(10) Determine the eigen values/vectors of the matrix43 -2

Consider the matrix A=243−2 First we will find eigen value of the matrix A The eigenvalues of A are…

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Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter7: Eigenvalues And Eigenvectors

Section7.1: Eigenvalues And Eigenvectors

Problem 66E: Show that A=[0110] has no real eigenvalues.

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Question

(10) Determine the eigen values/vectors of the matrix
4
3 -2

Expert Solution

Step 1

Consider the matrix $A = (\begin{matrix} 2 & 4 \\ 3 & - 2 \end{matrix})$

First we will find eigen value of the matrix $A$

The eigenvalues of $A$ are the roots of the characteristic equation $d e t (A - λ I) = 0$

Hence,

$\begin{matrix} \det (A - λ I) = 0 \\ \det ((\begin{matrix} 2 & 4 \\ 3 & - 2 \end{matrix}) - λ (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})) = 0 \\ \det ((\begin{matrix} 2 & 4 \\ 3 & - 2 \end{matrix}) - (\begin{matrix} λ & 0 \\ 0 & λ \end{matrix})) = 0 \\ \det ((\begin{matrix} 2 - λ & 4 \\ 3 & - 2 - λ \end{matrix})) = 0 \end{matrix}$

Now,

$\begin{matrix} (2 - λ) (- 2 - λ) - 12 = 0 \\ - 4 - 2 λ + 2 λ + λ^{2} = 0 \\ λ^{2} - 4 = 0 \\ λ = 4 and - 4 \end{matrix}$

Therefore, eigenvalues of the matrix $A = (\begin{matrix} 2 & 4 \\ 3 & - 2 \end{matrix})$ is $λ_{1} = 4 and λ_{2} = - 4$ .

Step 2

Now, we will find the eigenvector corresponding to eigenvalues $λ_{1} = 4 and λ_{2} = - 4$ of the given matrix $A$

We know that the eigenvector corresponding to every eigenvalue $λ$ is given by $(A - λ I) x = 0$

Hence, the eigenvector corresponding to every eigenvalue $λ_{1} = 4$ is given by $(A - 4 I) x = 0$

Now, solve $(A - 4 I) x = 0$

$\begin{matrix} (A - 4 I) x = 0 \\ ((\begin{matrix} 2 & 4 \\ 3 & - 2 \end{matrix}) - 4 (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})) (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \\ ((\begin{matrix} 2 & 4 \\ 3 & - 2 \end{matrix}) - (\begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix})) (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \\ (\begin{matrix} - 2 & 4 \\ 3 & - 6 \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \\ (\begin{matrix} - 2 x_{1} + 4 x_{2} \\ 3 x_{1} - 6 x_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \end{matrix}$

Hence, $\{\begin{array}{l} - 2 x_{1} + 4 x_{2} = 0 \\ 3 x_{1} - 6 x_{2} = 0 \end{array}\} ..... (1)$

Now, solve the system (1), we get

$x_{1} = 2 and x_{2} = 1$

Hence, the eigenvector corresponding to every eigenvalue $λ_{1} = 4$ is $x = (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} 2 \\ 1 \end{matrix})$

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