Find the Eigenvector of the matrix $A=\begin{pmatrix}3 & 1 & 1\\2&4&2\\1&1&3\end{pmatrix}$ corresponding to Eigenvalue $\lambda=2 $.
The Eigen values of $A$ are $\lambda =2,2,6$. It is easy to find the Eigenvector corresponding to Eigen value $\lambda =6$, the one i find using Cramer's rule is $(1, 2, 1)$. But i am unable to find the Eigenvector corresponding to Eigenvalue $\lambda =2$ (Cramer's rule fails here).
Best Answer
It's the kernel of $A-2I$:
$\begin{pmatrix}1 &1 &1 \\ 2 &2 &2 \\ 1 &1 &1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=0$
$x+y+z=0$
Example eigenvectors are $\begin{pmatrix}1\\-1\\0\end{pmatrix}, \begin{pmatrix}1\\0\\-1\end{pmatrix},$ and $\begin{pmatrix}0\\1\\-1\end{pmatrix}$.
(Only two of those are linearly independent.)