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Eigenvalues and eigenvectors of a non-symmetric matrix?

  1. Aug 14, 2014 #1
    I have a non symmetric matrix AB where A and B are symmetric matrices. How can I find the eigenvectors and eigenvalues of AB?

    In a paper( Fisher Linear Discriminant Analysis by M Welling), the author asks to find eigenvalues and eigenvectors of B^(1/2)* A *B^(1/2) which is a symmetric matrix. But how can I get eigenvalues and eigenvectors of $AB$ from eigenvalues and eigenvectors of B^(1/2) A B^(1/2)?

    Could someone please help me?

    I tried the solution suggested by others. But it doesn't work

    eigenvectors of AB=[0.516537330395033,-0.781188319935242,-0.0177964973702446;-0.710088559129181,-0.185707982205180,0.521054279012559;0.478501227273470,0.596034692062496,0.853337988726655];

    eigenvectors of B^(1/2)* A* B^(1/2)= [0.517933641073670,-0.855373946305353,-0.00895295628145857;-0.725005239437560,-0.444501257554748,0.526104585439372;0.453995755742558,0.265996323309655,0.850372747536919];

    product of B^(-1/2) and eigenvectors of B^(1/2)* A *B^(1/2) =[0.0692193234209673,-0.179045453961740,-0.160160577798552;-0.0951564325356731,-0.0425635779871867,4.68925725410385;0.0641222410443052,0.136608931923259,7.67966316565345];

    Both are different. Is there any mistake in what I did?
  2. jcsd
  3. Aug 14, 2014 #2
  4. Sep 3, 2014 #3
    Product of two symmetric matrices doesn’t possess nice properties unless the matrices commute (in that case the product is also symmetric). It can be non-diagonalizable. Even it is diagonalizable, there is, in general, no correspondence between eigenvectors of A B and of factors themselves.
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