= Solution
The <characteristic polynomial> is
$$
\det(A-\lambda I)=(2-\lambda)(10-\lambda)-9
=\lambda^2-12\lambda+11=(\lambda-1)(\lambda-11).
$$
Thus the <eigenvalues> are 1 and 11. Corresponding <eigenvectors> are $(3,-1)^T$ and $(1,3)^T$. Their dot product is zero and each has <Euclidean norm> $\sqrt{10}$. Normalizing gives
$$
\boxed{P=\frac1{\sqrt{10}}\begin{pmatrix}3&1\\-1&3\end{pmatrix},\qquad
D=\begin{pmatrix}1&0\\0&11\end{pmatrix}.}
$$
The columns are <orthonormal>, so $P^TP=I$. The <eigenvector> equations give $AP=PD$ and hence $P^TAP=D$, proving the claimed <orthogonal diagonalization of a real symmetric matrix>.
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