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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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Get some ideas from here: – Giuseppe Negro Jun 16 '12 at 12:17
The nominator (not denominator) has to be an $n\times n$ matrix. – user20266 Jun 16 '12 at 12:19
$u v^T$ is the outer product and is an $n \times n$ matrix, while $v^T u$ is the dot product and is a ($1 \times 1$) scalar. – TMM Jun 16 '12 at 12:28
@Thomas Please use "numerator" when discussing the top half of a fraction. – rschwieb Jun 16 '12 at 13:48
@Clash I think there's a typo in the condition for invertability. Should be $v^Tu \ne -1$. – Andrew Jun 16 '12 at 14:56
up vote 5 down vote accepted

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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So, multiplying by $(I + uv^T)$ for the left side I have the identiy matrix. On the right side I'm getting $I - \frac{(uv^T)^2}{1+v^Tu}$. This would only be the identity matrix if $\frac{(uv^T)^2}{1+v^Tu}$ is $0$. Or am I missing something? Thanks! – Clash Jun 16 '12 at 14:57
@Clash multiplying two binomials together (although these aren't exactly binomials) should produce an expression with 4 terms. You need two more terms. – Andrew Jun 16 '12 at 15:05
@Andrew, thanks! Are the two other terms $uv^T - \frac{uv^T}{1+v^Tu}$? – Clash Jun 16 '12 at 15:11
@Andrew, thanks! I got it now with your help and wikipedia! :) It would take me a while alone however to notice that there was a scalar in the middle of everything that I could factor out. – Clash Jun 16 '12 at 15:14
That's why I made a point of noting that $v^tu$ is a scalar. – Gerry Myerson Jun 16 '12 at 22:17

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 $$B(wI-B)=wB-B^2=wB-wB=0$$

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