Back to Finding eigenvectors numerically

Exercise 7

Avg difficulty 2.5

Suppose that we apply the power method to the matrix AA with an initial vector x0\xvec_0 and find the eigenvalue λ=3\lambda=3 and eigenvector v\vvec. Suppose that we then apply the power method again with a different initial vector and find the same eigenvalue λ=3\lambda=3 but a different eigenvector w\wvec. What can we conclude about the matrix AA in this case?

Austin, Understanding Linear Algebra, §5.2 Finding eigenvectors numerically, Exercise 7

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Adapted from Understanding Linear Algebra © 2023 David Austin. Licensed under CC BY 4.0. Source: Understanding Linear Algebra. Imported at revision 3e6dced.