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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!

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    @Andrew, you may be right! I just double checked it and this is how it's written in the exercise. I'm going to fire an email to the exercise creator.2012-06-16

2 Answers 2

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Why not multiply the right side by $I+uv^t$ and see if you get the identity matrix? Note that $v^tu$ is a scalar.

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    That's why I made a point of noting that $v^tu$ is a scalar.2012-06-16
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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 $B(wI-B)=wB-B^2=wB-wB=0$