= Solution
The <vector> $(1,-1,0)^T$ is an <eigenvector> with <eigenvalue> $4$. The complementary subspace of <vectors> with first two components equal is preserved by $A$: a vector $(s,s,t)^T$ maps to $(4s+2t,4s+2t,4s+2t)^T$. There, the <eigenvalue> equations reduce to the $2\times2$ <matrix> $\begin{pmatrix}4&2\\4&2\end{pmatrix}$, whose <characteristic polynomial> is $\lambda(\lambda-6)$. For $\lambda=6$, $t=s$; for $\lambda=0$, $t=-2s$.
Normalising by the <Euclidean norm>, a complete <orthonormal basis> of <eigenvectors> is
$$
\boxed{\begin{array}{c|c}
\lambda&\text{unit eigenvector}\\\hline
0&\mathbf e_0=(1,1,-2)^T/\sqrt6\\
4&\mathbf e_4=(1,-1,0)^T/\sqrt2\\
6&\mathbf e_6=(1,1,1)^T/\sqrt3
\end{array}}
$$
Sign changes of any of these <eigenvectors> are equally valid. The <eigenvalues> also check part (a): their sum is $10$ and the sum of their squares is $52$.
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