If $(PQ - I)$ is invertible, verify $(QP - I)$ is invertible with inverse:$ -I + Q(PQ-I)^{-1}P$

Question

Let P,Q be n x n matrices. Suppose that (PQ - I) is invertible. Verify that (QP - I) is also invertible with the following inverse:

$$ (QP-I)^{-1} = -I + Q(PQ-I)^{-1}P $$

I have some intuition for why it's true, namely $(AB)^{-1} = B^{-1}A^{-1}$ but beyond that I'm not sure why this is the case.

Answer 1

Let $A=-I+Q(PQ-I)^{-1}P$. Then $$ (QP-I)A = I-QP +(QP-I)Q(PQ-I)^{-1}P $$ It suffices to show that the the second two terms cancel. $$ (QP-I)Q(PQ-I)^{-1}P = (QPQ-Q)(PQ-I)^{-1}P = Q(PQ-I)(PQ-I)^{-1}P = QP, $$ and the result follows since if a square matrix has a one-sided inverse, the same matrix is a two-sided inverse for it.