Show that if $B\in \mathbb{C}^{p\times q}$ and $C\in \mathbb{C}^{q\times p}$, then $I_p -BC$ is invertible $\Longleftrightarrow I_q-CB$ is invertible.
Specifically, use the Schur complement: $$\begin{bmatrix} I_p-BC & B\\ C & I_q-CB \end{bmatrix} = \begin{bmatrix} I_p & O \\ C(I_p-BC)^{-1} & I_q \end{bmatrix} \begin{bmatrix} I_p-BC & O \\ O & (I_q-CB)-C(I_p-BC)^{-1}B\end{bmatrix} \begin{bmatrix} I_p & (I_p-BC)^{-1}B \\ O & I_q\end{bmatrix}$$
I believe the best way to go about solving this problem is to prove the term $(I_q-CB)-C(I_p-BC)^{-1}B$ is invertible, because we know the first and third matrices in our Schur decomposition are invertible (the diagonal elements are nonzero, and they are block triangular) and the $I_p-BC$ term is invertible by assumption.
Any help would be appreciated.
Or you could just check that if $X$ is the inverse of $I_p-BC$ then $Y=I_q+CXB$ is the inverse of $I_q-CB$.