<s>[INST] How to correct this diagonalization argument? 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 $(*)$?
 [/INST] 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.
</s>