Question

Obtain the eigenvalues and eigenvectors of the matrix:
M= [2 3 0]
[3 2 0]
[0 0 1]

Answer 1

Answer to Question #84479 - Math - Linear Algebra

Question:

Obtain the eigenvalues and eigenvectors of the matrix: $M = \begin{bmatrix} 2 & 3 & 0 \\ 3 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

Solution:

The characteristic equation is $|M - \lambda I| = 0$, where the roots $\lambda$ are called the eigenvalues of $M$.

$$\left| M - \lambda I \right| = \left| \begin{array}{ccc} 2 - \lambda & 3 & 0 \\ 3 & 2 - \lambda & 0 \\ 0 & 0 & 1 - \lambda \end{array} \right| = 0.$$ $$(1 - \lambda) \left\{ (2 - \lambda)^2 - 9 \right\} = 0 \Rightarrow (1 - \lambda) (4 - 4\lambda + \lambda^2 - 9) = 0 \Rightarrow (1 - \lambda) (\lambda^2 - 4\lambda - 5) = 0.$$ $$(1 - \lambda) (\lambda + 1) (\lambda - 5) = 0.$$

The eigenvalues are $\lambda = -1, \lambda = 1, \lambda = 5$.

The eigenvectors $X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$ are obtained by solving $(M - \lambda I)X = 0$ for each $\lambda$.

When $\lambda = -1$, $(M - \lambda I)X = \begin{pmatrix} 2 + 1 & 3 & 0 \\ 3 & 2 + 1 & 0 \\ 0 & 0 & 1 + 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \Rightarrow \begin{pmatrix} 3x_1 + 3x_2 \\ 3x_1 + 3x_2 \\ 2x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$.

$$x_1 + x_2 = 0, x_3 = 0. \text{ Take } x_2 = 1.$$

Hence the eigenvector corresponding to $\lambda = -1$ is $X = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$.

When $\lambda = 1$, $(M - \lambda I)X = \begin{pmatrix} 2 - 1 & 3 & 0 \\ 3 & 2 - 1 & 0 \\ 0 & 0 & 1 - 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \Rightarrow \begin{pmatrix} x_1 + 3x_2 \\ 3x_1 + x_2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$.

$$x_1 = 0, x_2 = 0, x_3 = 1.$$

Hence the eigenvector corresponding to $\lambda = 1$ is $X = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

When $\lambda = 5$ , $(M - \lambda I)X = \begin{pmatrix} 2 - 5 & 3 & 0 \\ 3 & 2 - 5 & 0 \\ 0 & 0 & 1 - 5 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \Rightarrow \begin{pmatrix} -3x_1 + 3x_2 \\ 3x_1 - 3x_2 \\ -4x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ .

$x_{1} - x_{2} = 0, x_{3} = 0$ . Take $x_{2} = 1$ .

Hence the eigenvector corresponding to $\lambda = 5$ is $X = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ .

Answer provided by https://www.AssignmentExpert.com