# Sherman–Morrison formula

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.

## 1 Answer

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.

• Thanks for the response. – Arthur Mar 7 '19 at 3:38