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## Q. 5.3

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

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

## 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.