Question

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

Answer 1

Answer:

Eigenvalues : 4 and 2

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

Step-by-step 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 $=\begin{bmatrix}y\\ y\end{bmatrix}\space\space\:y\ne \:0$

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 $=\begin{bmatrix}-y\\ y\end{bmatrix}\space\space\:y\ne \:0$

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