<s>[INST] Quick method for finding eigenvalues and eigenvectors in a symmetric $5 \times 5$ matrix? The matrix $B$:
$B = 
\pmatrix{
0 &  0 &  0 &  0 &  0 \cr 
0 &  8 &  0 & -8 &  0 \cr 
0 &  0 &  8 &  0 & -8 \cr 
0 & -8 &  0 &  8 &  0 \cr 
0 &  0 & -8 &  0 &  8 \cr 
}$
Which has nonzero eigenvalues $\lambda_1=16$ and $\lambda_2=16$ and corresponding eigenvectors:
v$_1 = 
\pmatrix{
0\cr 
\frac{1}{2} \sqrt2 \cr 
0 \cr 
-\frac{1}{2} \sqrt2 \cr 
0\cr 
}$  and  v$_2 =
\pmatrix{
0\cr 
0\cr 
\frac{1}{2} \sqrt2 \cr 
0 \cr 
-\frac{1}{2} \sqrt2 \cr 
}$  
What is the method for obtaining these eigenvalues and corresponding eigenvectors? 

It's a large matrix and I'm hoping there's some kind of easy trick to it. From what I can remember of eigen decomposition, normally I'd do:
$Ax = \lambda x \implies|A-\lambda I|x = 0$
$\implies 
\det \pmatrix{
0-\lambda &  0 &  0 &  0 &  0 \cr 
0 &  8-\lambda &  0 & -8 &  0 \cr 
0 &  0 &  8-\lambda &  0 & -8 \cr 
0 & -8 &  0 &  8-\lambda &  0 \cr 
0 &  0 & -8 &  0 &  8-\lambda \cr 
}$ $\pmatrix{
x_1 \cr 
x_2 \cr 
x_3 \cr 
x_4 \cr 
x_5 \cr 
}$ = $\pmatrix{
0 \cr 
0 \cr 
0 \cr 
0 \cr 
0 \cr 
}$
So the determinant is
$\implies -\lambda \det  \pmatrix{ 
8-\lambda &  0 & -8 &  0 \cr 
0 &  8-\lambda &  0 & -8 \cr 
-8 &  0 &  8-\lambda &  0 \cr 
0 & -8 &  0 &  8-\lambda \cr 
}$
$\implies -\lambda * [ (8- \lambda)\det  \pmatrix{ 
8-\lambda &  0 & -8 \cr 
0 &  8-\lambda &  0 \cr 
-8 &  0 &  8-\lambda \cr 
}-8 \det \pmatrix{ 
0 &  8-\lambda & -8 \cr 
-8 &  0 &  0 \cr 
0 & -8 &  8-\lambda \cr 
}]$
etc.
There's got to be an easier way?
 [/INST] Hint: if $A$ and $B$ are square of the same order, $$\det\begin{pmatrix} A & B \\ B & A\end{pmatrix}=\det(A-B)\det(A+B)$$
So setting 
$$A=\begin{pmatrix}8-\lambda & 0 \\ 0 & 8-\lambda\end{pmatrix}$$ 
and 
$$B=\begin{pmatrix}-8 & 0\\0&-8\end{pmatrix}$$
shows that the determinant of your $4\times 4$ matrix is $(16-\lambda)^2\lambda^2$, and therefore the determinant of your original matrix is $$\boxed{-\lambda^3(16-\lambda)^2}$$
</s>