= Solution
The <matrix> is symmetric under interchange of the first and third coordinates. For an antisymmetric <vector> $(1,0,-1)$, direct multiplication gives $A(1,0,-1)^{\mathsf T}=-6(1,0,-1)^{\mathsf T}$. For a symmetric <vector> $(p,q,p)$, the <linear transformation> reduces to
$$
\begin{pmatrix}p\\q\end{pmatrix}\longmapsto\begin{pmatrix}8&-4\\-8&4\end{pmatrix}\begin{pmatrix}p\\q\end{pmatrix}.
$$
The reduced <characteristic polynomial> is $\lambda(\lambda-12)$. At $\lambda=0$, its equations give $q=2p$; at $\lambda=12$, they give $q=-p$. Normalizing the three <eigenvectors> gives
$$
\boxed{\begin{aligned}
\lambda=-6:&\quad\mathbf v_-=(1,0,-1)^{\mathsf T}/\sqrt2,\\
\lambda=0:&\quad\mathbf v_0=(1,2,1)^{\mathsf T}/\sqrt6,\\
\lambda=12:&\quad\mathbf v_+=(1,-1,1)^{\mathsf T}/\sqrt3.
\end{aligned}}
$$
Their <norms> are one. Their unnormalized pairwise <dot products> are $1-1=0$, $1-1=0$ and $1-2+1=0$, proving mutual <orthogonality>. The <eigenvalues> sum to six and have product zero, agreeing with the <matrix trace> and <determinant> from part (a).
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