Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 5 - Eigenvalues and Eigenvectors - 5.1 Exercises - Page 274: 21

Answer

a) False b) True c) True d) True e) False

Work Step by Step

a) False. The equation will be true regardless of the value of $\lambda$ if x is the zero vector, so $\lambda$ is an eigenvalue of A if and only if the equation is true for a nonzero vector x b) True. If 0 is an eigenvalue of A, it means there is a nonzero vector in the kernel. This means the matrix is not invertible. c) True by definition d) True. We can determine if a vector is an eigenvector of a matrix by left multiplying it with the matrix. If the result is a scalar multiple of itself, it is an eigenvector. To find an eigenvector, we need to find nontrivial solution(s) to $(A-\lambda I)x=0$ e) Row reducing may change eigenvalues and eigenvectors. If the matrix is already diagonal, the eigenvalues are on the main diagonal, but row reducing or diagonalizing will change the eigenvalues and eigenvectors. This about dividing a row by a scalar.
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