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 5E from Chapter 7.SE from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
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Step-by-Step Solution
Given Information
We have to show that if $\mathbf{v}$ is an eigenvector of an $n \times n$ matrix $A$ and $\mathbf{v} $ corresponds to a nonzero eigenvalue of $A,$ then $\mathbf{v}$ is in $\operatorname{Col} A $
Step-1:
By the definition of eigenvalue and eigenvector:\[A \mathbf{v}=\lambda \mathbf{v}\]Multiply both sides by inverse of $\lambda$\[\begin{array}{l}{\lambda ^{ - 1}}A{\bf{v}} = {\lambda ^{ - 1}}\lambda {\bf{v}}\\{\lambda ^{ - 1}}A{\bf{v}} = {\bf{v}}\\{\bf{v}} = A\left( {{\lambda ^{ - 1}}{\bf{v}}} \right)\end{array}\]This follows that $\mathbf{v}=A\left(\lambda^{-1} \mathbf{v}\right) \text { is consistent }$, hence the vector v is linear combination of columns of matrix A. Therefore,
$\mathrm{v}$ is in $\mathrm{Col} A$
We have to show that if $\mathbf{v}$ is an eigenvector of an $n \times n$ matrix $A$ and $\mathbf{v} $ corresponds to a nonzero eigenvalue of $A,$ then $\mathbf{v}$ is in $\operatorname{Col} A $
Step-1:
By the definition of eigenvalue and eigenvector:\[A \mathbf{v}=\lambda \mathbf{v}\]Multiply both sides by inverse of $\lambda$\[\begin{array}{l}{\lambda ^{ - 1}}A{\bf{v}} = {\lambda ^{ - 1}}\lambda {\bf{v}}\\{\lambda ^{ - 1}}A{\bf{v}} = {\bf{v}}\\{\bf{v}} = A\left( {{\lambda ^{ - 1}}{\bf{v}}} \right)\end{array}\]This follows that $\mathbf{v}=A\left(\lambda^{-1} \mathbf{v}\right) \text { is consistent }$, hence the vector v is linear combination of columns of matrix A. Therefore,
$\mathrm{v}$ is in $\mathrm{Col} A$