A finite-dimensional vector space is useful because its elements can be named by scalars. The difficulty is that the naming system is not built into the space. In $\mathbb{R}^2$, the vector usually written as $(3,2)$ already looks like two numbers, but those numbers depend on having silently chosen the horizontal and vertical directions. Change the directions, and the same vector receives a different coordinate record. The central tension is that two opposite mistakes are easy to make. If we choose too few vectors, some elements cannot be reached. If we choose too many, the same element has several descriptions. A basis is exactly the point where spanning and uniqueness meet. [example: A Spanning Set with Ambiguous Coordinates] Let $V=\mathbb{R}^2$ over $\mathbb{R}$, and set $v_1=(1,0)$, $v_2=(0,1)$, and $v_3=(1,1)$. The set $\{v_1,v_2,v_3\}$ spans $V$ because every vector $(a,b)\in \mathbb{R}^2$ can be written using $v_1$ and $v_2$ alone: \begin{align*} av_1+bv_2+0v_3=a(1,0)+b(0,1)+0(1,1)=(a,0)+(0,b)+(0,0)=(a,b). \end{align*} Now consider the particular vector $(1,1)$. Using $v_1$ and $v_2$ gives \begin{align*} 1v_1+1v_2+0v_3=1(1,0)+1(0,1)+0(1,1)=(1,0)+(0,1)+(0,0)=(1,1). \end{align*} Using $v_3$ instead gives \begin{align*} 0v_1+0v_2+1v_3=0(1,0)+0(0,1)+1(1,1)=(0,0)+(0,0)+(1,1)=(1,1). \end{align*} The two coefficient triples are different, since $(1,1,0)\ne (0,0,1)$, but they produce the same vector. Thus this spanning set gives existence of coordinates without uniqueness; the extra vector $v_3$ is redundant because $v_3=v_1+v_2$. [/example] This example shows why a basis is not just a convenient list of generators. It is a coordinate system with no redundancy. The whole theory explains how such lists exist, how large they are, how they change, and what breaks when we leave vector spaces for general modules. ## Definition A basis is the exact compromise promised by the opening example: enough vectors to reach every element, but not so many that coordinates become ambiguous. We state the main definition first, then unpack the two ingredients that make it work. [definition: Basis] Let $V$ be a vector space over a field $k$. A basis of $V$ is a subset $B \subset V$ such that $B$ is linearly independent and $B$ is a spanning set for $V$. [/definition] The definition hides two separate demands. To understand why both are necessary, we first isolate reachability, then uniqueness. ### Spanning A set of vectors is useful only if the vectors we care about can be assembled from it using the allowed scalar operations. This is the algebraic version of having enough directions. [definition: Linear Combination] Let $V$ be a vector space over a field $k$, and let $S \subset V$. A linear combination of elements of $S$ is a vector of the form \begin{align*} a_1v_1 + \cdots + a_nv_n, \end{align*} where $n \in \mathbb{N}$, $v_1, \ldots, v_n \in S$, and $a_1, \ldots, a_n \in k$. [/definition] Linear combinations are finite by definition. This matters because vector spaces do not carry a default notion of infinite summation. Infinite expansions require additional topology, as in Hilbert space or Banach space theory, and are not part of the algebraic notion of basis. We also allow the empty linear combination, whose value is $0$; with this convention, the span of the empty set is $\{0\}$. Once linear combinations are available, the natural next question is what part of the ambient space they reach. We need a name for the set generated by a chosen family before we can ask whether the family reaches everything. [definition: Span] Let $V$ be a vector space over a field $k$, and let $S \subset V$. The span of $S$, denoted $\operatorname{span}_k(S)$, is the set of all linear combinations of elements of $S$. [/definition] Knowing the span of a set still leaves an existence problem: some vectors of the ambient space may lie outside all finite combinations of the chosen vectors. The useful case is when this obstruction disappears, so every vector in the space can be reached from the chosen set. [definition: Spanning Set] Let $V$ be a vector space over a field $k$. A subset $S \subset V$ is a spanning set for $V$ if \begin{align*} \operatorname{span}_k(S) = V. \end{align*} [/definition] A spanning set answers the existence question: can every vector be described using the chosen vectors? It does not answer the uniqueness question. The ambiguity in the opening example came from a nonzero relation among the chosen vectors. ### Independence The second half of the problem is uniqueness. If a combination of chosen vectors can vanish without all its coefficients vanishing, then the same vector can be written in more than one way. We first name the algebraic witness to that ambiguity. [definition: