Respuesta :
Answer:
The eigenvectors of the matrix are [tex]\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}[/tex],[tex]\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}[/tex] and [tex]\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}[/tex].
Step-by-step explanation:
The given matrix is
[tex]\begin{bmatrix}2&0&2\\ \:0&2&0\\ \:2&0&2\end{bmatrix}[/tex]
It is given that the matrix has eigenvalues 0, 2, and 4.
For [tex]\lambda=0[/tex]
[tex]\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}[/tex]
[tex]\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}[/tex]
Using Row operations we get
[tex]\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}[/tex]
[tex]x+z=0\\ y=0[/tex]
[tex]y=0\\ x=-z[/tex]
[tex]\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-z\\ 0\\ z\end{bmatrix}\space\space\:z\neq 0[/tex]
Let z=1,
[tex]\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}[/tex]
At [tex]\lambda=0[/tex] eigen vector is [tex]\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}[/tex].
Similarly,
For [tex]\lambda=2[/tex]
[tex]\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}[/tex]
[tex]\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}[/tex]
Using row operations.
[tex]\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}[/tex]
[tex]x=0\\ z=0[/tex]
At [tex]\lambda=2[/tex] eigen vector is [tex]\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}[/tex].
For [tex]\lambda=4[/tex]
[tex]\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}[/tex]
[tex]\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}[/tex]
Using row operations.
[tex]\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}[/tex]
[tex]x-z=0\\ y=0[/tex]
[tex]y=0\\ x=z[/tex]
[tex]\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}z\\ 0\\ z\end{bmatrix}\space\space\:z\neq 0[/tex]
at z=1
[tex]\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}[/tex]
At [tex]\lambda=4[/tex] eigen vector is [tex]\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}[/tex].
Therefore the eigenvectors of the matrix are [tex]\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}[/tex],[tex]\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}[/tex] and [tex]\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}[/tex].
