Mathematics: the matrix has eigenvalues 0, 2, and 4. (2 0 2 0 2 0 2 0 2) determine corresponding eigenvectors.

the matrix has eigenvalues 0, 2, and 4. (2 0 2 0 2

Ответ:

gehdmoe2

31.12.2021

The eigenvectors of the matrix are $\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}$ , $\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}$ and $\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}$ .

Step-by-step explanation:

The given matrix is

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

It is given that the matrix has eigenvalues 0, 2, and 4.

For $\lambda=0$

$\left(A-\lambda I\right)=\begin{bmatrix}2&0&2\\ 0&2&0\\ 2&0&2\end{bmatrix}-0\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}=\begin{bmatrix}2&0&2\\ 0&2&0\\ 2&0&2\end{bmatrix}$

$\begin{bmatrix}2&0&2\\ 0&2&0\\ 2&0&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

Using Row operations we get

$\begin{bmatrix}1&0&1\\ 0&1&0\\ 0&0&0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

$x+z=0\\ y=0$

$y=0\\ x=-z$

$\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-z\\ 0\\ z\end{bmatrix}\space\space\:z\neq 0$

Let z=1,

$\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}$

At $\lambda=0$ eigen vector is $\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}$ .

Similarly,

For $\lambda=2$

$\left(A-\lambda I\right)=\begin{bmatrix}2&0&2\\ 0&2&0\\ 2&0&2\end{bmatrix}-2\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}=\begin{bmatrix}0&0&2\\ 0&0&0\\ 2&0&0\end{bmatrix}$

$\begin{bmatrix}0&0&2\\ 0&0&0\\ 2&0&0\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

Using row operations.

$\begin{bmatrix}1&0&0\\ 0&0&1\\ 0&0&0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

$x=0\\ z=0$

At $\lambda=2$ eigen vector is $\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}$ .

For $\lambda=4$

$\left(A-\lambda \right)=\begin{bmatrix}2&0&2\\ 0&2&0\\ 2&0&2\end{bmatrix}-4\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}=\begin{bmatrix}-2&0&2\\ 0&-2&0\\ 2&0&-2\end{bmatrix}$

$\begin{bmatrix}-2&0&2\\ 0&-2&0\\ 2&0&-2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

Using row operations.

$\begin{bmatrix}1&0&-1\\ 0&1&0\\ 0&0&0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

$x-z=0\\ y=0$

$y=0\\ x=z$

$\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}z\\ 0\\ z\end{bmatrix}\space\space\:z\neq 0$

at z=1

$\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}$

At $\lambda=4$ eigen vector is $\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}$ .

Therefore the eigenvectors of the matrix are $\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}$ , $\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}$ and $\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}$ .