Answer
True
Work Step by Step
The given statement is a part of the Theory 7.6.1 with
$$(A-\lambda I)^pv=0$$
By setting $p = 1$, we see that every eigenvector of $A$ is a generalized eigenvector of $A$.
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 - 7.6 Jordan Canonical Forms - True-False Review - Page 486: a
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