# Thread: Diagonalizing Symmetric Bilinear Forms

1. ## 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)?

2. The scaling factor has nothing to do with the eigenvalue. It is the length of the eigenvector, which for both of these eigenvectors is $\sqrt{2^2+1^2} = \sqrt5$.