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

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I+vv'), have rank 1 and are not invertible. The inverse of (I+vv') is not pretty and can be represented by the equation det(A) = ||v||^2 + 1.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
I think I saw someone say that the matrix defined by an out product has rank 1 and therefore not invertible.

Mat
 

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

1. What is the inverse of (I+vv')?

The inverse of (I+vv') is (I-vv'), where I is the identity matrix and v is a vector. This can be derived using the Sherman-Morrison formula: (I+uv')^-1 = I - (u/(1+v'u))v', where u and v are vectors.

2. How do you compute the inverse of (I+vv')?

The inverse of (I+vv') can be computed using the Sherman-Morrison formula or by using the matrix inversion method. This involves finding the determinant of the matrix and then using it to calculate the inverse.

3. What is the importance of the norm of v being less than 1 in this problem?

In this problem, the norm of v being less than 1 ensures that the matrix (I+vv') is invertible. If the norm is greater than or equal to 1, the matrix will not have an inverse. This is because the determinant of (I+vv') is equal to 1-v'v, which becomes 0 if the norm of v is equal to 1.

4. Can the inverse of (I+vv') be solved analytically?

Yes, the inverse of (I+vv') can be solved analytically using the Sherman-Morrison formula or the matrix inversion method. However, for larger matrices, it may be more efficient to use numerical methods.

5. What are the applications of the inverse of (I+vv')?

The inverse of (I+vv') has various applications in mathematics, engineering, and computer science. It is used in solving systems of linear equations, computing determinants, and in machine learning algorithms such as linear regression and principal component analysis.

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