Linear Relation] Let $V$ be a vector space over a field $k$, and let $v_1, \ldots, v_n \in V$. A linear relation among $v_1, \ldots, v_n$ is an equality \begin{align*} a_1v_1 + \cdots + a_nv_n = 0, \end{align*} where $a_1, \ldots, a_n \in k$. [/definition] Relations are not bad by themselves; the relation with all coefficients zero is unavoidable. To rule out ambiguous coordinates, we need a condition saying that no finite chosen list admits a nonzero relation. This condition is linear independence. [definition: Linearly Independent Set] Let $V$ be a vector space over a field $k$. A subset $S \subset V$ is linearly independent if for every $n \in \mathbb{N}$, every choice of distinct vectors $v_1, \ldots, v_n \in S$, and every choice of scalars $a_1, \ldots, a_n \in k$, the equality \begin{align*} a_1v_1 + \cdots + a_nv_n = 0 \end{align*} implies \begin{align*} a_1 = \cdots = a_n = 0. \end{align*} [/definition] Linear independence is the uniqueness half of coordinates. A nonzero relation is exactly a way of changing coefficients without changing the resulting vector. ### Basis Now the two requirements meet. We have already named their intersection: a basis is a spanning set with no nonzero finite relation among its distinct elements. The definition is short, but it carries a strong consequence: a basis is a coordinate system. This is often the first result to use after finding a basis. [quotetheorem:372] This theorem is why bases are useful rather than decorative. The vector has not changed, but the basis gives it a finite algebraic address. [example: The Standard Basis of $k^n$] Let $k$ be a field, and let $V=k^n$. For $1 \le i \le n$, let $e_i\in k^n$ be the vector whose $i$-th coordinate is $1$ and whose other coordinates are $0$. We show that $E=\{e_1,\ldots,e_n\}$ is a basis by checking spanning and linear independence. First let $x=(x_1,\ldots,x_n)\in k^n$. The $j$-th coordinate of $x_1e_1+\cdots+x_ne_n$ is \begin{align*} x_1(e_1)_j+\cdots+x_n(e_n)_j=x_j(e_j)_j=x_j, \end{align*} because $(e_j)_j=1$ and $(e_i)_j=0$ for $i\ne j$. Hence every coordinate agrees with the corresponding coordinate of $x$, so \begin{align*} x=x_1e_1+\cdots+x_ne_n. \end{align*} Thus $E$ spans $k^n$. To prove linear independence, suppose \begin{align*} a_1e_1+\cdots+a_ne_n=0 \end{align*} for scalars $a_1,\ldots,a_n\in k$. Taking the $j$-th coordinate of both sides gives \begin{align*} a_1(e_1)_j+\cdots+a_n(e_n)_j=0. \end{align*} All terms with $i\ne j$ vanish, and $(e_j)_j=1$, so this equation becomes \begin{align*} a_j=0. \end{align*} Since this holds for every $1\le j\le n$, all coefficients are zero. Therefore $E$ is linearly independent, and hence $E$ is a basis of $k^n$. The expansion above shows that the coefficients relative to this basis are exactly the usual coordinates of $x$. [/example] ## Coordinates and Linear Maps ### Coordinate Vectors A basis turns an abstract vector into a tuple of scalars. This tuple is not the vector itself; it is a coordinate record relative to a specified ordered basis. Without naming the basis, the tuple has no reliable meaning. [definition: Ordered Basis] Let $V$ be a finite-dimensional vector space over a field $k$. An ordered basis of $V$ is a finite sequence $(b_1, \ldots, b_n)$ whose underlying set $\{b_1, \ldots, b_n\}$ is a basis of $V$. [/definition] To use an ordered basis computationally, we need a notation for the tuple of coefficients attached to a vector. This is the coordinate vector, and it is the object that lets abstract vectors be multiplied by matrices. [definition: Coordinate Vector] Let $V$ be a finite-dimensional vector space over a field $k$, and let $B = (b_1, \ldots, b_n)$ be an ordered basis of $V$. If \begin{align*} v = a_1b_1 + \cdots + a_nb_n, \end{align*} then the coordinate vector of $v$ relative to $B$ is \begin{align*} [v]_B = (a_1, \ldots, a_n) \in k^n. \end{align*} [/definition] The order of the basis matters here. The same unordered basis gives many ordered bases, and permuting the order permutes the coordinate entries. [example: Coordinates in a Nonstandard Basis] Let $V=\mathbb{R}^2$, let $b_1=(1,1)$ and $b_2=(1,-1)$, and set $B=(b_1,b_2)$. To see that $B$ is a basis, suppose first that \begin{align*} a_1b_1+a_2b_2=(0,0). \end{align*} Expanding the left side gives \begin{align*} a_1(1,1)+a_2(1,-1)=(a_1+a_2,a_1-a_2). \end{align*} Thus $a_1+a_2=0$ and $a_1-a_2=0$. Adding these two equations gives $2a_1=0$, so $a_1=0$; then $a_1+a_2=0$ gives $a_2=0$. Hence $b_1,b_2$ are linearly independent. Since two linearly independent vectors in $\mathbb{R}^2$ span $\mathbb{R}^2$, $B$ is a basis. For $v=(3,1)$, write \begin{align*} (3,1)=a_1b_1+a_2b_2. \end{align*} Expanding the right side gives \begin{align*} a_1b_1+a_2b_2=a_1(1,1)+a_2(1,-1)=(a_1+a_2,a_1-a_2). \end{align*} Therefore the two coordinates must satisfy \begin{align*} a_1+a_2=3. \end{align*} \begin{align*} a_1-a_2=1. \end{align*} Adding the equations gives $2a_1=4$, so $a_1=2$. Substituting $a_1=2$ into $a_1+a_2=3$ gives $2+a_2=3$, hence $a_2=1$. Therefore \begin{align*} v=2b_1+1b_2. \end{align*} So $[v]_B=(2,1)$. The vector $(3,1)$ and its coordinate vector $(2,1)$ live in different coordinate systems, even though both are written as ordered pairs. [/example] ### Matrices from Bases Coordinates become most powerful when they turn linear maps into arrays of scalars. To do this, the domain and codomain both need ordered bases, since the input coordinates and output coordinates may live in different spaces. [definition: Matrix of a Linear Map] Let $V$ and $W$ be finite-dimensional vector spaces over a field $k$. Let $B = (b_1, \ldots, b_n)$ be an ordered basis of $V$, and let $C = (c_1, \ldots, c_m)$ be an ordered basis of $W$. For a linear map $T: V \to W$, the matrix of $T$ relative to $B$ and $C$ is the matrix $[T]_{C \leftarrow B} \in k^{m \times n}$ whose $j$-th column is $[T(b_j)]_C$. [/definition] This column rule only specifies what happens to basis vectors. The possible obstruction is that an arbitrary vector is a linear combination of many basis vectors, so the matrix must respect those coefficients exactly if it is to represent the whole linear map rather than just a table of selected values. [quotetheorem:7837] [citeproof:7837] [example: Differentiation on Polynomials] Let $V$ be the vector space over $\mathbb{R}$ of polynomials of degree at most $3$, and let $W$ be the vector space over $\mathbb{R}$ of polynomials of degree at most $2$. Use the ordered bases $B=(1,x,x^2,x^3)$ and $C=(1,x,x^2)$. Let $T:V\to W$ be differentiation, so $T(p)=p'$. On the basis vectors of $B$, \begin{align*} T(1)=0. \end{align*} \begin{align*} T(x)=1. \end{align*} \begin{align*} T(x^2)=2x. \end{align*} \begin{align*} T(x^3)=3x^2. \end{align*} Writing these outputs in the basis $C=(1,x,x^2)$ gives \begin{align*} [T(1)]_C=(0,0,0). \end{align*} \begin{align*} [T(x)]_C=(1,0,0). \end{align*} \begin{align*} [T(x^2)]_C=(0,2,0). \end{align*} \begin{align*} [T(x^3)]_C=(0,0,3). \end{align*} By the definition of the matrix of a linear map, these four coordinate vectors are the columns of $[T]_{C\leftarrow B}$. Thus $[T]_{C\leftarrow B}$ is the $3\times 4$ matrix whose first column is $(0,0,0)$, second column is $(1,0,0)$, third column is $(0,2,0)$, and fourth column is $(0,0,3)$; equivalently, its only nonzero entries are $m_{12}=1$, $m_{23}=2$, and $m_{34}=3$. For example, if $p=a+bx+cx^2+dx^3$, then \begin{align*} [p]_B=(a,b,c,d). \end{align*} Differentiating term by term gives \begin{align*} T(p)=b+2cx+3dx^2. \end{align*} Hence \begin{align*} [T(p)]_C=(b,2c,3d). \end{align*} The matrix records exactly this coordinate rule; it is not the derivative itself, but the derivative written relative to the chosen polynomial bases. [/example] ## Size and Dimension ### Exchange A basis has two jobs, so it is not immediate that all bases of the same vector space have the same number of elements. If coordinate systems can be chosen in many ways, why should their lengths agree? The exchange principle is the mechanism behind the answer: independent vectors can be inserted into a spanning list, but each insertion forces an old vector to be removed. [quotetheorem:373] A counting principle is not yet an invariant: two different bases could still, in principle, have different lengths. To attach a number to the space itself, we need to rule out that dependence on the chosen coordinate system. [quotetheorem:915] Now the number of basis vectors becomes an invariant of the space rather than an accident of a chosen coordinate system. This invariant deserves its own name because it measures how many independent directions the space contains. [definition: Dimension] Let $V$ be a vector space over a field $k$. If $V$ has a finite basis $B$, the dimension of $V$ over $k$ is \begin{align*} \dim_k V = |B|. \end{align*} [/definition] If $V$ has no finite basis, then $V$ is called infinite-dimensional. The finite definition deliberately separates the computable case from the set-theoretic questions that appear later for infinite bases. [example: Polynomial Spaces] Let $P_{\le d}(k)$ be the vector space of polynomials over a field $k$ of degree at most $d$, and set \begin{align*} S=\{1,x,x^2,\ldots,x^d\}. \end{align*} We show that $S$ is a basis by checking spanning and linear independence. First let $p\in P_{\le d}(k)$. By the definition of $P_{\le d}(k)$, there are scalars $a_0,a_1,\ldots,a_d\in k$ such that \begin{align*} p=a_0+a_1x+\cdots+a_dx^d. \end{align*} This