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Eigenvalue Factorization and Matrix Substitution

  1. Mar 21, 2010 #1
    In my literature reviews I found a few things that I can't quite understand.
    1. The problem statement, all variables and given/known data

    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!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 21, 2010 #2


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    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.


    You simply square the inverse of the matrix.
  4. Mar 21, 2010 #3
    I'll give it a try gabbagabbahey. Thanks.

    Any idea what the matrix "U" is?
  5. Mar 21, 2010 #4


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    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.
  6. Mar 21, 2010 #5
    Yes I recall now. Thanks!
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