= Solution
The symmetry between the last two coordinates gives the <eigenvector> $(0,1,-1)^T$ with <eigenvalue> $2$. On vectors $(s,t,t)^T$, the remaining <eigenvalue equation> is
$$
\begin{pmatrix}3&2\\1&2\end{pmatrix}\binom st=\lambda\binom st,
$$
whose <characteristic polynomial> is $\lambda^2-5\lambda+4=(\lambda-1)(\lambda-4)$. Corresponding <eigenvectors> are $(1,-1,-1)^T$ and $(2,1,1)^T$. Thus \b[all <eigenvalues> and <eigenspaces> are]
$$
\boxed{\lambda=1:\ \operatorname{span}\{(1,-1,-1)^T\};\quad
\lambda=2:\ \operatorname{span}\{(0,1,-1)^T\};\quad
\lambda=4:\ \operatorname{span}\{(2,1,1)^T\}.}
$$
Nonzero scalar multiples give all <eigenvectors> in each <eigenspace>. The three displayed vectors are mutually <orthogonal> and have squared norms $3,2,6$, respectively.
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