Is my alternate answer correct?

Question

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

@GitGud, its as I explained. I am composing $TX = T$, doesn't $X = I$ have to be the identity map? — jacob smith, Aug 18 '16 at 10:58
@GitGud Then I guess I can't use my method...I guess a even convincing argument would be if $T = 0$ is the $0$ map. Rookie mistake, got it. — jacob smith, Aug 18 '16 at 11:01
here are some proofs of that fact; maybe you'll find an argument you like better. — Ben Grossmann, Aug 18 '16 at 13:27

score 1 · Answer 1 · answered Aug 18 '16 at 11:02

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 $TT\not=I$.

score 0 · Answer 2 · answered Aug 18 '16 at 14:03

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 $$ -\frac{1}{a_0}(T^{k-1}+\cdots+a_1 I)T=I \\ \implies -\frac{1}{a_0}(T^{k-1}+\cdots+a_1 I)=S \\ \implies TS=ST \\ \implies ST=I. $$

Is my alternate answer correct?

2 Answers2