 Solution

Let $u_1,\dots ,u_m$ and $v_1,\dots ,v_n$ be two finite bases of the vector space $V$. We prove the underlying exchange argument rather than assuming that dimension is already well-defined.

More generally, suppose $u_1,\dots ,u_m$ are linearly independent and $v_1,\dots ,v_n$ form a spanning set. Starting with the spanning list of $v$'s, replace one of its members by $u_1$. Indeed, express $u_1$ as a linear combination of that list; at least one coefficient is nonzero, so we can solve for its corresponding $v_j$ and retain a spanning list. Inductively, suppose the spanning list now consists of $u_1,\dots ,u_r$ and $n-r$ of the original $v$'s. Express $u_{r+1}$ in that list. Some coefficient of a remaining $v$ must be nonzero, since otherwise $u_{r+1}$ would lie in the span of the preceding $u$'s, contradicting linear independence. Solve for that $v$ and replace it by $u_{r+1}$.

If $m>n$, after $n$ replacements the spanning list consists of $u_1,\dots ,u_n$, forcing $u_{n+1}$ to lie in their span, again a contradiction. Thus every finite linearly independent list has at most as many members as every finite spanning set. This proves the Steinitz exchange lemma in the form required here. Applying it to the two bases in both orders gives $m\le n$ and $n\le m$, hence $m=n$. The zero vector space has only the empty basis, which also fits the argument.