Finite-dimensional vector space, and its dimension $\dim_F V$; infinite-dimensional means having no finite basis

Let $V$ be a vector space over a field $F$ (Vector space over a field).

$V$ is finite-dimensional over $F$ when it has a finite basis (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Finite, countably infinite, countable, uncountable): some basis $B$ of $V$ satisfies $B \approx n$ for some $n \in \mathbb{N}$ (Equinumerous sets, $A \approx B$ and $A \preceq B$).

For such a $V$, the dimension of $V$ over $F$, written $\dim_F V$, is that $n$:

$\dim_F V \;:=\; \text{the unique } n \in \mathbb{N} \text{ such that } V \text{ has a basis } B \text{ with } B \approx n .$

This is well defined. Existence of such an $n$ is the hypothesis, together with the fact that a finite set is equinumerous with exactly one natural number (The pigeonhole principle on $\mathbb{N}$, claim 3). Uniqueness is If $V$ has a basis with $n$ elements and a basis with $m$ elements then $n = m$; and if $V$ has one finite basis then every basis of $V$ is finite: two bases of $V$ with $n$ and with $m$ elements force $n = m$. That theorem is therefore a prerequisite of this definition, not a later justification of it, and it is listed in deps.

$V$ is infinite-dimensional over $F$ when it is not finite-dimensional over $F$, that is, when $V$ has no finite basis. No number is attached to such a space here: the symbol $\dim_F V$ is defined only in the finite-dimensional case, and the expression $\dim_F V = \infty$ is not used.

The zero space. $\varnothing$ is a basis of $\{0_V\}$ (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis) and $\varnothing \approx 0$, so $\{0_V\}$ is finite-dimensional with $\dim_F \{0_V\} = 0$. Conversely a space of dimension $0$ has a basis $B \approx 0$, that is $B = \varnothing$, and then $V = \operatorname{span}(\varnothing) = \{0_V\}$ (Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$).

• The subscript $F$ is not ornamental. By A field is a vector space over itself, and over any subfield $K \subseteq F$ every $F$-vector space is a $K$-vector space by restricting the scalars the same set with the same addition is a vector space over any subfield $K \subseteq F$ (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations), and for a proper subfield the two structures can have different bases and different dimensions. The companion page's basis of $\mathbb{R}$ over $\mathbb{Q}$ is the extreme case: $\mathbb{R}$ is a vector space both over itself and over the embedded copy of $\mathbb{Q}$ inside it, and it is infinite-dimensional over the latter. So "the dimension of $V$" is incomplete language in exactly the way that "the vector space $V$" is, and both the space and the field are part of the statement of every result below.

• Infinite-dimensional is defined as a negation, deliberately. Assigning a size to an infinite basis would require knowing that any two infinite bases of a space are equinumerous, which If $V$ has a basis with $n$ elements and a basis with $m$ elements then $n = m$; and if $V$ has one finite basis then every basis of $V$ is finite does not prove and this page does not claim; the standard argument for it is cardinal arithmetic, developed much later in the library. The companion page therefore records a proper subspace with an equinumerous basis rather than any statement of the form $\dim U = \dim V$ for infinite-dimensional spaces.

• Dimension counts a basis, not a spanning set and not an independent set. A spanning set may be larger than $\dim_F V$ and an independent set smaller; If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $\mathbb{N}$ is what bounds the second by the first, and For $B \subseteq V$ the following are equivalent: $B$ is a basis; $B$ is a maximal linearly independent subset of $V$; $B$ is a minimal spanning subset of $V$ — maximality and minimality being in the inclusion order is what says a basis is exactly where the two meet.