A vector space supplies addition and scalar multiplication, but it does not automatically supply coordinates. The central problem of finite-dimensional linear algebra is to decide when every vector can be described by a finite list of scalars, in a way that respects addition, scalar multiplication, subspaces, and linear maps.

The failure appears as soon as we leave Euclidean space. Polynomial spaces and function spaces obey the axioms of a [vector space](/page/Vector%20Space), but most of them cannot be controlled by finitely many vectors. Finite-dimensional vector spaces are the part of linear algebra where abstract vectors can be replaced by finite coordinate data.

[example: Polynomial Space Is Not Controlled by Finitely Many Polynomials] Let $k$ be a field and let $k[x]$ be the vector space of polynomials with coefficients in $k$. Take any finite list $p_1,\ldots,p_m \in k[x]$. If every $p_i$ is the zero polynomial, then every linear combination $a_1p_1+\cdots+a_mp_m$ is $0$, so $x$ is not in its span. Otherwise, let $d$ be the maximum degree among the nonzero polynomials in the list. For each $i$, write \begin{align*} p_i=c_{i0}+c_{i1}x+\cdots+c_{id}x^d \end{align*} where $c_{ir}=0$ when the actual degree of $p_i$ is less than $r$. Then for scalars $a_1,\ldots,a_m \in k$, \begin{align*} a_1p_1+\cdots+a_mp_m=a_1(c_{10}+c_{11}x+\cdots+c_{1d}x^d)+\cdots+a_m(c_{m0}+c_{m1}x+\cdots+c_{md}x^d). \end{align*} Collecting coefficients of equal powers of $x$ gives \begin{align*} a_1p_1+\cdots+a_mp_m=(a_1c_{10}+\cdots+a_mc_{m0})+(a_1c_{11}+\cdots+a_mc_{m1})x+\cdots+(a_1c_{1d}+\cdots+a_mc_{md})x^d. \end{align*} This polynomial has no $x^{d+1}$ term, so it cannot equal $x^{d+1}$, whose coefficient of $x^{d+1}$ is $1$. Therefore no finite list of polynomials spans $k[x]$, even though $k[x]$ is a vector space. [/example]

This example explains why finite-dimensionality is not a decorative adjective. It separates vector spaces where linear algebra is governed by finite systems of scalar equations from vector spaces where additional structure is needed.

## Definition

The relevant finiteness condition is not that the vector space has finitely many elements. If $k$ is infinite, even the line $k$ has infinitely many vectors. What must be finite is the amount of data needed to produce every vector by linear combination.

[definition: Finite-Dimensional Vector Space] Let $k$ be a field and let $V$ be a vector space over $k$. The vector space $V$ is finite-dimensional if there exists a finite subset $S \subset V$ such that every $v \in V$ is a linear combination of elements of $S$. [/definition]

The zero vector space is finite-dimensional, generated by the empty set. This convention matters because kernels, intersections, and quotient constructions can naturally produce the zero space.

Finite-dimensionality depends on producing vectors from a finite list, but that phrase is meaningless until the allowed algebraic recipe is fixed. We need a precise name for the finite scalar recipes that are allowed to build new vectors from chosen ones; no limits, topology, or infinite sums are being used.

[definition: Linear Combination] Let $k$ be a field, let $V$ be a vector space over $k$, and let $v_1, \ldots, v_m \in V$. A linear combination of $v_1, \ldots, v_m$ is a vector of the form $a_1v_1+\cdots+a_mv_m$, where $a_1, \ldots, a_m \in k$. [/definition]

Once a set of candidate vectors has been chosen, we need to separate the vectors it actually produces from the vectors that may still be unreachable. This reachable part should itself be a subset of the ambient space, so generation can later be expressed as equality with the whole space.

[definition: Span] Let $k$ be a field, let $V$ be a vector space over $k$, and let $S \subset V$. The span of $S$ is the subset $\operatorname{span}(S) \subset V$ consisting of all finite linear combinations of elements of $S$. For $S=\varnothing$, set $\operatorname{span}(S)=\{0\}$. [/definition]

Span tells us what a proposed list can reach, but it does not say whether the list has repeated information. We next name lists that reach everything, because finite-dimensionality begins with the existence of such a finite reaching set.

[definition: Generating Set] Let $k$ be a field and let $V$ be a vector space over $k$. A subset $S \subset V$ is a generating set for $V$ if $\operatorname{span}(S)=V$. A finite generating set is a generating set with finitely many elements. [/definition]

A generating set can be too large, as happens in $k^2$ with $(1,0)$, $(0,1)$, and $(1,1)$. The defect is that one vector can be recovered from the others, so the same vector may have competing recipes. To make coordinates unique, we must forbid nontrivial scalar relations among the chosen vectors.

[definition: Linearly Independent Set] Let $k$ be a field, let $V$ be a vector space over $k$, and let $S \subset V$. The subset $S$ is linearly independent if every equality $a_1v_1+\cdots+a_mv_m=0$, with $v_1, \ldots, v_m$ distinct elements of $S$ and $a_1, \ldots, a_m \in k$, forces $a_1=\cdots=a_m=0$. [/definition]

Spanning alone guarantees that every vector has at least one recipe, while independence prevents hidden redundancy in those recipes. A usable coordinate system needs both properties at once: every vector must be reachable, and its scalar expansion must not depend on arbitrary choices.

[definition: Basis] Let $k$ be a field and let $V$ be a vector space over $k$. A basis of $V$ is a linearly independent generating set for $V$. [/definition]

A basis should let us write each vector by coordinates, but this has to be justified from the two halves of the definition. The next theorem is the reason a basis is more than a convenient spanning list: it gives both existence and uniqueness of scalar expansion.

[quotetheorem:372]