Answer
See solution
Work Step by Step
$Ax=\lambda x$
$A(\mu x)=\mu Ax=(\mu\lambda) x$
Therefore, for each nonzero complex scalar $\mu$, the vector $\mu x$ is an eigenvector of A corresponding to eigenvalue $\mu\lambda$.
Linear Algebra and Its Applications (5th Edition)
by Lay, David C.; Lay, Steven R.; McDonald, Judi J.
Published by
Pearson
ISBN 10:
032198238X
ISBN 13:
978-0-32198-238-4
Chapter 5 - Eigenvalues and Eigenvectors - 5.5 Exercises - Page 302: 22
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