# A little proof

1. Feb 25, 2008

### John Smith

I need help with this proof.
We have u and v of size n*1. It is giving that I of size n*n.
A = I + u*v^Transpose

Proof that if u^T*v is not = -1
then A is reverseble and that A is
A^-1 = I - (1 / (1+u^T*V))*uv^T

2. Feb 25, 2008

### HallsofIvy

Staff Emeritus
The obvious thing to do is to look at
$$(I+uv^T)(I- \frac{1}{1+ u^Tv}uv^T)$$
and
$$(I- \frac{1}{1+ u^Tv}uv^T)(I+ uv^T)$$

What do you get when you multiply those out?

(Are you certain that one of those $uv^T$ is not supposed to be $vu^T$?)

Last edited: Feb 25, 2008
3. Feb 26, 2008

### John Smith

I did multiply the equation out and find out that ((v^T)* u)^T = u^T * v
But I was wondering if this was enough to show out the proof.

4. Feb 26, 2008

### HallsofIvy

Staff Emeritus
No, that is not the point at all!
The products of both
$$(I+uv^T)(I- \frac{1}{1+ u^Tv}uv^T)$$
and
$$(I- \frac{1}{1+ u^Tv}uv^T)(I+ uv^T)$$
should be the identity matrix!

5. Feb 27, 2008

### John Smith

Yes thank you I think that I got it now.