Proof that finite-dimensional spaces are isomorphic iff they have equal dimension

Proof

We show that an isomorphism sends a basis to a basis, and that a bijection between bases extends to an isomorphism.

1. Suppose $V$ and $W$ are isomorphic, and let $T \colon V \to W$ be a linear isomorphism. Let $n = \dim V$, the dimension, and let $(b_1, \dots, b_n)$ be a basis of $V$.
2. The family $(T(b_1), \dots, T(b_n))$ is linearly independent. Suppose $\sum_i a_i T(b_i) = 0$ with $a_i \in F$. Then $T\left(\sum_i a_i b_i\right) = 0$ by linearity.
3. $T$ is injective, so its kernel is $\{0\}$ by the kernel criterion. Hence $\sum_i a_i b_i = 0$, and all $a_i = 0$ because $(b_i)$ is linearly independent.
4. The family $(T(b_1), \dots, T(b_n))$ spans $W$. Let $w \in W$. Since $T$ is surjective, $w = T(v)$ for some $v = \sum_i a_i b_i$, and then $w = \sum_i a_i T(b_i)$.
5. So $(T(b_1), \dots, T(b_n))$ is a basis of $W$ with $n$ vectors. By the dimension theorem, $\dim W = n = \dim V$.
6. Conversely, suppose $\dim V = \dim W = n$. Let $(b_1, \dots, b_n)$ be a basis of $V$ and $(c_1, \dots, c_n)$ a basis of $W$.
7. By the extension of values on a basis, there is a linear map $S \colon V \to W$ with $S(b_i) = c_i$ for all $i$. There is also a linear map $R \colon W \to V$ with $R(c_i) = b_i$ for all $i$.
8. The composite $R \circ S \colon V \to V$ is linear, since $R(S(u + av)) = R(S(u)) + a R(S(v))$ for $u, v \in V$ and $a \in F$. It sends each $b_i$ to $R(c_i) = b_i$.
9. The identity map of $V$ also sends each $b_i$ to $b_i$. By the uniqueness part of the same result, $R \circ S = \mathrm{id}_V$.
10. In the same way, $S \circ R \colon W \to W$ fixes each $c_i$, so $S \circ R = \mathrm{id}_W$.
11. By the two-sided inverse criterion, $S$ is bijective. So $S$ is a linear isomorphism, and $V$ and $W$ are isomorphic.