I'm kind of confused on some linear algebra. On a previous question I was given a quadratic form and I had to find a matrix $P$ such that $P^TAP$ is diagonal. I did this by using a suitable change of co-ordinates and letting $P$ be the change of basis matrix and this worked fine.
Now I am given a (symmetric) matrix and I have to find an orthogonal matrix $P$ such that $P^TAP$ is diagonal. In my notes, they work out the eigenvectors and let $P$ be the matrix of eigenvectors. What I want to know is why the same process used for the first matrix doesn't work here?
For example, I have the matrix $\begin{pmatrix}-5 & 12 \\ 12 & 5\end{pmatrix}$. I want to try and find an orthogonal matrix $P$ such that $P^TAP$ is diagonal. I know that I can do this by finding the eigenvectors, but I want to try it using the other method. So I get that this corresponds to the quadratic form $q = -5x^2+24xy+5y^2=-5(x-\frac{12}{5}y)^2-\frac{169}{5}y^2$. Using the change of variable $x'=x-\frac{12}{5}y, y'=y$, I get $q = -5(x')^2-\frac{169}{5}(y')^2$ with change of basis matrix $P = \begin{pmatrix}1 & 0 \\ \frac{-12}{5} & 1\end{pmatrix}$. But this matrix is not orthogonal and does not satisfy $P^TAP$ being a diagonal matrix. Could anyone explain why it doesn't work for this matrix, but does work for one like $q=wz-xy$?
i will take your quadratic form $-5x^2 + 24xy + 5y^2$ transform into $ax_1^2 + by_1^2$ with the orthogonal transformation $x = x_1 \cos \theta - \sin \theta y_1, y = x_1\sin \theta + y_1 \cos \theta.$
$-5x^2 + 24 xy + 5y^2 = x_1^2(-5\cos^2 \theta + 24 \sin \theta \cos \theta + 5 \sin^2 \theta) + 2x_1y_1(10 \sin \theta \cos \theta + 12 \cos^2 \theta - 12 \sin^2 \theta) + y_1^2(-5 \sin^2 \theta -24 \sin \theta \cos \theta + 5 \cos^2 \theta).$
now choose $\theta$ so that coefficient of $x_1y_1$ is zero. that is $\tan(2\theta) = {12 \over 5}.$ putting this value of $\theta$ gives you $0 = x_1^2(-5*5 + 12*12) + y_1^2(5*5 - 12*12) = x_1^2 - y_1^2$
i hope i did all the computation right.