B$ can be obtained from $A$ by a finite sequence of elementary row operations. [/definition] The point of defining row equivalence is to ask which invariants survive row reduction. Rank is the invariant needed for solving systems, so it must be preserved by this relation. [quotetheorem:7992] This is what makes Gaussian elimination compatible with the intrinsic notion of rank: elementary row operations change the equations, but not the number of independent constraints they impose. In practice, it permits rank to be read from any row echelon form instead of from the original matrix. The limitation is that this only accounts for row operations; to compare the row-based computation with the column-based definition of rank, we still need a theorem identifying row rank and column rank. There is still a conceptual mismatch to resolve: row reduction exposes independent pivot rows, while rank was defined as the dimension of the span of the columns. To make row reduction a genuine rank computation rather than just a row-space procedure, we need the row count and column count to be the same for every matrix over a field. [quotetheorem:389] Rows count independent constraints and columns count reachable outputs. This raises a classification problem: once row and column coordinates may both be changed, what remains of a matrix besides its rank? [remark: Canonical Rank Form] If $T:V\to W$ is a linear map of rank $r$ between finite-dimensional vector spaces over a field, suitable bases of $V$ and $W$ put its matrix in canonical rank form: the upper-left block is the $r\times r$ identity matrix $I_r$, and every remaining block is zero. [/remark] This normal form says that, after choosing suitable coordinates, every rank $r$ map is an identity on $r$ directions and zero on the rest. This is the cleanest geometric meaning of rank over a field, and it is the reason rank is the only invariant left once both domain and codomain coordinates may be changed. ### Products and Sums Rank also controls how information moves through algebraic operations. A composition cannot recover directions lost by either factor, and a sum cannot have image larger than the span of the two original images. [quotetheorem:7993] The product inequality says that lost information stays lost. The simplest nonzero maps show the opposite extreme: a rank-one map preserves only one scalar measurement of the input. [example: Rank-One Matrices] Let $u\in k^m$ and $v=(v_1,\dots,v_n)\in k^n$ be nonzero vectors, and define $A:k^n\to k^m$ by \begin{align*} A(x_1,\dots,x_n)=u(v_1x_1+\cdots+v_nx_n). \end{align*} For every $x\in k^n$, the scalar \begin{align*} \lambda=v_1x_1+\cdots+v_nx_n \end{align*} lies in $k$, so \begin{align*} A(x)=\lambda u\in \operatorname{span}_k\{u\}. \end{align*} Thus $\operatorname{im}(A)\subseteq \operatorname{span}_k\{u\}$. Because $v\neq 0$, there is an index $j$ with $v_j\neq 0$. Let $e_j\in k^n$ be the $j$-th standard basis vector. Then \begin{align*} A(e_j)=u(v_1\cdot 0+\cdots+v_j\cdot 1+\cdots+v_n\cdot 0)=v_j u. \end{align*} Since $v_j\neq 0$ in the field $k$, it has an inverse $v_j^{-1}$. For any $\mu\in k$, \begin{align*} A(\mu v_j^{-1}e_j)=u(v_j\mu v_j^{-1})=\mu u. \end{align*} Hence every vector in $\operatorname{span}_k\{u\}$ lies in $\operatorname{im}(A)$, so \begin{align*} \operatorname{im}(A)=\operatorname{span}_k\{u\}. \end{align*} The vector $u$ is nonzero, so $\{u\}$ is a one-element basis of this image. Therefore \begin{align*} \operatorname{rank}(A)=\dim_k\operatorname{im}(A)=1. \end{align*} This rank-one map keeps only the single scalar measurement $v_1x_1+\cdots+v_nx_n$ of the input and then places that scalar in the fixed output direction $u$. [/example] ## Rank Beyond Vector Spaces ### Free Modules Over a field, every finite-dimensional vector space has a basis and all bases have the same size. Over a ring, modules may have relations or torsion, so rank first has to be restricted to modules that really behave like coordinate spaces. [definition: Free Module] Let $R$ be a ring. A left $R$-module $M$ is free if there exists a set $I$ and elements $(e_i)_{i \in I}$ of $M$ such that every $m \in M$ has a unique expression \begin{align*} m = \sum_{i \in F} r_i e_i, \end{align*} where $F \subset I$ is finite and $r_i \in R$ for each $i \in F$. [/definition] The remaining problem is numerical: once a module has a basis, rank should count basis elements. Over arbitrary rings this count is not automatically intrinsic, because a ring may fail invariant basis number and allow the same free module to have bases of different sizes. The next definition is therefore conditional: it defines rank only in the cases where the basis size is independent of the chosen basis. [definition: Rank of a Free Module] Let $R$ be a ring, and let $M$ be a free left $R$-module with a basis $(e_i)_{i \in I}$. If all bases of $M$ have the same cardinality, the rank of $M$ is \begin{align*} \operatorname{rank}_R(M) := |I|. \end{align*} [/definition] This recovers dimension when $R$ is a field. For many common rings, including commutative rings and the rings that usually appear in first examples, free modules have invariant basis number, so the basis size is well-defined. The qualification matters because arbitrary rings can fail this property, and then a free module may admit bases of different cardinalities. For lattices over $\mathbb{Z}$, the definition gives the usual count of independent integral directions. [example: Rank of a Lattice] In $\mathbb{Z}^3$, let $e_1=(1,0,0)$, $e_2=(0,1,0)$, and $e_3=(0,0,1)$. Every element $(a,b,c)\in\mathbb{Z}^3$ has the expression \begin{align*} (a,b,c)=ae_1+be_2+ce_3. \end{align*} If also $(a,b,c)=a'e_1+b'e_2+c'e_3$, then comparing coordinates gives $a=a'$, $b=b'$, and $c=c'$. Thus $e_1,e_2,e_3$ form a basis of the free $\mathbb{Z}$-module $\mathbb{Z}^3$, so $\operatorname{rank}_{\mathbb{Z}}(\mathbb{Z}^3)=3$. Now let $H$ be the subgroup generated by $(1,0,0)$ and $(0,2,0)$. Its elements are exactly \begin{align*} a(1,0,0)+b(0,2,0)=(a,2b,0) \end{align*} with $a,b\in\mathbb{Z}$. Define $\varphi:\mathbb{Z}^2\to H$ by \begin{align*} \varphi(a,b)=(a,2b,0). \end{align*} This map is surjective by the displayed description of $H$. If $\varphi(a,b)=(0,0,0)$, then $(a,2b,0)=(0,0,0)$, so $a=0$ and $2b=0$. Since $b\in\mathbb{Z}$, $2b=0$ gives $b=0$. Hence $\varphi$ is injective, so $H\cong\mathbb{Z}^2$ and $\operatorname{rank}_{\mathbb{Z}}(H)=2$. The subgroup generated by the first two standard basis vectors is \begin{align*} \{(a,b,0):a,b\in\mathbb{Z}\}. \end{align*} It contains $(0,1,0)$, but $(0,1,0)\notin H$, because membership in $H$ would require $1=2b$ for some $b\in\mathbb{Z}$. Thus $H$ has rank $2$ without being the full coordinate plane subgroup generated by $e_1$ and $e_2$. [/example] ### Torsion and Generic Rank Many important modules are not free, but over an integral domain there is still a way to measure the number of independent non-torsion directions. Passing to the fraction field removes torsion and turns the module into a vector space. [definition: Rank over an Integral Domain] Let $R$ be an integral domain with fraction field $K=\operatorname{Frac}(R)$, and let $M$ be