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Math Help - Diagonalizing Symmetric Bilinear Forms

  1. #1
    Super Member Deadstar's Avatar
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    Diagonalizing Symmetric Bilinear Forms

    Find a 2x2 orthogonal matrix P such that P^{-1} S P is diagonal where

    S=\left(\begin{array}{cc}7&-6\\-6&-2\end{array}\right)

    So here's my working...

    The eigenvalues of S are 10 and -5.
    Thus the eigenvectors are \left(\begin{array}{cc}2\\-1\end{array}\right) and \left(\begin{array}{cc}1\\2\end{array}\right).

    Now i know these are supposed to be scaled by \frac{1}{\sqrt{|\lambda|}} which would surely give me,
    P=\left(\begin{array}{cc}\frac{2}{\sqrt{10}}&\frac  {1}{\sqrt{5}}\\\frac{-1}{\sqrt{10}}&\frac{2}{\sqrt{5}}\end{array}\right)

    But the answer has it as all of them are divided by \sqrt{5}... Why is that? Do you scale all of the entries by the same eigenvalues and just choose any eigenvalue to use?

    Also, am i right in saying S is a type (1,1) matrix?
    And is this because there is a +ve and a -ve eigenvalue, or is it because you compute the determinant of the 1x1 matrix (i.e, the number 7), and the determinant of the 2x2 matrix (i.e, S) and since one is +ve and one is -ve that means its type (1,1)?
    Last edited by Deadstar; April 8th 2009 at 09:14 AM.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    The scaling factor has nothing to do with the eigenvalue. It is the length of the eigenvector, which for both of these eigenvectors is \sqrt{2^2+1^2} = \sqrt5.
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