If $V,W$ are finite-dimensional and $T: V \to W$ , $S: W \to V$ are injective linear maps then $T$ is an isomorphism Question: If $V,W$ are finite-dimensional and $T: V \to W$ , $S: W \to V$ are injective linear maps then $T$ is an isomorphism.
We know that $V,W$ are finite-dimensional vector spaces.
$\implies \dim(Ker(T)) + \dim(Im(T)) = \dim(V)$ and $\dim(Ker(S)) + \dim(Im(S)) = \dim(W)$.
Since we also know that that $T,S$ are both injective linear maps.
This further implies the following:
$\dim(Im(T)) = \dim(V)$ and $\dim(Im(S)) = \dim(W)$.
$\implies Im(T) = V$ and $Im(S) = W$.
Also, we know that $S$ is surjective if $\dim(V) \leq \dim(W).$ Thus, we also know that $S$ is surjective, i.e. $Im(S) = V$.
This means we have $V=W=Im(S)$.
So we can re-write an equivalent definition of $T$ as follows: $T: V \to W \iff T:Im(S) \to Im(S).$ 
Since, the dimension of domain and codomain is the same for $T$, implies that $T$ is injective if and only if $T$ is surjective. Thus, $T$ is surjective, which means $T$ is an isomorphism. 
Confirmation: Please correct me if anything has gone wrong in my proof! Please do not suggest an alternative way of proving this until after where my problems (unless my proof is correct) are advised and fixed in MY WAY of proof. Thank you in advance.
 A: There are some issues to fix, but you largely have the right idea. Rank-nullity theorem will be central to this proof.

This further implies the following:
$\dim(Im(T)) = \dim(V)$ and $\dim(Im(S)) = \dim(W)$.
$\implies Im(T) = V$ and $Im(S) = W$.

As pointed out in the comments above, $Im(T)\subseteq W,$ so there's nothing in the hypotheses that would allow us to conclude that $Im(T)=V.$ Similarly, we can't conclude that $Im(S)=W.$

Also, we know that $S$ is surjective if $\dim(V) \leq \dim(W).$

True! And a very useful observation!

Thus, we also know that $S$ is surjective, i.e. $Im(S) = V$.

Unfortunately, this (and all the rest) is the fruit of the poisoned tree. That is, it's true (by the Principle of Explosion), as long as you assume the false statement that you assumed (without realizing it) above.

For ideas on how to fix your proof, hover over the spoiler box below.

Now, instead, we should consider $ST:V\to V.$ Since $S$ and $T$ are both injective, then it can readily be shown that $ST$ is, too. As a result, by rank-nullity, we have that $\dim(V)=\dim(Im(ST)).$ Since $V$ is finite-dimensional and $Im(ST)$ is a subspace of $V,$ then we have that $Im(ST)=V$. Since $Im(ST)=S\bigl[T[V]\bigr]$ is a subspace of $S[W],$ which is in turn a subspace of $V,$ then we're done.