an $R$-module. The rank of $M$ over $R$ is \begin{align*} \operatorname{rank}_R(M) := \dim_K(K \otimes_R M), \end{align*} provided this dimension is defined as a cardinal number. [/definition] This definition agrees with free rank for free finite-rank modules, but it ignores torsion. The next example shows exactly what disappears after tensoring with the fraction field. [example: Torsion Does Not Contribute to Rank] Let $M=\mathbb{Z}^2 \oplus \mathbb{Z}/6\mathbb{Z}$ as a $\mathbb{Z}$-module. We first show that the torsion summand becomes zero after tensoring with $\mathbb{Q}$. Let $\overline{1}$ be the class of $1$ in $\mathbb{Z}/6\mathbb{Z}$. Since $6\overline{1}=\overline{0}$, we have \begin{align*} 1\otimes \overline{1}=6\cdot \frac{1}{6}\otimes \overline{1}=\frac{1}{6}\otimes 6\overline{1}=\frac{1}{6}\otimes \overline{0}=0. \end{align*} For any $q\in\mathbb{Q}$ and any class $\overline{r}\in\mathbb{Z}/6\mathbb{Z}$, \begin{align*} q\otimes \overline{r}=q\otimes r\overline{1}=qr\otimes \overline{1}=qr(1\otimes \overline{1})=0. \end{align*} Hence \begin{align*} \mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Z}/6\mathbb{Z}=0. \end{align*} Now define \begin{align*} \Phi:\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Z}^2\to \mathbb{Q}^2 \end{align*} by \begin{align*} \Phi(q\otimes(a,b))=(qa,qb). \end{align*} Its inverse is the map \begin{align*} \Psi(x,y)=x\otimes(1,0)+y\otimes(0,1). \end{align*} Indeed, for $q\in\mathbb{Q}$ and $(a,b)\in\mathbb{Z}^2$, \begin{align*} \Psi(\Phi(q\otimes(a,b)))=qa\otimes(1,0)+qb\otimes(0,1)=q\otimes(a,0)+q\otimes(0,b)=q\otimes(a,b). \end{align*} Also, for $(x,y)\in\mathbb{Q}^2$, \begin{align*} \Phi(\Psi(x,y))=\Phi(x\otimes(1,0)+y\otimes(0,1))=(x,0)+(0,y)=(x,y). \end{align*} Thus $\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Z}^2\cong\mathbb{Q}^2$, while the $\mathbb{Z}/6\mathbb{Z}$ summand contributes zero. Therefore \begin{align*} \mathbb{Q}\otimes_{\mathbb{Z}}M\cong \mathbb{Q}^2. \end{align*} By the definition of rank over an integral domain, \begin{align*} \operatorname{rank}_{\mathbb{Z}}(M)=\dim_{\mathbb{Q}}(\mathbb{Q}\otimes_{\mathbb{Z}}M)=\dim_{\mathbb{Q}}\mathbb{Q}^2=2. \end{align*} The finite summand changes the abelian group $M$, but it disappears after tensoring with $\mathbb{Q}$ and so does not affect the rank. [/example] In commutative algebra, rank often describes the size of a module away from special primes. This motivates a separate phrase for the fraction-field rank of a finitely generated module. [definition: Generic Rank] Let $R$ be an integral domain with fraction field $K=\operatorname{Frac}(R)$, and let $M$ be a finitely generated $R$-module. The generic rank of $M$ is \begin{align*} \dim_K(K \otimes_R M). \end{align*} [/definition] Generic rank is the stable size seen at the generic point. Special fibers may gain torsion or lose freeness, but the generic rank captures the part visible before specializing. [example: Zero Divisors Break the Vector-Space Picture] Let $R=\mathbb{Z}/6\mathbb{Z}$, and write elements as residue classes $0,1,2,3,4,5$. Define the $R$-linear map $T:R\to R$ by \begin{align*} T(x)=2x. \end{align*} Evaluating $T$ on all six elements gives \begin{align*} T(0)=0,\quad T(1)=2,\quad T(2)=4,\quad T(3)=6=0,\quad T(4)=8=2,\quad T(5)=10=4 \end{align*} in $\mathbb{Z}/6\mathbb{Z}$. Therefore \begin{align*} \operatorname{im}(T)=\{0,2,4\}. \end{align*} The kernel consists of the residue classes $x$ satisfying $2x=0$ in $\mathbb{Z}/6\mathbb{Z}$, equivalently $6$ divides $2x$. This is the same as saying $3$ divides $x$, so among the residue classes modulo $6$ the possibilities are $x=0$ and $x=3$. Hence \begin{align*} \ker(T)=\{0,3\}. \end{align*} The obstruction is already visible in the scalar ring: $2\neq 0$ and $3\neq 0$ in $R$, but \begin{align*} 2\cdot 3=6=0 \end{align*} in $R$, so $R$ has zero divisors and is not a field. Thus $\ker(T)$ and $\operatorname{im}(T)$ are not being measured as vector spaces over the same ground field. Their cardinalities are \begin{align*} |R|=6,\quad |\operatorname{im}(T)|=3,\quad |\ker(T)|=2, \end{align*} and the additive dimension count suggested by rank-nullity would read $6=3+2$, which is false. This is why module rank is usually defined through freeness, localization, or hypotheses on the ring rather than by copying the vector-space picture without modification. [/example] ## Tensor and Bilinear Rank Rank can also mean the number of simple pieces needed to build an object. For matrices the simple pieces are rank-one maps. For higher tensors the same idea remains, but the behavior is less rigid. [definition: Tensor Rank] Let $k$ be a field, and let $V_1, \dots, V_d$ be finite-dimensional $k$-vector spaces. The zero tensor in $V_1 \otimes_k \cdots \otimes_k V_d$ has tensor rank $0$. For a nonzero tensor $T \in V_1 \otimes_k \cdots \otimes_k V_d$, the tensor rank of $T$ is the least $r \in \mathbb{N}$ such that there exist vectors $v_{j,i} \in V_i$ with \begin{align*} T = \sum_{j=1}^r v_{j,1} \otimes \cdots \otimes v_{j,d}. \end{align*} [/definition] For two tensor factors this is matrix rank after choosing bases. With three or more tensor factors, rank becomes harder because there is no universal row-reduction procedure. [example: A Rank-Two Tensor] Let $V=\mathbb{R}^2$ with basis $e_1,e_2$, and let $e_1^*,e_2^*$ be the [dual basis](/theorems/414). In $V\otimes_{\mathbb{R}}V\otimes_{\mathbb{R}}V$, set \begin{align*} T=e_1\otimes e_1\otimes e_1+e_2\otimes e_2\otimes e_2. \end{align*} This is a sum of two simple tensors, so by the definition of tensor rank, \begin{align*} \operatorname{rank}_{\otimes}(T)\leq 2. \end{align*} It remains to show that $T$ is not simple. Contract the third tensor factor by defining, for $\varphi\in V^*$, \begin{align*} C_\varphi(u\otimes v\otimes w)=\varphi(w)(u\otimes v) \end{align*} and extending linearly. Since $e_1^*(e_1)=1$, $e_1^*(e_2)=0$, $e_2^*(e_1)=0$, and $e_2^*(e_2)=1$, we get \begin{align*} C_{e_1^*}(T)=1(e_1\otimes e_1)+0(e_2\otimes e_2)=e_1\otimes e_1. \end{align*} Similarly, \begin{align*} C_{e_2^*}(T)=0(e_1\otimes e_1)+1(e_2\otimes e_2)=e_2\otimes e_2. \end{align*} The tensors $e_1\otimes e_1$ and $e_2\otimes e_2$ are distinct elements of the basis $\{e_i\otimes e_j: i,j\in\{1,2\}\}$ of $V\otimes_{\mathbb{R}}V$, so they are linearly independent. Now suppose, for contradiction, that $T$ were simple, say \begin{align*} T=u\otimes v\otimes w. \end{align*} Then \begin{align*} C_{e_1^*}(T)=e_1^*(w)(u\otimes v) \end{align*} and \begin{align*} C_{e_2^*}(T)=e_2^*(w)(u\otimes v). \end{align*} Both contractions would therefore lie in the one-dimensional subspace $\operatorname{span}_{\mathbb{R}}\{u\otimes v\}$, so they could not be linearly independent. This contradicts the two contractions computed above. Hence $T$ is not simple, so its tensor rank is not $1$. Since $T\neq 0$ and $\operatorname{rank}_{\otimes}(T)\leq 2$, the tensor rank of $T$ is exactly $2$. [/example] Some algebraic operations are bilinear rather than linear. To measure their complexity, we decompose the operation into scalar measurements of each input followed by fixed output vectors. [definition: Bilinear Rank] Let $k$ be a field, let $U,V,W$ be finite-dimensional $k$-vector spaces, and let $\mu: U \times V \to W$ be a bilinear map. The bilinear rank of $\mu$ is the least $r \in \mathbb{N}$ such that there exist linear functionals $f_j: U \to k$, linear functionals $g_j: V \to k$, and vectors $w_j \in W$ satisfying \begin{align*} \mu(u,v)=\sum_{j=1}^r f_j(u)g_j(v)w_j \end{align*} for all $u \in U$ and $v \in V$. [/definition] The zero bilinear map is assigned bilinear rank $0$. This definition turns multiplication algorithms into rank decompositions. A shorter decomposition means fewer scalar products are needed. [example: Multiplication of Complex Numbers] Regard $\mathbb{C}$ as the real vector space with basis $1,i$. For [real numbers](/page/Real%20Numbers) $a,b,c,d$, multiplication is \begin{align*} (a+bi)(c+di)=ac+adi+bci+bd\,i^2. \end{align*} Since $i^2=-1$, this becomes \begin{align*} (a+bi)(c+di)=(ac-bd)+(ad+bc)i. \end{align*} Set \begin{align*} p_1=ac,\quad p_2=bd,\quad p_3=(a+b)(c+d). \end{align*} Expanding $p_3$ over $\mathbb{R}$ gives \begin{align*} p_3=ac+ad+bc+bd. \end{align*} Therefore \begin{align*} p_3-p_1-p_2=(ac+ad+bc+bd)-ac-bd=ad+bc. \end{align*} Substituting $p_1=ac$, $p_2=bd$, and $p_3-p_1-p_2=ad+bc$ into the usual product formula gives \begin{align*} (a+bi)(c+di)=(p_1-p_2)+(p_3-p_1-p_2)i. \end{align*} The three quantities $p_1$, $p_2$, and $p_3$ are three real scalar products: $a\cdot c$, $b\cdot d$, and $(a+b)\cdot(c+d)$. The displayed formula reconstructs the complex product from these three scalar products using only real additions and subtractions, so complex multiplication has bilinear rank at most $3$. [/example] ## Beyond and Connected Topics Rank is one of the organizing invariants in [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra), where it appears with bases, dimension, determinants, eigenvalues, and normal forms. The finite-dimensional theory treats rank as image dimension and uses it to control systems of equations and linear maps. In [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra), rank becomes subtler because modules over rings need not be vector spaces. Localization, torsion, projective modules, and generic rank replace some of the simpler field-based tools. In [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions), rank appears through free resolutions, Euler characteristics, and alternating sums of ranks. Exactness governs how ranks add, while homology measures the obstruction to exactness. In [Lie Algebras II: Structure and Classification](/page/Lie%20Algebras%20II%3A%20Structure%20and%20Classification), rank has a specialized meaning: the rank of a semisimple Lie algebra is the dimension of a Cartan subalgebra. This is not matrix rank, but it follows the same broad pattern of measuring the independent directions controlling a structure. Rank also connects to determinants and minors. For a square matrix $A \in k^{n \times n}$, the condition $\det A \neq 0$ is equivalent to $\operatorname{rank}(A)=n$. More generally, the largest size of a nonzero minor detects rank and leads to determinantal ideals in algebraic geometry and commutative algebra. ## References Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Androma, [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). Androma, [Lie Algebras II: Structure and Classification](/page/Lie%20Algebras%20II%3A%20Structure%20and%20Classification). Axler, *Linear Algebra Done Right* (2015). Dummit and Foote, *Abstract Algebra* (2004). Eisenbud, *Commutative Algebra with a View Toward Algebraic Geometry* (1995). Landsberg, *Tensors: Geometry and Applications* (2012).