# Is this proof of “$ST = I \iff TS = I$” correct?

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 $$ST = I$$.

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

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 $$T$$.)