Question

The matrix has eigenvalues 0, 2, and 4.

(2 0 2

0 2 0

2 0 2)

Determine corresponding eigenvectors.

Answer 1

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}`.

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\\eq 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\\eq 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}`.