Orthogonal diagonalisation of a $4\times 4$ matrix Can somebody help me to orthogonally diagonalise the matrix $\begin{bmatrix}0 & 0 & 0 & 1\\0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}$?

The assiocated quadratic form is : $$q(a,b,c,d)=ad+bc+cb+da=2ad+2bc=\frac{1}{2}(a+d)^2-\frac{1}{2}(a-d)^2+\frac{1}{2}(b+c)^2-\frac{1}{2}(b-c)^2$$
So you can deduce that the eigen values are $1$ and $-1$, and by coming back to a matrix notation, you have : 
$$\begin{bmatrix}0 & 0 & 0 & 1\\0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}=P\begin{bmatrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix}P^t$$
$$P=\frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 & 0 & 0\\0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \\ 1 & -1 & 0 & 0\end{bmatrix}$$
