Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 7 - Eigenvalues and Eigenvectors - 7.3 Diagonalization - Problems - Page 461: 35

Answer

See below

Work Step by Step

We have: $S^{-1}AS=J_\lambda$ $\rightarrow AS=SJ_\lambda \\ \rightarrow A\begin{bmatrix} v_1,v_2 \end{bmatrix}=\begin{bmatrix} v_1,v_2 \end{bmatrix}\begin{bmatrix} \lambda & 1\\ 0 & \lambda \end{bmatrix} \\ \rightarrow \begin{bmatrix} Av_1, Av_2 \end{bmatrix}=\begin{bmatrix} \lambda v_1,v_1+\lambda v_2 \end{bmatrix} \\ \rightarrow Av_1=\lambda v_1\\ Av_2=v_1+\lambda v_2 \\ \rightarrow (A-\lambda)v_1=0\\ (A-\lambda)v_2=v_1$
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