is a linear combination of the elements of $S$, so every polynomial in $P_{\le d}(k)$ lies in $\operatorname{span}_k(S)$. Conversely, any linear combination \begin{align*} a_0\cdot 1+a_1x+\cdots+a_dx^d \end{align*} is a polynomial of degree at most $d$, so it lies in $P_{\le d}(k)$. Hence $\operatorname{span}_k(S)=P_{\le d}(k)$. To prove linear independence, suppose \begin{align*} a_0\cdot 1+a_1x+\cdots+a_dx^d=0 \end{align*} as a polynomial. Equality with the zero polynomial means that every coefficient is zero, so \begin{align*} a_0=a_1=\cdots=a_d=0. \end{align*} Thus $S$ is linearly independent. Since $S$ is both spanning and linearly independent, it is a basis of $P_{\le d}(k)$. The list $1,x,x^2,\ldots,x^d$ has $d+1$ elements, namely one element for each exponent $0,1,\ldots,d$. Therefore \begin{align*} \dim_k P_{\le d}(k)=d+1. \end{align*} The extra $1$ is the constant term, which is the coefficient of $x^0$. [/example] ### Extending and Trimming Dimension is useful because it turns basis construction into a controlled process. Starting from independent vectors, we want to add enough vectors to span. Starting from spanning vectors, we want to remove redundancy without losing the whole space. The first issue is existence: an independent set is locally consistent, but it may not yet describe every vector. The extension theorem says that, in finite dimension, no independent starting set is stranded before it becomes a basis. [quotetheorem:374] This result is used whenever coordinates are chosen under constraints. For example, if two independent vectors have already been selected because they are geometrically meaningful, the theorem permits us to complete them to a full coordinate system instead of starting over. Its finite-dimensional hypothesis is doing real work: the statement is a controlled finite construction, not yet the set-theoretic existence theorem needed for arbitrary infinite spaces. The same control also constrains subspaces. A subspace may sit inside a larger vector space in a complicated way, but it cannot contain more independent directions than the ambient space has available. This raises a separate question from extending independent sets: if $W \subseteq V$ is itself a vector space inside $V$, how does the size of a basis of $W$ compare with the size of a basis of $V$? The next result supplies that comparison. It is what makes "inside" compatible with dimension: inclusion of subspaces forces an inequality of sizes. [quotetheorem:375] Together, the two results make dimension a practical invariant rather than just a definition. The extension theorem explains how independent data can be completed to a basis, while the subspace dimension theorem explains why passing to a smaller linear world cannot increase dimension. Later change-of-basis arguments depend on this flexibility: once bases exist and their sizes are controlled, the remaining question is how different valid bases communicate with each other. ## Change of Basis ### Coordinate Translation Coordinates are useful only if we know how to translate between coordinate systems. The same vector may have two coordinate records, and the relation between them is controlled by an invertible matrix. [definition: Change of Basis Matrix] Let $V$ be a finite-dimensional vector space over a field $k$, and let $B = (b_1,\ldots,b_n)$ and $C = (c_1,\ldots,c_n)$ be ordered bases of $V$. The change of basis matrix from $C$-coordinates to $B$-coordinates is the matrix $P_{B \leftarrow C} \in k^{n \times n}$ whose $j$-th column is $[c_j]_B$. [/definition] The notation $P_{B \leftarrow C}$ is directional because coordinate vectors are not the vectors themselves: the same vector has different coordinate records in the bases $B$ and $C$. The next question is operational: once the columns of $P_{B \leftarrow C}$ have been built from the $C$-basis vectors, what does this matrix do to an arbitrary coordinate column? With the convention used in this page, multiplication by $P_{B \leftarrow C}$ sends $[v]_C$ to $[v]_B$. Some sources name the same direction differently. In particular, the quoted theorem below uses arrow notation $P_{B \to C}$ for the matrix that translates from $B$-coordinates to $C$-coordinates. Under this page's notation, that matrix is $P_{C \leftarrow B}$. Thus the theorem should be read as the same coordinate-translation principle with the source and target bases indicated by the surrounding convention, not as a contradiction of the definition above. [quotetheorem:3275] [example: Changing Coordinates in $\mathbb{R}^2$] Let $B=(e_1,e_2)$ be the standard ordered basis of $\mathbb{R}^2$, so $e_1=(1,0)$ and $e_2=(0,1)$. Let $C=(c_1,c_2)$ with $c_1=(1,1)$ and $c_2=(1,-1)$. Since $P_{B \leftarrow