Products

Holooly Rewards

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

Holooly Ads. Manager

Advertise your business, and reach millions of students around the world.

Holooly Tables

All the data tables that you may search for.

Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Holooly Sources

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Holooly Help Desk

Need Help? We got you covered.

Chapter 5

Q. 5.3

Let us find the eigenvalues and eigenvectors of a matrix of the second order

\pmb{A}=\begin{bmatrix} 2&1\\ 3&4 \end{bmatrix} .

Step-by-Step

Verified Solution

The characteristic polynomial is also of order two:

\det \left\{\lambda \begin{bmatrix} 1&0\\0&1 \end{bmatrix}-\begin{bmatrix} 2&1\\3&4 \end{bmatrix} \right\} =\det \begin{bmatrix} \lambda -2&-1\\-3&\lambda -4 \end{bmatrix}=\lambda^2-6\lambda +5

=(\lambda -5)(\lambda -1)=g(\lambda ).

Thus, \lambda ^{2}-6\lambda +5=0 is the characteristic equation of the matrix. The roots of the characteristic equation, or the eigenvalues, are

\lambda_1=5      and     \lambda _2=1.

To obtain the eigenvector corresponding to the eigenvalue \lambda _1=5, we solve equation 5.94 by using the given matrix \pmb{A}. Thus

[\lambda \pmb{1-A}]\pmb{X}=\pmb{0}.                                      (5.94)

\left\{\begin{bmatrix} 5&0\\05 \end{bmatrix}-\begin{bmatrix}2&1\\3&4 \end{bmatrix} \right\}\begin{bmatrix} x_1\\x_2 \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}

or

\begin{bmatrix} 3&-1\\-3&1 \end{bmatrix} \begin{bmatrix} x_1\\x_2 \end{bmatrix}= \begin{bmatrix} 0\\0 \end{bmatrix}           and       x_2 = 3 x_1.

Therefore

\begin{bmatrix} x_1\\x_2 \end{bmatrix}=\begin{bmatrix}x_1\\3x_1 \end{bmatrix} =\begin{bmatrix} 1\\3 \end{bmatrix} \begin{bmatrix} x_1 \end{bmatrix} for any value of x_1.

The eigenvector corresponding to the eigenvalue \lambda_2=1  is obtained similarly.

\begin{bmatrix} -1&-1\\-3&-3 \end{bmatrix} \begin{bmatrix}x_1\\x_2 \end{bmatrix}=\begin{bmatrix}0\\0 \end{bmatrix}

from which

\begin{bmatrix} x_1\\x_2 \end{bmatrix}=\begin{bmatrix} x_1\\-x_1 \end{bmatrix}= \begin{bmatrix}1\\-1 \end{bmatrix} \begin{bmatrix}x_1 \end{bmatrix} for any value of x_1.

The first method to be discussed for finding functions of a matrix is based on the Caley-Hamilton theorem.