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Found in: Page 323

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Is $\stackrel{\mathbf{⇀}}{\mathbf{v}}$ an eigenvector of ${{\mathbf{A}}}^{\mathbf{-}\mathbf{1}}$? If so, what is the eigenvalue?

Yes, the required given value is $\frac{1}{\lambda }$ .

See the step by step solution

## Step 1: Definition of eigenvalue

Eigenvalue:An eigenvalue of A is a scalar ${\mathbit{\lambda }}$ such that the equation ${\mathbf{Av}}{\mathbf{=}}{\mathbf{\lambda v}}$ has a nontrivial solution.

## Step 2: Given data

Consider a $n×n$ invertible matrix A and $\stackrel{⇀}{v}$ is an eigenvector of the matrix A with respect to the eigenvalue $\lambda$ .

The objective is to find whether the vector $\stackrel{⇀}{v}$ is eigenvector for is ${A}^{-1}$ or not. If yes find the eigenvalue.

## Step 3:Checking whether v⇀an eigenvector of A-1

Since $\stackrel{⇀}{v}$ is an eigenvector of the matrix A with respect to the eigenvalue $\lambda$ .

$A\stackrel{⇀}{v}=\lambda \stackrel{⇀}{v}$

Find whether the vector $\stackrel{⇀}{v}$ is eigenvector for is ${A}^{-1}$ or not as,

Consider

role="math" localid="1659528893977" $A\stackrel{⇀}{v}=\lambda \stackrel{⇀}{v}{A}^{-1}\left(A\stackrel{⇀}{v}\right)\phantom{\rule{0ex}{0ex}}={A}^{-1}\left(\lambda \stackrel{⇀}{v}\right)\phantom{\rule{0ex}{0ex}}{A}^{-1}\left({A}^{-1}A\right)\stackrel{⇀}{v}=\lambda \left({A}^{-1}\stackrel{⇀}{v}\right)I\stackrel{⇀}{v}\phantom{\rule{0ex}{0ex}}=\lambda \left({A}^{-1}\stackrel{⇀}{v}\right)\mathrm{sin}ce{A}^{-1}A=I\lambda \left({A}^{-1}\stackrel{⇀}{v}\right)\phantom{\rule{0ex}{0ex}}=\stackrel{⇀}{v}{A}^{-1}\stackrel{⇀}{v}=\frac{1}{\lambda }\stackrel{⇀}{v}$

Therefore, from equation (1) the vector $\stackrel{⇀}{v}$ is eigenvector for is ${A}^{-1}$ with respect to the eigenvalue ${\lambda }^{-1}=\frac{1}{\lambda }$.