How to prove the following about linear transformations, composition and equal transformation dimension?

Question

Consider the following question: Let $V$ and $W$ be finite-dimensional vector spaces. Suppose $T: V \rightarrow W$ and $S: W \rightarrow V$ linear maps such that $T \circ S = id_W$. As well, suppose $\dim V = \dim W$. Prove that $S \circ T = id_V$

I'm a bit stuck on this proof. This is what I've done so far.

Take, $w \in W, v \in V$. Since, $T \circ S = id_W$, $T \circ S(w) = id_w \cdot w = w$.

I'm not entirely sure now how to use the fact that $\dim V = \dim W$. I'd like to somehow connect it to injectivity. Then say that since the linear transformations are injective then we have the same mapping to the identity under the composition above. But I'm missing a step to reach this conclusion, or there may be a better way I'm not seeing. Any advice to prove this would be much appreciated!

Answer 1

Let us be a bit more general for a moment. Suppose $V$, $W$ are finite-dimensional vector spaces with $\dim V = \dim W$.
Let $T : V \to W$ be a linear transformation.

The following are equivalent:

1. $T$ is a bijection.
2. $T$ is one-to-one.
3. $T$ is onto.

And the proof is simple: the rank-nullity theorem.

Namely, you get that $$\dim(\operatorname{image}(T)) + \dim(\ker(T)) = \dim(V) = \dim(W).$$ The last equality follows from our hypothesis. (The $\ker$ stands for the kernel, you may know it as "null space".)

Now, you can prove the equivalence by recalling the following facts:

1. $T$ is one-to-one $\iff$ $\ker(T) = \{0\}$.
2. $T$ is onto $\iff$ $\dim(\operatorname{image}(T)) = \dim(W)$.

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Now, back to your question. You are given that $S \circ T = \operatorname{id}_V$. From this, it follows that $S$ is onto and $T$ is one-to-one. (This is a general fact about functions.)
Thus, we see that $S$ and $T$ are both bijections. In particular, they have an inverse. Usual set theory now tells us that $T \circ S$ must also be the (appropriate) identity map.