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

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Homework Help Overview

The problem involves finding the inverse of the matrix (I + vv'), where v is a vector with a norm less than 1 and I is the identity matrix. The discussion centers around the properties of this matrix and its invertibility.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand the structure of the matrix A = I + vv' and its inverse. One participant expresses difficulty in simplifying the expression for A and questions the aesthetic of the inverse. Another participant updates with a determinant calculation, expressing uncertainty about its validity while noting its potential appeal. A concern is raised regarding the rank of the matrix defined by the outer product, suggesting it may not be invertible.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the matrix and its properties. Some guidance has been offered regarding the determinant, but there is no explicit consensus on the invertibility of the matrix or the form of its inverse.

Contextual Notes

Participants are working under the constraint that the norm of vector v is less than 1, which may influence the properties of the matrix in question. There is also a mention of potential assumptions regarding the rank of the matrix defined by the outer product.

tennishaha
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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

[tex]A \ = \ I \ + \ vv'[/tex] then

[tex]A_{ij} \ = \ v_iv_j[/tex] when [tex]i \neq j[/tex] and

[tex]A_{ij} \ = \ v_iv_j + 1[/tex] when [tex]i = j[/tex].

I can't see how [tex]A^{-1}[/tex] can turn out pretty. But maybe I'm just missing something...it is getting pretty late.
 
UPDATE:

All I can seem to find is

[tex]det(A) \ = \ ||\textbf{v}||^2 \ + \ 1.[/tex]

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 [tex]A^{-1}[/tex] is pretty.
 
I think I saw someone say that the matrix defined by an out product has rank 1 and therefore not invertible.

Mat
 

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