Respuesta :
Answer:
Eigenvalues : 4 and 2
Eigenvectors : <1,1> and <-1,1>.
Step-by-step explanation:
The given matrix is
[tex]A=\begin{bmatrix}3&1\\1&3\end{bmatrix}[/tex]
We need to find the eigenvectors and eigenvalues for the matrix.
[tex]|A-\lambda I|=0[/tex]
λ represents the eigen values.
[tex]\det \left(\begin{bmatrix}3&1\\1&3\end{bmatrix}-\lambda\begin{bmatrix}1&0\\ 0&1\end{bmatrix}\right)=0[/tex]
[tex]\lambda^2-6\lambda+8=0[/tex]
[tex]\lambda=4,2[/tex]
For [tex]\lambda=4[/tex]
[tex](A-\lambda I)X=0[/tex]
[tex]\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}[/tex]
[tex]\begin{bmatrix}-1&1\\ 1&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}[/tex]
[tex]R_2\rightarrow R_2-R_1[/tex]
[tex]\begin{bmatrix}1&-1\\ 0&0\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}[/tex]
[tex]x-y=0[/tex]
[tex]x=y[/tex]
Eigenvector [tex]=\begin{bmatrix}y\\ y\end{bmatrix}\space\space\:y\ne \:0[/tex]
Eigenvector [tex]=\begin{bmatrix}1\\ 1\end{bmatrix}[/tex]
Similarly,
For [tex]\lambda=2[/tex]
[tex](\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[/tex]
[tex]\begin{bmatrix}1&1\\ 1&1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}[/tex]
[tex]R_2\rightarrow R_2-R_1[/tex]
[tex]\begin{bmatrix}1&1\\ 0&0\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}[/tex]
[tex]x+y=0[/tex]
[tex]x=-y[/tex]
Eigenvector [tex]=\begin{bmatrix}-y\\ y\end{bmatrix}\space\space\:y\ne \:0[/tex]
Eigenvector [tex]=\begin{bmatrix}-1\\ 1\end{bmatrix}[/tex]
