Is my alternate answer correct? I found this answer from my book which is completely different from mine. 

If $\text{dim } V < \infty$ and $T,S$ are operators on the vector space $V$ then $TS = I$ iff $ST = I$.

My book went through some argument with the inverse, but i basically just said that 
$$TST = T \implies T(ST) = T \implies ST = I$$
Is that okay? I am basically saying for an operator to map with something to get back itself must be identity
 A: No, actually not, since in general you don't have the implication
$$
T(ST)=T\Rightarrow ST=I\; (\text{so }T(ST)=T\not\Rightarrow ST=I \text{ holds})
$$
As an example, look at 
$$
T=
\begin{pmatrix}
1&0\\ 0&0
\end{pmatrix}
$$
and $$
T(TT)=T
$$
and we have $TT\not=I$.
A: Somewhere dimension has to enter the picture because the result is not true for infinite-dimensional spaces. The minimal polynomial is one way to do this. A minimal polynomial exists for $T$ because $V$ is finite-dimensional. Suppose $TS=I$. It is then shown that $ST=I$. If
$$
           m(z) = z^{k}+a_{k-1}z^{k-1}+\cdots+a_1z+a_0
$$
is the minimal polynomial for $T$, then $a_0 \ne 0$ because, otherwise, $m$ cannot be minimal:
$$
        (T^{k-1}+\cdots+a_1I)T=0 \\
    \implies (T^{k-1}+\cdots+a_1I)TS=0 \\
    \implies T^{k-1}+\cdots+a_1I = 0.
$$
Because $a_0\ne 0$, then
$$
         -\frac{1}{a_0}(T^{k-1}+\cdots+a_1 I)T=I \\
      \implies -\frac{1}{a_0}(T^{k-1}+\cdots+a_1 I)=S \\
      \implies TS=ST \\
      \implies ST=I.
$$
