= Solution
The $y$ coordinate is an independent eigen-direction, while the $xz$ block is $\begin{pmatrix}2&1\\1&2\end{pmatrix}$. Thus
$$
\det(A-\lambda I)=(1-\lambda)^2(3-\lambda),
$$
which gives $\det(A-I)=0$. The <eigenspaces> are
$$
\boxed{\lambda=1:\ \operatorname{span}\{(1,0,-1),(0,1,0)\};\qquad
\lambda=3:\ \operatorname{span}\{(1,0,1)\}.}
$$
Divide the two diagonal $xz$ vectors by $\sqrt2$ to obtain an <orthonormal eigenbasis>. The eigenvalue one has multiplicity two; giving only the $y$ axis would omit an independent eigenvector.
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