Solution

An orthonormal eigenbasis is

v_{9} = \frac{1}{3} (1, 1, 1)^{T}, v_{6} = \frac{1}{6} (1, - 2, 1)^{T}, v_{2} = \frac{1}{2} (1, 0, - 1)^{T} .

The starting vector has the orthogonal decomposition

x_{0} = 0 v_{9} + \frac{3}{2} v_{6} + \frac{1}{2} v_{2} .

Hence its component in the dominant eigenspace is exactly zero. Normalizing

A^{k} x_{0}

gives

x_{k} = \frac{6 ^{k} ( 1 , - 2 , 1 ) ^{T} + 2 ^{k} ( 1 , 0 , - 1 ) ^{T}}{2 3 \cdot 3 6 ^{k} + 4 ^{k}} .

Since the two remaining eigenvectors are orthogonal,

r (x_{k}) = \frac{18 \cdot 3 6 ^{k} + 2 \cdot 4 ^{k}}{3 \cdot 3 6 ^{k} + 4 ^{k}} .

Therefore

k \to \infty lim r (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 (x_{k}) - 6 = - \frac{4 \cdot 4 ^{k}}{3 \cdot 3 6 ^{k} + 4 ^{k}} = O (3^{- 2 k}),

in agreement with the squared spectral-ratio estimate.

Solved by gpt-5.6-sol high.

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