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An orthonormal eigenbasis isThe starting vector has the orthogonal decompositionHence its component in the dominant eigenspace is exactly zero. Normalizing $A^k{\mathbf{x}}_0$ givesSince the two remaining eigenvectors are orthogonal,ThereforeThe 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 $2$; indeedin agreement with the squared spectral-ratio estimate.