C}$ is defined by using $[c_1]_B$ and $[c_2]_B$ as its columns, we first write each $C$-basis vector in the basis $B$: \begin{align*} c_1=(1,1)=1e_1+1e_2. \end{align*} \begin{align*} c_2=(1,-1)=1e_1-1e_2. \end{align*} Thus $[c_1]_B=(1,1)$ and $[c_2]_B=(1,-1)$, so $P_{B \leftarrow C}$ is the matrix with first column $(1,1)$ and second column $(1,-1)$. For $v=(3,1)$, the earlier computation gave $[v]_C=(2,1)$, meaning \begin{align*} v=2c_1+1c_2. \end{align*} Expanding this expression in standard coordinates gives \begin{align*} 2c_1+1c_2=2(1,1)+1(1,-1). \end{align*} \begin{align*} 2(1,1)+1(1,-1)=(2,2)+(1,-1). \end{align*} \begin{align*} (2,2)+(1,-1)=(3,1). \end{align*} The same calculation is encoded by the change of basis matrix: \begin{align*} P_{B \leftarrow C}[v]_C=2(1,1)+1(1,-1). \end{align*} \begin{align*} 2(1,1)+1(1,-1)=(3,1). \end{align*} Therefore $[v]_B=(3,1)$. The matrix $P_{B \leftarrow C}$ translates the $C$-coordinate record $(2,1)$ back into the standard-coordinate record $(3,1)$ for the same vector. [/example] ### Similarity The matrix of a linear operator changes when the basis changes, but the underlying operator remains the same. Here $\operatorname{End}(V)$ denotes the set of linear maps $V \to V$, and $\operatorname{GL}_n(\mathbb{F})$ denotes the group of invertible $n \times n$ matrices over the field $\mathbb{F}$. To compare two matrices that describe the same operator, we need the change of basis matrices on both sides of the operator matrix. [quotetheorem:400] The inverse appears because changing the input coordinates and changing the output coordinates go in opposite directions: one translation moves a coordinate vector into the basis where the operator matrix acts, and the other translates the result back. Thus similarity does not say that the entries are unchanged; it says that the two matrices encode the same linear transformation in different coordinate languages. Quantities depending only on the operator, such as invertibility and the characteristic polynomial, are therefore candidates for similarity invariants. This is why similarity becomes the organizing relation for studying operators by choosing a convenient basis. ## Bases Beyond Vector Spaces ### Free Modules This subsection is a brief comparison, not a full introduction to module theory. A ring $R$ is like a field except that nonzero elements need not have multiplicative inverses; a left $R$-module is an additive abelian group whose elements can be multiplied on the left by elements of $R$. A free $R$-module is a module with a basis in the module sense: every element is written uniquely as a finite $R$-linear combination of chosen basis elements. An ideal is an additive subgroup of a ring closed under multiplication by ring elements, and quotient notation such as $R/I$ means that elements differing by an element of the ideal $I$ are identified. Invariant basis number means that if $R^m \cong R^n$ as free $R$-modules, then $m=n$. When algebras appear below, an algebra is a vector space with a compatible multiplication. A Lie algebra is an algebra whose product is written as a bracket $[x,y]$ and measures infinitesimal symmetry. The operator notation introduced above is reused in this broader setting. These notions are included only to show how the idea of a basis survives in richer settings; the main model remains ordinary vector-space coordinates. The vector space story is tidy because fields allow division by nonzero scalars. In module theory over a general ring, the same words become more delicate. A module may have generators without having a basis, and a linearly independent generating set may fail to exist. [definition: Free Module] Let $R$ be a ring. A left $R$-module $M$ is free if there exists a subset $B \subset M$ such that every $m \in M$ has a unique expression \begin{align*} m = \sum_{b \in B} r_b b, \end{align*} where $(r_b)_{b \in B}$ is a finitely supported family of elements of $R$. [/definition] Freeness asserts that some coordinate system exists, but many arguments need to refer to the particular set of coordinate directions being used. That set must encode both generation and uniqueness, since generators alone can have relations over a ring. [definition: Module Basis] Let $R$ be a ring, and let $M$ be a left $R$-module. A module basis of $M$ is a subset $B \subset M$ such that every $m \in M$ has a unique expansion \begin{align*} m = \sum_{b \in B} r_b b, \end{align*} where $(r_b)_{b \in B}$ is a finitely supported family of elements of $R$. [/definition] [example: A Module with No Basis] Consider the abelian group $\mathbb{Z}/2\mathbb{Z}$ as a $\mathbb{Z}$-module. It is generated by $\bar{1}$ because the only two elements are reached by integer multiples