Sherman–Morrison formula

Question

Sherman–Morrison formula states that if $A\in\mathbb{R}^{n\times n}$ is an invertible square matrix and $u,v\in \mathbb{R}^n$. Then $A+uv^\top$ is invertible iff $1+v^\top A^{-1}u \ne 0$. Consider the generalization: if $A\in\mathbb{R}^{n\times n}$ is an invertible square matrix and $U\in \mathbb{R}^{n\times k}$ and $V\in \mathbb{R}^{k\times n}$. Then $A+UV$ is invertible iff $I_k+VA^{-1} U$ invertible. Is this generalization also true? I know that the "if" direction holds according to Woodbury matrix identity. Does the other direction also holds?

Any comment is greatly appreciated.

Answer 1

Yes, it is. In fact, you already have all the components to prove it. The 'if' as well as the 'only if' both parts are provable using Woodbury Matrix identity. As,

$$ (A + UV)^{-1} = A^{-1} - A^{-1}U(I + VA^{-1}U)^{-1}VA^{-1} $$

And,

$$ (I + VA^{-1}U)^{-1} = I - V(A + UV)^{-1}U $$

Both of the above identities are easily derivable from Woodbury Matrix identity by appropriate substitution.