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Just wanted to find out if this proof is valid:

Suppose that $V$ is finite dimensional and $S,T \in \mathcal{L}(V)$, that is both $S$ and $T$ are operators in $V$. Prove that $ST = I \iff TS = I$.

Proof:

Assume $ST = I$.

$TS = T(I)S = T(ST)S = TS(TS) \implies TS - TSTS = 0 \implies TS(I - TS) = 0 \implies $ either $TS = 0$ or $I - TS = 0 $. If it were true that $TS = 0$, we would have that $ST = ST(I) = ST(ST) = S(0)T = 0 \neq I$. Thus only $I - TS = 0$ is true and $TS = I$.

To prove the implication in the other direction we only need to reverse the roles of $S$ and $T$ showing that if $TS = I$ then $ST = I$.

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    $\begingroup$ Be careful: a product of matrices may be zero, even if none of the factors is zero. $\endgroup$ Commented Apr 27, 2022 at 18:40

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No, this proof has a flaw: $TS(I-TS)=0$ does not imply either $TS=0$ or $I-TS=0$. (Suppose for example that $V=\Bbb R^2$ and $TS(x,y) = (x,0)$.)

Note that there's another warning bell: nowhere in the proof did you use the assumption that $V$ is finite-dimensional—and the statement is false for infinite-dimensional vector spaces. (Example: let $V$ be the set of all infinite sequences of real numbers, and let $S$ be the operator that deletes the first element while $T$ is the operator that prepends a $0$ before the other elements.)

(You're definitely correct that you only need to prove one implication, by reversing the roles of $S$ and $T$.)

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