Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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.

share|cite|improve this answer
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 $$B(wI-B)=wB-B^2=wB-wB=0$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.