Answer to Question #227528 in Linear Algebra for umme habiba sumi

Question

Find the eigen value and eigen vector of A = [ 3 3 1 5 ] 

Answer 1

$$A=\begin{bmatrix} 3 & 3 \\ 1 & 5 \end{bmatrix}$$

$$\det(A-\lambda I)=\begin{vmatrix} 3-\lambda & 3 \\ 1 & 5-\lambda \end{vmatrix}=0$$

$$( 3-\lambda)( 5-\lambda)-3(1)=0$$

$$15-8\lambda+\lambda^2-3=0$$

$$\lambda^2-8\lambda+12=0$$

$$(\lambda-2)(\lambda-6)=0$$

The eigenvalues are $\lambda_1=2, \lambda_2=6.$

$\lambda=2$

$$A-\lambda I=\begin{bmatrix} 3-\lambda & 3 \\ 1 & 5-\lambda \end{bmatrix}=\begin{bmatrix} 1 & 3 \\ 1 & 3 \end{bmatrix}$$

$R_2=R_2-R_1$

$$\begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix}$$

Solve

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

If we take $x_2=t,$ then $x_1=-3t.$

The eigenvector is $\vec x=\begin{bmatrix} -3 \\ 1 \end{bmatrix}$

$\lambda=6$

$$A-\lambda I=\begin{bmatrix} 3-\lambda & 3 \\ 1 & 5-\lambda \end{bmatrix}=\begin{bmatrix} -3 & 3 \\ 1 & -1 \end{bmatrix}$$

$R_1=-R_1/3$

$$\begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}$$

$R_2=R_2-R_1$

$$\begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}$$

Solve

$$\begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} y_1 \\ y_2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

If we take $y_2=t,$ then $y_1=t.$

The eigenvector is $\vec y=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Eigenvalue: $2,$ eigenvector: $\begin{bmatrix} -3 \\ 1 \end{bmatrix}.$

Eigenvalue: $6,$ eigenvector: $\begin{bmatrix} 1 \\ 1 \end{bmatrix}.$