Question

Find the Eigen values and Eigen vectors of the following matrices. Do some problems by hand to be sure you understand what the process means. Then check your results by computer.

$\left(\begin{array}{cc}1& 3\\ 2& 2\end{array}\right)$

Short answer

The Eigen values of matrix A are, $\lambda =-1,\lambda =4$and the corresponding Eigen vectors $X_1=\left(\begin{array}{c}3\\ -2\end{array}\right)$and$X_2=\left(\begin{array}{c}1\\ 1\end{array}\right)$

Step-by-step solution

Given information

The given matrix is $A=\left(\begin{array}{cc}1& 3\\ 2& 2\end{array}\right)$.

Definition of Eigen values and Eigen vectors

Eigen values are the special set of scalar values that is associated with the set of linear equations most probably in the matrix equations

An Eigen vector or characteristic vector of a linear transformation is a nonzero vector that changes at most by a scalar factor when that linear transformation is applied to it.

The roots of characteristic equation$\left|A-\lambda l\right|\mathbf{=}\mathbf{0}$of matrix A are known as the Eigen values of matrix A(I the unit matrix of same order as of A. The eigenvector corresponding to Eigen value$\mathit{\lambda }$is given by $\mathbf{|}\mathbf{A}\mathbf{-}\mathbf{\lambda }\mathbf{,}\mathbf{l}\mathbf{|}{\mathbf{X}}_{\mathbf{i}}\mathbf{=}\mathbf{0}$where Ois null matrix.

Find the Eigen values of the given function 

The characteristic matrix of matrix is$A\mathrm{is}(A-\lambda l)$

$$\begin{gathered} A=\left(\begin{array}{cc}1& 3\\ 2& 2\end{array}\right)-\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}1-\lambda & 3\\ 2& 2-\lambda \end{array}\right) \end{gathered}$$

$\left(\begin{array}{cc}1-\lambda & 3\\ 2& 2-\lambda \end{array}\right)=0$

The characteristic equation of the matrix Ais

$$\begin{gathered} (1-\lambda )(2-\lambda )-6=0 \\ {\lambda }^2-3\lambda -4=0 \\ (\lambda -4)(\lambda +1)=0 \\ \lambda =4,-1 \end{gathered}$$

Hence, the Eigen values of matrix A are $\lambda =-1$and $\lambda =4$.

Find the Eigen vectors of the given function

Let $X_1=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)$be the eigenvector corresponding to Eigen value ${\lambda }_1=-1$then $\left(a-x_{1}l\right)$

$$\begin{gathered} (A-(-1\left)l\right)X_1=0\left(\left(\begin{array}{cc}1& 3\\ 2& 2\end{array}\right)-(-1)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right) \\ \left(\begin{array}{cc}2& 3\\ 2& 3\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{c}2x_1+3x_2\\ 2x_1+3x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

This gives

role="math" localid="1658813802279" $$\begin{gathered} 2x_1+3x_2=0 \\ 2x_1=-3x_2 \end{gathered}$$

And

$$\begin{gathered} 2x_1+3x_2=0 \\ 2x_1=-3x_2 \end{gathered}$$

If $x_1=3$then $x_2=-2$

Hence be the eigen vector corresponding to Eigen value $\lambda =-1$is $X_1=\left(\begin{array}{c}3\\ .2\end{array}\right)$.

Let $X_2=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)$be the Eigen vector corresponding to Eigen value ${\lambda }_2=4$

Then $\left(A-{\lambda }_{2}l\right)X_2=0$

role="math" localid="1658814279291" $$\begin{gathered} \left(A-3l\right)x_2=0\left[\left(\begin{array}{cc}1& 3\\ 2& 2\end{array}\right)-4\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right] \\ \left(\begin{array}{c}{x}_1\\ x_0\end{array}\right)=\left(\begin{array}{cc}-3& 3\\ 2& -2\end{array}\right) \\ \left(\begin{array}{c}{x}_1\\ x_0\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{c}-3x_1+3x_2\\ 2x_1-2x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

Hence

$-3x_1+3x_2=0$

$$\begin{gathered} x_1=x_2\mathrm{and}{}_{x_1=x_2}^{2x_1=2x_2=0} \\ \mathrm{If}{x}_1=l\mathrm{then}{x}_2=1 \end{gathered}$$

Hence be the eigenvector corresponding to Eigen value $\lambda =4$is $X_2=\left(\begin{array}{c}1\\ 1\end{array}\right)$.