Show that $v^tA^{-1}u\ne -1$ if and only if $A+uv^t$ is invertible

Question

Show that $v^tA^{-1}u\ne -1$ if and only if $A+uv^t$ is invertible, where $A$ square matrix and $u,v \in \mathbb{R}^n$

I feel I want to use eigenvalues. I know, $A+uv^t$ is invertible if and only if all eigenvalues are non-zero. Any type of proof please.

Answer 1

Assuming $A$ is invertible, you can use this result (details here) to say that $$\det(A+uv^T)=(1+v^TA^{-1}u)\det A$$

And $A+uv^T$ is invertible iff $\det(A+uv^T)\ne 0$.