of $\bar{1}$: \begin{align*} 0\cdot \bar{1}=\bar{0} \end{align*} and \begin{align*} 1\cdot \bar{1}=\bar{1}. \end{align*} We show that this module cannot have a $\mathbb{Z}$-module basis. Suppose a candidate module basis contains a nonzero element $b$. Since the only nonzero element of $\mathbb{Z}/2\mathbb{Z}$ is $\bar{1}$, we have $b=\bar{1}$. Then \begin{align*} 2b=2\bar{1}=\bar{2}=\bar{0}. \end{align*} Thus the zero element has the expansion with all coefficients equal to $0$, but it also has the expansion using coefficient $2$ on $b$. These coefficient choices are different in $\mathbb{Z}$ because $2\ne 0$, so uniqueness of coefficients fails. Therefore a module basis cannot contain any nonzero element. If every chosen element is $\bar{0}$, then every integer multiple of every chosen element is still $\bar{0}$: \begin{align*} n\bar{0}=\bar{0} \end{align*} for every $n\in\mathbb{Z}$. Such a set spans only $\{\bar{0}\}$, not all of $\mathbb{Z}/2\mathbb{Z}$. Hence $\mathbb{Z}/2\mathbb{Z}$ has no $\mathbb{Z}$-module basis, even though it is generated by one element. [/example] Because modules may fail to be free, freeness is already meaningful information. When a module is free, the next structural question is how many coordinate directions its basis contains. This leads to rank. [definition: Rank of a Free Module] Let $R$ be a ring for which free modules have invariant basis number. If $M$ is a free left $R$-module with a finite module basis $B$, the rank of $M$ is \begin{align*} \operatorname{rank}_R M = |B|. \end{align*} [/definition] Rank counts basis elements, but coordinates should also give an explicit model for the module. For a finite free module, the remaining question is whether choosing a basis turns the module into the standard module of coordinate columns, with no extra relations hidden in the ring action. [quotetheorem:7835] [example: Generators of an Ideal Need Not Be a Basis] Let $R=k[x,y]$ for a field $k$, and consider the ideal $I=(x,y)$ as an $R$-module. By definition, \begin{align*} I=(x,y)=\{rx+sy:r,s\in R\}, \end{align*} so every element of $I$ is an $R$-linear combination of $x$ and $y$. Thus $\{x,y\}$ generates $I$ as an $R$-module. However, the generators do not have unique coefficients. Using the coefficient pair $(y,-x)$ gives \begin{align*} y\cdot x+(-x)\cdot y=yx-xy. \end{align*} Since multiplication in $k[x,y]$ is commutative, $yx=xy$, so \begin{align*} yx-xy=xy-xy=0. \end{align*} The zero element also has the coefficient pair $(0,0)$: \begin{align*} 0\cdot x+0\cdot y=0+0=0. \end{align*} These two coefficient pairs are different because $y\ne 0$ and $-x\ne 0$ in the polynomial ring $k[x,y]$. Hence the same element $0\in I$ has two different expressions in terms of $x$ and $y$, so $\{x,y\}$ is not a module basis of $I$. This only proves that this generating set is not a basis; proving that $I$ is not free as an $R$-module requires a separate argument. [/example] ### Algebra Bases and Structure Constants The language of bases also appears in algebras over fields and in Lie algebras. There the underlying vector space still has a basis, but the basis may be chosen to make multiplication or brackets readable. [definition: Algebra Basis] Let $A$ be an algebra over a field $k$. An algebra basis of $A$ is a basis of the underlying vector space of $A$ over $k$. [/definition] The word algebra basis does not mean that products of basis elements remain basis elements. To compute multiplication from a basis, we need to record how every product expands back into that same basis. [definition: Structure Constants] Let $V$ be a finite-dimensional vector space over a field $k$, let $B=(b_1,\ldots,b_n)$ be an ordered basis of $V$, and let $\mu: V \times V \to V$ be a bilinear map. The structure constants of $\mu$ relative to $B$ are the scalars $c_{ijr}\in k$ defined by \begin{align*} \mu(b_i,b_j) = \sum_{r=1}^n c_{ijr}b_r \end{align*} for $1 \le i,j \le n$. [/definition] [example: Structure Constants for $\mathfrak{sl}(2)$] Let $\mathfrak{sl}(2)$ be the Lie algebra over $\mathbb{C}$ with basis $e,f,h$, where $e_{12}=1$, $f_{21}=1$, $h_{11}=1$, $h_{22}=-1$, and all unspecified matrix entries are $0$. We compute the Lie bracket $\mu(X,Y)=[X,Y]=XY-YX$ in this basis. For $[h,e]$, the only possibly nonzero entry of $he$ in the position of $e$ is \begin{align*} (he)_{12}=h_{11}e_{12}+h_{12}e_{22}=1\cdot 1+0\cdot 0=1. \end{align*} All other entries of $he$ are $0$, so $he=e$. Similarly, \begin{align*} (eh)_{12}=e_{11}h_{12}+e_{12}h_{22}=0\cdot 0+1\cdot(-1)=-1. \end{align*} All other entries of $eh$ are $0$, so $eh=-e$. Therefore \begin{align*} [h,e]=he-eh=e-(-e)=2e. \end{align*} For $[h,f]$, the only possibly nonzero entry of $hf$ in the position of $f$ is \begin{align*} (hf)_{21}=h_{21}f_{11}+h_{22}f_{21}=0\cdot 0+(-1)\cdot 1=-1. \end{align*} Thus $hf=-f$. Also \begin{align*} (fh)_{21}=f_{21}h_{11}+f_{22}h_{21}=1\cdot 1+0\cdot 0=1. \end{align*} Thus $fh=f$, and hence \begin{align*} [h,f]=hf-fh=(-f)-f=-2f. \end{align*} For $[e,f]$, the product $ef$ has \begin{align*} (ef)_{11}=e_{11}f_{11}+e_{12}f_{21}=0\cdot 0+1\cdot 1=1, \end{align*} and its other entries are $0$. The product $fe$ has \begin{align*} (fe)_{22}=f_{21}e_{12}+f_{22}e_{22}=1\cdot 1+0\cdot 0=1, \end{align*} and its other entries are $0$. Therefore $ef-fe$ has diagonal entries $1$ and $-1$, with off-diagonal entries $0$, so \begin{align*} [e,f]=ef-fe=h. \end{align*} Relative to the ordered basis $(e,f,h)$, these identities say that the nonzero structure constants include $c_{311}=2$, $c_{322}=-2$, and $c_{123}=1$, together with the constants forced by skew-symmetry of the Lie bracket. The basis gives coordinates for the vector space and turns the bracket into explicit coefficient data. [/example] ## Infinite Bases and Choice ### Hamel Bases Finite bases are familiar because they can be found by row reduction or by repeatedly adding and removing vectors. Infinite-dimensional spaces require a different level of set-theoretic support. The definition of basis is still algebraic and finite in each expansion, but the basis itself may be infinite. [definition: Hamel Basis] Let $V$ be a vector space over a field $k$. A Hamel basis of $V$ is a basis of $V$ in the algebraic sense: every vector of $V$ has a unique finite linear combination in terms of the basis. [/definition] [example: Polynomials Versus Power Series] Let $k$ be a field, and consider the monomials \begin{align*} 1,x,x^2,x^3,\ldots \end{align*} in the polynomial ring $k[x]$. Every polynomial $p\in k[x]$ has the form \begin{align*} p=a_0+a_1x+\cdots+a_dx^d \end{align*} for some $d\ge 0$ and scalars $a_0,\ldots,a_d\in k$, so $p$ is a finite linear combination of the monomials. If \begin{align*} a_0+a_1x+\cdots+a_dx^d=0 \end{align*} as a polynomial, then equality with the zero polynomial means each coefficient is zero: \begin{align*} a_0=a_1=\cdots=a_d=0. \end{align*} Thus the monomials span $k[x]$ and are linearly independent, so they form a Hamel basis of $k[x]$ over $k$. The same monomials do not form a Hamel basis of the formal power series ring $k[[x]]$. The power series \begin{align*} s=1+x+x^2+x^3+\cdots \end{align*} has coefficient $1$ on every power $x^i$. If $s$ were a finite linear combination of the monomials, then for some $N\ge 0$ there would be scalars $a_0,\ldots,a_N\in k$ such that \begin{align*} s=a_0+a_1x+\cdots+a_Nx^N. \end{align*} The coefficient of $x^{N+1}$ on the right is $0$, because no $x^{N+1}$ term occurs there. The coefficient of $x^{N+1}$ in $s$ is $1$. Since $1\ne 0$ in a field, the two formal power series are not equal. Hence $s$ is not in the span of the monomials by finite linear combinations. Algebraic bases allow only finite expansions; under the axiom of choice, $k[[x]]$ still has some Hamel basis over $k$, but the monomials alone are not that basis. [/example] ### Existence and Dimension The existence of bases in full generality is not proved by row reduction. It requires a maximality argument, usually Zorn’s lemma, and is tied to the axiom of choice. [quotetheorem:1229] For infinite-dimensional spaces, dimension should still measure the number of coordinate directions, but the number may now be an infinite cardinal rather than an integer. Here a cardinal is the size of a set, so two indexing sets with the same cardinality are being treated as having the same number of basis vectors even when that number is infinite. Since different infinite bases may look very different, the definition records the cardinality of a basis and relies on the standard vector-space fact that all bases of a vector space have the same cardinality. [definition: Infinite Dimension] Let $V$ be a vector space over a field $k$, and let $B$ be a basis of $V$. If $B$ is infinite, the dimension of $V$ is the cardinality \begin{align*} \dim_k V = \operatorname{card}(B). \end{align*} [/definition] The invariant here is still a vector-space invariant: it is not about the appearance of a particular basis, but about the number of independent coordinate directions needed to span the space. The proof of cardinal invariance for arbitrary bases uses set-theoretic choice principles beyond the finite exchange arguments above, so this page uses the result as background rather than developing that machinery. For example, the polynomial ring $k[x]$ as a vector space over $k$ has countable basis $1,x,x^2,\ldots$, and any other vector-space basis of $k[x]$ has the same cardinality. ## Bases as Presentations