Orthogonal diagonalisation of a $4\times 4$ matrix Can somebody help me to orthogonally diagonalise the matrix $\begin{bmatrix}0 & 0 & 0 & 1\\0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}$?
 A: An Eigenvector $(v_1,v_2,v_3,v_4)$ associated to an eigenvalue $\lambda$ satisfies $$
\begin{cases}
x_4 =\lambda x_1\\
x_3 = \lambda x_2\\
x_2 = \lambda x_3\\
x_1 =\lambda x_4
\end{cases}$$
Hence $\lambda^2=1$ proving that the eigenvalues are $-1$ and $1$. From there you can easily find an orthonormal basis of eigenvectors.
A: 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 : 
$$\begin{bmatrix}0 & 0 & 0 & 1\\0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}=P\begin{bmatrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix}P^t$$
$$P=\frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 & 0 & 0\\0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \\ 1 & -1 & 0 & 0\end{bmatrix}$$
A: You can see that the matrix squared is the identity matrix, so eigenvalues must fulfil ${\lambda_i}^2$ = 1 for all $i$. There are two possibilities over the most common fields $\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}$: $1$ and $-1$.
So we solve for first case $\lambda=1$:
$$\left[\begin{array}{cccc|c}
-1&0&0&1&0\\
0&-1&1&0&0\\
0&1&-1&0&0\\
1&0&0&-1&0
\end{array}\right]$$
We see we get $\{[1,0,0,1],[0,1,1,0]\}$
And for $\lambda=-1$:
$$\left[\begin{array}{cccc|c}
1&0&0&1&0\\
0&1&1&0&0\\
0&1&1&0&0\\
1&0&0&1&0
\end{array}\right]$$
We see we get $\{[1,0,0,-1],[0,1,-1,0]\}$
We can verify these vectors spanning are pairwise orthogonal, so only thing left is normalization, in our case will be a same factor of $\frac{1}{\sqrt{2}}$ as both $1^2+1^2$ and $1^2+ (-1)^2$ evaluate to $2$. Now what is left is to stuff it all into matrices, but that should be an good exercise.
A: Also, if you are curious about more advanced concepts you can consider your matrix as a product of two other matrices:
$$\left[\begin{array}{cccc}
0&1&0&0\\
1&0&0&0\\
0&0&0&1\\
0&0&1&0
\end{array}\right]\left[\begin{array}{cccc}
0&0&1&0\\
0&0&0&1\\
1&0&0&0\\
0&1&0&0
\end{array}\right] = (C_2 \otimes I_2) (I_2 \otimes C_2) $$
Where $C_2 = \left[\begin{array}{cc}0&1\\1&0\\\end{array}\right]$, $I_2 = \left[\begin{array}{cc}1&0\\0&1\\\end{array}\right]$ and $\otimes$ denotes a Kronecker product. There are many results derived for properties of these products, including their eigenvalue spectra in terms of the individual factors spectrum. Not really needed to solve the problem, but great to get more in-depth understanding.
