Invertibility of $I + CA^{-1}B$

Question

Suppose that $C$ and $B$ are $n \times p$ and $p \times n$, respectively, and that $A + BC$ and $A$ are full rank. Then why is $(I + CA^{-1}B)$ also full rank?

I've tried to manipulate things around, and I wanted to show that $(I + CA^{-1}B)x = 0$ implies $x =0$. So far, all I can see is that if $(I + CA^{-1}B)x = 0$, then $(IB + BCA^{-1}B)x = 0$ and thus $(A + BC)(A^{-1}B)x = 0$, so $A^{-1}Bx = 0$ by the non singularity of $A + BC$, and hence $x \in \mathcal{N}(B)$. But that isn't good enough, since $B$ is not full rank.

Answer 1

Using the Weinstein-Aronszajn determinant identity, since matrices $\rm A$ and $\rm A + B C$ have full rank,

$$0 \neq \det(\rm A + B C) = \det (\rm A) \cdot \det (\rm I + A^{-1} B \, C) = \underbrace{\det (\rm A)}_{\neq 0} \cdot \det (\rm I + C \, A^{-1} B)$$

and, thus, $\det (\rm I + C \, A^{-1} B) \neq 0$, i.e., matrix $\rm I + C \, A^{-1} B$ has ful rank.

Answer 2

I think you nearly got it. $(I+CA^{-1}B)x=0$ also implies that $CA^{-1}Bx=-x$. But you got $A^{-1}Bx=0$. I guess you see the problem now.