Question

Find the eigenvectors and eigenvalues for the matrix.
A =
[3 1
1 3]

Answer 1

Eigenvalues : 4 and 2

Eigenvectors : <1,1> and <-1,1>.

Explanation:

The given matrix is

`A=\begin{bmatrix}3&1\\1&3\end{bmatrix}`

We need to find the eigenvectors and eigenvalues for the matrix.

`|A-\lambda I|=0`

λ represents the eigen values.

`\det \left(\begin{bmatrix}3&1\\1&3\end{bmatrix}-\lambda\begin{bmatrix}1&0\\ 0&1\end{bmatrix}\right)=0`

`\lambda^2-6\lambda+8=0`

`\lambda=4,2`

For
`\lambda=4`

`(A-\lambda I)X=0`

`\left(\begin{bmatrix}3&1\\1&3\end{bmatrix}-4\begin{bmatrix}1&0\\ 0&1\end{bmatrix}\right)\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}`

`\begin{bmatrix}-1&1\\ 1&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}`

`R_2\rightarrow R_2-R_1`

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

`x-y=0`

`x=y`

Eigenvector
`=\begin{bmatrix}y\\ y\end{bmatrix}\space\space\:y\\e \:0`

Eigenvector
`=\begin{bmatrix}1\\ 1\end{bmatrix}`

Similarly,

For
`\lambda=2`

`(\begin{bmatrix}3&1\\ 1&3\end{bmatrix}-2\begin{bmatrix}1&0\\ 0&1\end{bmatrix})\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}0\\ 0\end`

`\begin{bmatrix}1&1\\ 1&1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}`

`R_2\rightarrow R_2-R_1`

`\begin{bmatrix}1&1\\ 0&0\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}`

`x+y=0`

`x=-y`

Eigenvector
`=\begin{bmatrix}-y\\ y\end{bmatrix}\space\space\:y\\e \:0`

Eigenvector
`=\begin{bmatrix}-1\\ 1\end{bmatrix}`