= Solution
The symmetry between the first and third coordinates makes $(1,0,-1)^{\mathsf T}$ an <eigenvector>, with <eigenvalue> $-6$. In the remaining invariant <plane>, write a <vector> as $(s,t,s)^{\mathsf T}$. Its image is $(8s-4t,-8s+4t,8s-4t)^{\mathsf T}$, so the reduced <eigenvalue> equation has <determinant>
$$
\det\begin{pmatrix}8-\lambda&-4\\-8&4-\lambda\end{pmatrix}
=\lambda(\lambda-12).
$$
The corresponding directions have $t=2s$ for $\lambda=0$ and $t=-s$ for $\lambda=12$. Normalizing gives
$$
\boxed{\begin{array}{c|c}
\lambda=-6&\mathbf e_-=(1,0,-1)^{\mathsf T}/\sqrt2\\[2pt]
\lambda=0&\mathbf e_0=(1,2,1)^{\mathsf T}/\sqrt6\\[2pt]
\lambda=12&\mathbf e_+=(1,-1,1)^{\mathsf T}/\sqrt3.
\end{array}}
$$
Their squared unnormalized norms are $2,6,3$. The pairwise unnormalized <dot products> are $1-1=0$, $1-1=0$, and $1-2+1=0$, so these unit <eigenvectors> are mutually orthogonal. The <eigenvalues> sum to $6$ and have product zero, agreeing with part (a).
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