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Math Help - matrix multiplied by its transpose on the left or right = same eigenvalues?

  1. #1
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    matrix multiplied by its transpose on the left or right = same eigenvalues?

    If P is an m x n matrix

    does P.P^T have the same eigenvalues as P^T.P?

    this is what I seem to be getting when I do my calculations although I sometimes end up with a 0 eigenvalue in one and not the other.. is there an explanation of this?

    many thanks
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  2. #2
    Super Member PaulRS's Avatar
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    We can easily show it holds for the case of matrixes P of <br />
2 \times 2<br />
, however, it doesn't hold in general ( take <br />
P = \left( {\begin{array}{*{20}c}<br />
   3  \\<br />
   1  \\<br /> <br />
 \end{array} } \right)<br />
you'll find that one of the matrixes has an additional eigenvalue)

    Let's see the 2x2 matrixes. Remember that the characteristic polynomial of a matrix <br />
A<br />
in this case is given by: <br />
\chi _A \left( \lambda  \right) = \lambda ^2  - {\text{tr}}\left( A \right) \cdot \lambda  + \det \left( A \right) = <br />
\det \left( {A - \lambda  \cdot I} \right)<br />

    For any 2 matrixes A and B we have: <br />
{\text{tr}}\left( {AB} \right) = {\text{tr}}\left( {BA} \right)<br />

    So if  P is a 2x2 matrix we have: <br />
{\text{tr}}\left( {P^t P} \right) = {\text{tr}}\left( {PP^t } \right)<br />
and <br />
{\text{det}}\left( {P^t P} \right) = {\text{det}}\left( {PP^t } \right)<br />
and therefore <br />
\chi _{P^t P} \left( \lambda  \right) = \chi _{PP^t } \left( \lambda  \right)<br />
for all <br />
\lambda <br />
.

    Thus the assertion is valid when P is a 2x2 matrix
    Last edited by PaulRS; February 11th 2009 at 06:04 PM.
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  3. #3
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    Opalg's Avatar
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    Quote Originally Posted by James0502 View Post
    If P is an m x n matrix

    does P.P^T have the same eigenvalues as P^T.P?

    this is what I seem to be getting when I do my calculations although I sometimes end up with a 0 eigenvalue in one and not the other.. is there an explanation of this?
    It is true more generally that if A is an mn matrix and B is an nm matrix then AB and BA have the same nonzero eigenvalues.

    In fact, suppose that \lambda\ (\ne0) is an eigenvalue for AB, with eigenvector x. Then BA(Bx) = B(ABx) = B(\lambda x) = \lambda(Bx). Thus \lambda is also an eigenvector for BA, with eigenvector Bx.

    The argument breaks down if \lambda = 0, because of the possibility that Bx = 0 in that case.
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