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 - 7.6 Jordan Canonical Forms - Problems - Page 487: 31

Answer

See below

Work Step by Step

1. Find eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$ $\begin{bmatrix} -\lambda & 0 & 0 & 1 & 4\\ 0 & -\lambda & 1 & 1 & 1\\ 0 & 0 & -\lambda & 6 & 0\\ 0 & 0 & 0 & -\lambda & 0\\ 0 & 0 & 0 & 0 & -\lambda\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \\ v_4 \\ v_5 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \\0 \\ 0 \end{bmatrix}$ $\begin{bmatrix} -\lambda & 0 & 0 & 1 & 4\\ 0 & -\lambda & 1 & 1 & 1\\ 0 & 0 & -\lambda & 6 & 0\\ 0 & 0 & 0 & -\lambda & 0\\ 0 & 0 & 0 & 0 & -\lambda\end{bmatrix}=0$ $\lambda^5=0$ $\lambda_1=0$ with multiplicity 5. 2. Find eigenvectors: For $\lambda=0$ let $B=A-\lambda_1I$ $B=\begin{bmatrix} -\lambda & 0 & 0 & 1 & 4\\ 0 & -\lambda & 1 & 1 & 1\\ 0 & 0 & -\lambda & 6 & 0\\ 0 & 0 & 0 & -\lambda & 0\\ 0 & 0 & 0 & 0 & -\lambda\end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 & 1 & 4\\ 0 & 0 & 1 & 1 & 1\\ 0 & 0 & 0 & 6 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \\0 \\ 0\end{bmatrix} $ Obtain $\begin{bmatrix} 0 & 0 & 0 & 1 & 4\\ 0 & 0 & 1 & 1 & 1\\ 0 & 0 & 0 & 6 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \approx \begin{bmatrix} 0 & 0 & 1 & 1 & 1\\ 0 & 0 & 0 & 1 & 4\\ 0 & 0 & 0 & 0 & 24\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ $\rightarrow dim(E_2)=3$ Hence, $J=\begin{bmatrix} 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$
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