TITLE: Matrix inverse property, show that $(I + uv^T)^{-1} = I - \frac{uv^T}{1+v^Tu}$
QUESTION [6 upvotes]: Let $u, v \in \mathbb{R}^N, u^Tv \neq -1$. Thereby $I +uv^T \in \mathbb{R}^{N \times N}$ is invertible. Show that:
$$(I + uv^T)^{-1} = I - \frac{uv^T}{1+v^Tu}$$

I'm lost, why did the denominator get $uv^T$ as $v^T u$? Where did this $1$ come from? Any hints are very appreciated!

REPLY [3 votes]: Some ideas:  Let us put $$B:=uv^T\,\,,\,\,w:=v^Tu$$Now, let us do the matrix product$$(I+B)\left(I-\frac{1}{1+w}B\right)=\frac{1}{1+w}(I+B)(I-B+wI)=\frac{1}{1+w}(I-B^2+wI+wB)=$$$$=\frac{1}{1+w}\left[(1+w)I+B(wI-B)\right]$$Well, now just check the product $$B(wI-B)...:>)$$
*Added*$$B^2=\left(uv^T\right)\left(uv^T\right)=u\left(v^Tu\right)v^T=uwv^T=wuv^T=wB$$ so we get $$B(wI-B)=wB-B^2=wB-wB=0$$