# Diagonalizing Symmetric Bilinear Forms

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• Apr 8th 2009, 07:26 AM
Deadstar
Diagonalizing Symmetric Bilinear Forms
Find a 2x2 orthogonal matrix $\displaystyle P$ such that $\displaystyle P^{-1} S P$ is diagonal where

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

So here's my working...

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

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

But the answer has it as all of them are divided by $\displaystyle \sqrt{5}$... 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)?
• Apr 8th 2009, 11:10 AM
Opalg
The scaling factor has nothing to do with the eigenvalue. It is the length of the eigenvector, which for both of these eigenvectors is $\displaystyle \sqrt{2^2+1^2} = \sqrt5$.