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Eigenvalue Factorization and Matrix Substitution

  • #1
In my literature reviews I found a few things that I can't quite understand.

Homework Statement



I have the following equation:
http://img717.yfrog.com/img717/6416/31771570.jpg [Broken]

I'm told that by using the eigenvalue factorization:
http://img89.yfrog.com/img89/760/83769756.jpg [Broken]

, I can change the first equation to:
http://img28.imageshack.us/img28/5023/84802099.jpg [Broken]

2. The attempt at a solution

I tried changing Equation 2 to just be (A^T)A and then subbing into the first equation, but I can't quite do anything with those inverses.

Also, what does the exponent of '-2' mean in the context of a 4x4 matrix? Lastly, what is matrix U?

Thank you!
 
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Answers and Replies

  • #2
gabbagabbahey
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I tried changing Equation 2 to just be (A^T)A and then subbing into the first equation, but I can't quite do anything with those inverses.
I think it's probably easiest to start from [itex]\textbf{p}^{T}(\mathbf{\Lambda}+\lambda\textbf{I})^{-2}\textbf{q}=0[/itex] and work your way backwards instead.

Also, what does the exponent of '-2' mean in the context of a 4x4 matrix? Lastly, what is matrix U?
[tex]\textbf{C}^{-2}\equiv\textbf{C}^{-1}\textbf{C}^{-1}[/tex]

You simply square the inverse of the matrix.
 
  • #3
I'll give it a try gabbagabbahey. Thanks.

Any idea what the matrix "U" is?
 
  • #4
gabbagabbahey
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Any idea what the matrix "U" is?
It's the invertible matrix which relates the matrix [itex]\textbf{A}^{T}\textbf{A}\mathbf{\Sigma}[/itex] to the diagonal matrix [itex]\mathbf{\Lambda}[/itex] via a similarity transform. Its columns will be the eigenvectors of [itex]\textbf{A}^{T}\textbf{A}\mathbf{\Sigma}[/itex].

See http://en.wikipedia.org/wiki/Diagonalizable_matrix for a refresher on matrix diagonalization.
 
  • #5
Yes I recall now. Thanks!
 

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