Answer
See below
Work Step by Step
Let $(x,y,z) \in R^3$
Since any vector of the form $(0,y,0)$ is mapped to itself, then $\lambda_1=1$
Any vector of the form $(x,0,0)$ is mapped to itself, then $\lambda_2=0$
Differential Equations and Linear Algebra (4th Edition)
by Goode, Stephen W.; Annin, Scott A.
Published by
Pearson
ISBN 10:
0-32196-467-5
ISBN 13:
978-0-32196-467-0
Chapter 7 - Eigenvalues and Eigenvectors - 7.1 The Eigenvalue/Eigenvector Problem - Problems - Page 443: 11
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