Answer to Question #197045 in Linear Algebra for Abhishek

Question

Find the eigenvalues and eigenvectors of the matrices

1) [ 9 3 ]

2 9

2) 2 0 1

[ 0 2 0 ]

1 0 2

Answer 1

(1) $\begin{bmatrix} 9 & 3 \\ 2 & 9 \end{bmatrix}$

Start from forming a new matrix by subtracting λ

 from the diagonal entries of the given matrix: 

$\left[\begin{array}{cc}9 - \lambda & 3\\2 & 9 - \lambda\end{array}\right]$

The determinant of the obtained matrix is $\lambda^{2} - 18 \lambda + 75$

Solve the equation

The roots are comes to be $\lambda_{1} = 9 - \sqrt{6}$ , $\lambda_{2} = \sqrt{6} + 9$

These are the eigenvalues.

Next, find the eigenvectors.

a. $\lambda = 9 - \sqrt{6}$

$\left[\begin{array}{cc}9 - \lambda & 3\\2 & 9 - \lambda\end{array}\right] = \left[\begin{array}{cc}\sqrt{6} & 3\\2 & \sqrt{6}\end{array}\right]$

The null space of this matrix is$\left\{\left[\begin{array}{c}- \frac{\sqrt{6}}{2}\\1\end{array}\right]\right\}$

This is the eigenvector.

b. $\lambda = \sqrt{6} + 9$

$\left[\begin{array}{cc}9 - \lambda & 3\\2 & 9 - \lambda\end{array}\right] = \left[\begin{array}{cc}- \sqrt{6} & 3\\2 & - \sqrt{6}\end{array}\right]$

The null space of this matrix is $\left\{\left[\begin{array}{c}\frac{\sqrt{6}}{2}\\1\end{array}\right]\right\}$

This is the eigenvector.

(2)

$\begin{bmatrix} 2 & 0 & 1 \\ 0& 2 & 0 \\ 1 & 0 & 2 \end{bmatrix}$

as above done we well proceed the same

Start from forming a new matrix by subtracting λ

λ from the diagonal entries of the given matrix: 

$\left[\begin{array}{ccc}2 - \lambda & 0 & 1\\0 & 2 - \lambda & 0\\1 & 0 & 2 - \lambda\end{array}\right].$

The determinant of the obtained matrix is 

$- \lambda^{3} + 6 \lambda^{2} - 11 \lambda + 6$

Solve the equation $\lambda_{1} = 3$ , $\lambda_{2} = 2$ , $\lambda_{3} = 1$

These are the eigenvalues.

Next, find the eigenvectors.

a. $\lambda = 3$

$\left[\begin{array}{ccc}2 - \lambda & 0 & 1\\0 & 2 - \lambda & 0\\1 & 0 & 2 - \lambda\end{array}\right] = \left[\begin{array}{ccc}-1 & 0 & 1\\0 & -1 & 0\\1 & 0 & -1\end{array}\right]$

The null space of this matrix is $\left\{\left[\begin{array}{c}1\\0\\1\end{array}\right]\right\}$

This is the eigenvector.

b. $\lambda = 2$

$\left[\begin{array}{ccc}2 - \lambda & 0 & 1\\0 & 2 - \lambda & 0\\1 & 0 & 2 - \lambda\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 1\\0 & 0 & 0\\1 & 0 & 0\end{array}\right]$

The null space of this matrix is$\left\{\left[\begin{array}{c}0\\1\\0\end{array}\right]\right\}$

This is the eigenvector.

c. $\lambda = 1$

$\left[\begin{array}{ccc}2 - \lambda & 0 & 1\\0 & 2 - \lambda & 0\\1 & 0 & 2 - \lambda\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 0\\1 & 0 & 1\end{array}\right]$

The null space of this matrix is$\left\{\left[\begin{array}{c}-1\\0\\1\end{array}\right]\right\}$

This is the eigenvector.