Vector spaces of the same dimension are isomorphic, but is the converse true? If two vector spaces $V$ and $W$ are isomorphic, do they always have the same dimension?
In other words: $dim V = dimW $ is a sufficient condition for isomorphism, is it also necessary ?
 A: The answer is yes.
The key point is the following:

Lemma: if $f:V\rightarrow W$ is a linear map and $B$ spans $V$, then $f[B]$ spans $f[V]$. Here $f[B]=\{f(b): b\in B\}$ and $f[V]=\{f(v): v\in V\}$; in particular, $f[V]$ is the image of $f$.

Proof: Fix $w\in f[V]$. Since $w\in f[V]$ there is some $v\in V$ with $f(v)=w$ (there may be more than one, since $f$ might not be injective, but that's fine). Since $B$ spans $V$ there are scalars $s_1,..., s_n$ and vectors $b_1,...,b_n\in B$ such that $$s_1b_1 + ... + s_nb_n=v.$$ But then by linearity of $f$ we have $$s_1f(b_1)+...+s_nf(b_n)=f(v)=w.$$ So $w$ is a linear combination of elements of $f[B]$. $\quad\Box$
Now we can prove the desired result. If $f:V\rightarrow W$ is an isomorphism, let $B$ be a basis for $V$. Then:


*

*$f[B]=\{f(b): b\in B\}$ is a spanning set for $W$, by the above, and since $f$ is injective we have $\vert f[B]\vert=\vert B\vert$.

*If $f[B]$ is linearly independent in $W$, then $f[B]$ is a basis for $W$, and so $dim(V)=dim(W)$ since they have bases of the same cardinality.

*So suppose $f[B]$ is not linearly independent in $W$. Then we can find a $C\subsetneq B$ which spans $W$.

*But then - since $f^{-1}$ is also a bijective linear map - we have that $f^{-1}[C]=\{f^{-1}(c): c\in C\}$ spans $V$ and is a proper subset of $B$. Since no proper superset of a spanning set can be linearly independent, this contradicts the fact that $B$ was a basis for $V$.

There's a subtle issue in the above, around the existence of bases. It's easy to show that if $V$ has a finite spanning set then $V$ has a basis and any two bases of $V$ have the same cardinality, but the situation is more complicated with respect to general vector spaces: it's consistent with ZF (= set theory without the axiom of choice) that $(1)$ there are vector spaces which have no bases at all and $(2)$ there are vector spaces with bases of different cardinalities! So without AC, we might have a vector space with no dimension or multiple dimensions.
When we look carefully at the above argument, the "choice-free" theorem is the following:

Suppose $f:V\rightarrow W$ is an isomorphism of vector spaces. Then letting "$Bases(U)$" be the set of bases of $U$, the map $$F: Bases(V)\rightarrow Bases(W): B\mapsto f[B]$$ is a bijection and for all $B\in Bases(V)$ we have $\vert B\vert=\vert F(B)\vert$.

