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I want to diagonalize (with a orthogonal change of coordinates) the quadratic form $F(x,y)=x^2-\frac{n-2}{\sqrt{n-1}}xy-y^2$. I already know that the system $(*)$ bellow \begin{eqnarray} x=&\frac{(\sqrt{n-1}+1)u+(\sqrt{n-1}-1)v}{\sqrt{2n}}\\ y=&\frac{(-\sqrt{n-1}+1)u+(\sqrt{n-1}+1)v}{\sqrt{2n}}, \end{eqnarray} does the work but, I'm trying to find this answer by myself. What I did was the following.

The symmetric matrix associated to $F$ is $$ A=\begin{pmatrix}1 & \frac{2-n}{2\,\sqrt{n-1}}\cr \frac{2-n}{2\,\sqrt{n-1}} & -1\end{pmatrix} $$ with eigenvalues $\pm\frac{n}{2\sqrt{n-1}}$ and corresponding eigenvectors \begin{align} v_1=&(1,\frac{\left( n-2\right) \,\sqrt{n-1}}{\sqrt{n-1}\,n-2\,n+2})\\ v_2=&(1,-\frac{\left( n-2\right) \,\sqrt{n-1}}{\sqrt{n-1}\,n+2\,n-2}). \end{align}

As many textbooks teach, we define $$ P= \begin{pmatrix}1 & 1\cr \frac{\left( n-2\right) \,\sqrt{n-1}}{\sqrt{n-1}\,n-2\,n+2} & -\frac{\left( n-2\right) \,\sqrt{n-1}}{\sqrt{n-1}\,n+2\,n-2}\end{pmatrix}. $$ The inverse of $P$ is $$ P^{-1}= \begin{pmatrix}\frac{\sqrt{n-1}\,\left( {n}^{2}-4\,n+4\right) }{2\,\sqrt{n-1}\,{n}^{2}+4\,{n}^{2}-4\,n} & \frac{n-2}{2\,n}\cr \frac{\sqrt{n-1}\,\left( {n}^{2}-4\,n+4\right) }{2\,\sqrt{n-1}\,{n}^{2}-4\,{n}^{2}+4\,n} & \frac{2-n}{2\,n}\end{pmatrix}, $$ so we have $$P^{-1}\cdot A\cdot P=\begin{pmatrix}-\frac{n}{2\sqrt{n-1}} & 0\cr 0 & \frac{n}{2\sqrt{n-1}}\end{pmatrix}.$$ That is, the matrix $P$ gives a change of coordinates that diagonalizes $A$ but it is far from being orthogonal.

So, how could I find the change of coordinates $(*)$?

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I've edited your post to say "quadratic" instead of "bilinear". –  Alex Youcis Dec 10 '12 at 13:17
    
It's ok Alex, but is it wrong to call that function a bilinear form? –  Son Dec 10 '12 at 13:27
    
Technically yes. This is because "bilinear" means "linear in entry seperately"--something your function is not. For example, $F(2,0)=4$ which is not equal to $2F(1,0)=2$, but for a bilinear form $F(\lambda x,y)=\lambda F(x,y)$ for all variables $\lambda,x,y$. I hope this helps! :) –  Alex Youcis Dec 10 '12 at 13:38
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1 Answer 1

It's not far from being orthogonal; you just failed to normalize the columns. Being the eigenvectors of a symmetric matrix, they're already orthogonal; if you normalize them, $P$ will be orthogonal and $P^{-1}AP$ will still be diagonal.

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Ok Joriki, but how about the change of coordinates (*). How could I find it? A hint would be very nice. –  Son Dec 10 '12 at 17:12
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