Is the Following Proof Correct?
Theorem. Given that $S,T\in\mathcal{L}(V)$ where $V$ is finite-dimensional $ST=I$ if and only if $TS=I$.
Proof. ($\Rightarrow$). Assume that $ST=I$ it is apparent that $ST
\in\mathcal{L}(V)$ now let $v$ be arbitrary and assume that $v\in\operatorname{null}ST$ consequently
$$STv = Iv=v=0$$
this implies that $\operatorname{null} ST=\{0\}$ and therefore $ST$ is injective moreover from the equivalence of injectivity and is surjectivity for linear operators on a finite dimensional vector space $V$ it follows that $ST$ is invertible this together with $\mathbf{3-D-9}$ implies that both $S$ and $T$ are invertible consequently
$$ST = I \implies TST = TI=T\implies TS(TT^{-1})=TT^{-1}\implies TS=I.$$
$(\Leftarrow).$ The converse will involve a similar argument to that present above.
NOTE: $\mathbf{3-D-9}$ is the following result
Given that $V$ is finite-dimensional and $S,T\in\mathcal{L}(V)$. $ST$ is invertible if and only if both $S$ and $T$ are invertible.
$\blacksquare$
Best Answer
The identity operator $I:V\to V$ is invertible, and so is $ST.$ Now $ST$ is invertible, so both $S$ and $T$ is invertible. Assume that $ST=I.$ Then $T=S^{-1}\implies TS=S^{-1}S=I.$