of Structure A basis is often the beginning rather than the end of a classification problem. Once a basis is chosen, extra structure becomes a table of coefficients, a matrix, or a family of relations. The quality of the basis determines how readable that structure becomes. When a subspace is important, a random basis may hide it. An adapted basis is chosen so that the subspace occupies the first coordinate directions. [definition: Adapted Basis] Let $V$ be a finite-dimensional vector space over a field $k$, and let $U \subset V$ be a subspace. An ordered basis $(b_1,\ldots,b_n)$ of $V$ is adapted to $U$ if there exists $r$ with $0 \le r \le n$ such that $(b_1,\ldots,b_r)$ is an ordered basis of $U$. [/definition] Adapted bases make subspaces visible inside coordinates by turning an inclusion $U \subset V$ into the standard inclusion of the first $r$ coordinates. The possible obstruction is that a basis chosen for $U$ might not extend to a basis of all of $V$, so the coordinate picture would depend on an unavailable choice. [quotetheorem:7836] [example: Kernel-Adapted Basis] Let $T:V\to W$ be a linear map of finite-dimensional vector spaces over a field $k$, and suppose $\ker(T)$ has basis $(u_1,\ldots,u_r)$. Choose an ordered basis \begin{align*} B=(u_1,\ldots,u_r,v_{r+1},\ldots,v_n) \end{align*} of $V$ extending this kernel basis, and choose an ordered basis $C=(c_1,\ldots,c_m)$ of $W$. For each $1\le i\le r$, the vector $u_i$ lies in $\ker(T)$, so by the definition of kernel, \begin{align*} T(u_i)=0_W. \end{align*} The coordinate vector of $0_W$ in the basis $C$ is \begin{align*} [0_W]_C=(0,\ldots,0), \end{align*} because \begin{align*} 0_W=0c_1+\cdots+0c_m. \end{align*} Therefore \begin{align*} [T(u_i)]_C=(0,\ldots,0) \end{align*} for every $1\le i\le r$. By the definition of the matrix of a linear map, the $j$-th column of $[T]_{C\leftarrow B}$ is the coordinate vector of the image of the $j$-th basis vector of $B$. Hence the first $r$ columns are \begin{align*} [T(u_1)]_C,\ldots,[T(u_r)]_C, \end{align*} and each of these columns is the zero column. For the remaining basis vectors, the columns are \begin{align*} [T(v_{r+1})]_C,\ldots,[T(v_n)]_C. \end{align*} Thus the matrix of $T$ in the basis $B$ has the block form \begin{align*} [T]_{C\leftarrow B}=\bigl(0\ \cdots\ 0\ [T(v_{r+1})]_C\ \cdots\ [T(v_n)]_C\bigr), \end{align*} where the first $r$ columns are zero. If \begin{align*} x=a_1u_1+\cdots+a_ru_r+a_{r+1}v_{r+1}+\cdots+a_nv_n, \end{align*} then linearity gives \begin{align*} T(x)=a_1T(u_1)+\cdots+a_rT(u_r)+a_{r+1}T(v_{r+1})+\cdots+a_nT(v_n). \end{align*} Since $T(u_i)=0_W$ for $1\le i\le r$, this becomes \begin{align*} T(x)=a_{r+1}T(v_{r+1})+\cdots+a_nT(v_n). \end{align*} So the coordinates along the kernel basis do not affect the value of $T(x)$; the nonzero part of the matrix records exactly how $T$ acts on the chosen directions complementary to the kernel. [/example] ## Beyond and Connected Topics The first continuation is finite-dimensional linear algebra. There, bases support determinants, eigenvalues, diagonalisation, Jordan normal form, dual spaces, and bilinear forms. The page [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra) is the natural course-level reference for seeing bases used systematically in matrix and vector space arguments. The second continuation is commutative algebra. Vector spaces always have bases, but modules over rings often do not. This failure motivates free modules, projective modules, localisations, presentations, and exact sequences. The page [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra) develops the algebraic setting where basis-like behavior becomes a serious property rather than a default guarantee. The third continuation is homological algebra. When a module has no basis, one often replaces it by a resolution built from free modules. The bases of those free modules make maps explicit, while exactness records how the replacement remembers the original object. See [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions) for the systematic version of this idea. The fourth continuation is Lie theory. A Lie algebra is a vector space with a bracket, so a basis gives coordinates, while structure constants record the bracket. In semisimple Lie theory, carefully chosen bases reflect roots and weights. The page [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations) is the natural next reference for this direction. ## 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 I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations). Sheldon Axler, *Linear Algebra Done Right* (2015). Serge Lang, *Algebra* (2002). David Eisenbud, *Commutative Algebra with a View Toward Algebraic Geometry* (1995).