Let $A, B : V \to V$ be linear transformations, where V is a vector space. Prove that $I - ST$ is invertible iff $I - TS$ is invertible.

Question

It's very similar to Solve the matrix equation $[I+X(I-ST)^{-1}S](I-TS) = I$, but there's no answer given to that one.

A hint given is to consider $$I + T(I - ST)^{-1}S.$$ I tried to expand it, but things didn't cancel .

Answer 1

Simply note that if $I-ST$ is invertible, then the expression you are given to "consider" is actually a candidate for the inverse of $I-TS$, since: $$ (I - TS)(I + T(I-ST)^{-1}S) = I - TS + T(I-ST)^{-1}S - TST(I-ST)^{-1}S \\ = I - T\bigg(I - (I-ST)^{-1} + ST(I-ST)^{-1}\bigg)S \\ = I - T(I + (-I+ST)(I-ST)^{-1})S = I - T0S = I$$

which shows that $I-TS$ is invertible, and furthermore we have a formula for the inverse.

The converse is very similar : just switch $S$ and $T$ in the derivation above.

Answer 2

If $ I-ST $ is not invertible, let $ v $ be non-zero such that $ STv=v $. Then $ TS-I $ vanishes on the non-zero vector $ Tv $.