Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 3E from Chapter 5.SE from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 3E
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Step-by-Step Solution
Given Information
Given that $\mathbf{x}$ is an eigenvector of $A$ corresponding to an eigen value $\lambda .$
(a) We have to show that x is an eigenvector of $5I-A$ and
(b) We have to show that x is an eigenvector of $5I-3A+A^2$ and
We also have to find the corresponding eigenvectors.
Step-1: (a)
Simplify the equation: using $A \mathbf{x}=\lambda \mathbf{x}$ and $I(\mathbf{x})=\mathbf{x}$\[\begin{aligned}(5 I-A) \mathbf{x} &=5 I(\mathbf{x})-A \mathbf{x} \\ &=5 \mathbf{x}-\lambda \mathbf{x} \\ &=(5-\lambda) \mathbf{x} \end{aligned}\]Therefore
$x$ is an eigenvector of $5 I-A$ and the corresponding eigenvalue is $[5-\lambda]$
Step-2: (b)
\[\begin{array}{l}\left( {5I - 3A + {A^2}} \right){\bf{x}} = 5I({\bf{x}}) - 3A{\bf{x}} + {A^2}{\bf{x}}\\ = 5{\bf{x}} - 3\lambda {\bf{x}} + A(A{\bf{x}})\\ = 5{\bf{x}} - 3\lambda {\bf{x}} + A(\lambda {\bf{x}})\\ = 5{\bf{x}} - 3\lambda {\bf{x}} + \lambda (A{\bf{x}})\\ = 5{\bf{x}} - 3\lambda {\bf{x}} + \lambda (\lambda {\bf{x}})\\ = 5{\bf{x}} - 3\lambda {\bf{x}} + {\lambda ^2}{\bf{x}}\\ = \left( {5 - 3\lambda + {\lambda ^2}} \right){\bf{x}}\end{array}\]Therefore,
$x$ is an eigenvector of $5 I-3 A+A^{2}$ and the corresponding eigenvalue is $\left(5-3 \lambda+\lambda^{2}\right)$
Given that $\mathbf{x}$ is an eigenvector of $A$ corresponding to an eigen value $\lambda .$
(a) We have to show that x is an eigenvector of $5I-A$ and
(b) We have to show that x is an eigenvector of $5I-3A+A^2$ and
We also have to find the corresponding eigenvectors.
Step-1: (a)
Simplify the equation: using $A \mathbf{x}=\lambda \mathbf{x}$ and $I(\mathbf{x})=\mathbf{x}$\[\begin{aligned}(5 I-A) \mathbf{x} &=5 I(\mathbf{x})-A \mathbf{x} \\ &=5 \mathbf{x}-\lambda \mathbf{x} \\ &=(5-\lambda) \mathbf{x} \end{aligned}\]Therefore
$x$ is an eigenvector of $5 I-A$ and the corresponding eigenvalue is $[5-\lambda]$
Step-2: (b)
\[\begin{array}{l}\left( {5I - 3A + {A^2}} \right){\bf{x}} = 5I({\bf{x}}) - 3A{\bf{x}} + {A^2}{\bf{x}}\\ = 5{\bf{x}} - 3\lambda {\bf{x}} + A(A{\bf{x}})\\ = 5{\bf{x}} - 3\lambda {\bf{x}} + A(\lambda {\bf{x}})\\ = 5{\bf{x}} - 3\lambda {\bf{x}} + \lambda (A{\bf{x}})\\ = 5{\bf{x}} - 3\lambda {\bf{x}} + \lambda (\lambda {\bf{x}})\\ = 5{\bf{x}} - 3\lambda {\bf{x}} + {\lambda ^2}{\bf{x}}\\ = \left( {5 - 3\lambda + {\lambda ^2}} \right){\bf{x}}\end{array}\]Therefore,
$x$ is an eigenvector of $5 I-3 A+A^{2}$ and the corresponding eigenvalue is $\left(5-3 \lambda+\lambda^{2}\right)$