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?
Best Answer
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 $$\boxed{-\lambda^3(16-\lambda)^2}$$