= Solution
An orthonormal eigenbasis is
$$
\mathbf v_9=\frac1{\sqrt3}(1,1,1)^T,
\qquad
\mathbf v_6=\frac1{\sqrt6}(1,-2,1)^T,
\qquad
\mathbf v_2=\frac1{\sqrt2}(1,0,-1)^T.
$$
The starting vector has the <orthogonal decomposition by a closed subspace>[orthogonal decomposition]
$$
\mathbf x_0
=0\mathbf v_9+\frac{\sqrt3}{2}\mathbf v_6+\frac12\mathbf v_2.
$$
Hence its component in the dominant eigenspace is exactly zero. Normalizing $A^k\mathbf x_0$ gives
$$
\boxed{
\mathbf x_k
=
\frac{
6^k(1,-2,1)^T+2^k(1,0,-1)^T
}{
\sqrt2\sqrt{3\cdot36^k+4^k}
}
}.
$$
Since the two remaining eigenvectors are orthogonal,
$$
\boxed{
r(\mathbf x_k)
=\frac{18\cdot36^k+2\cdot4^k}
{3\cdot36^k+4^k}
}.
$$
Therefore
$$
\boxed{\lim_{k\to\infty}r(\mathbf x_k)=6}.
$$
The limit is not the leading eigenvalue $9$ because the nonorthogonality assumption in part (a) fails. On the <active spectrum of the power method>, the leading eigenvalue is $6$ and the next is $2$; indeed
$$
r(\mathbf x_k)-6
=-\frac{4\cdot4^k}{3\cdot36^k+4^k}
=O(3^{-2k}),
$$
in agreement with the squared spectral-ratio estimate.
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