Inverse of (I+vv') with Norm(v)<1 | Matrix Homework Question

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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

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

A \ = \ I \ + \ vv&#039; 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.
 
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.
 
I think I saw someone say that the matrix defined by an out product has rank 1 and therefore not invertible.

Mat
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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