Inverse of (I+vv') where norm(v)<1

[tennishaha]

Homework Statement

v is a vector with norm(v)<1
what is the inverse of (I+vv') where I is a identity matrix

Homework Equations

The Attempt at a Solution

 

[Raskolnikov]

I don't see how this simplifies at all. If

$$A \ = \ I \ + \ vv'$$ then

$$A_{ij} \ = \ v_iv_j$$ when $$i \neq j$$ and

$$A_{ij} \ = \ v_iv_j + 1$$ when $$i = j$$.

I can't see how $$A^{-1}$$ can turn out pretty. But maybe I'm just missing something...it is getting pretty late.

 

[Raskolnikov]

UPDATE:

All I can seem to find is

$$det(A) \ = \ ||\textbf{v}||^2 \ + \ 1.$$

I'm not 100% certain on this since I didn't really construct a fool proof but instead made a few "gut" leaps, but I think (and hope) it holds. This result is sort of pretty; however, I still stand by what I said earlier: I highly doubt $$A^{-1}$$ is pretty.

 

[hunt_mat]

I think I saw someone say that the matrix defined by an out product has rank 1 and therefore not invertible.

Mat