\[For~matrix~A = \left[ \begin{array}{ccc} 2 & 1 & 1 \\ 1 & 2 & 1\\1 & 1 &2\end{array}\right],~the~characteristic~polynomial~of~A~is~p_A(x)=(4-x)(1-x)^2.~Find~the~eigenvalues~and~corresponding~eigenspaces~of~A.\]
\[Using~the~rule~of~Sarrus\left[\begin{array}{ccc} \lambda-2 & -1 & -1 \\ -1 & \lambda-2 & -1\\-1 & -1 &\lambda-2\end{array}\right]\left[ \begin{array}{ccc} \lambda-2 & -1 \\ -1 & \lambda-2\\-1 & -1 \end{array}\right]\]
\[(\lambda-2)(\lambda-2)(\lambda-2)-1-1-(\lambda-2)-(\lambda-2)-(\lambda-2)\] \[Simplfies~to:~p(\lambda)=\lambda^3-6\lambda^2+9\lambda-2=0\] \[With~eigenvalues~of~\lambda=2,~\lambda=2-\sqrt3,~\lambda=2+\sqrt3\]
\[Now~solve~for~eigenvectors~where~\lambda=3,\left[ \begin{array}{ccc} 1 & -1 & -1 \\ -1 & 1 & -1\\-1 & -1 &1\end{array}\right]\vec{v}=\vec{0}\]
\[Row~Reduce\left[ \begin{array}{ccc} 1 & -1 & -1 \\ 0 & 0 & 0\\0 & 0 &0\end{array}\right]\]
\[v_1-v_2-v_3=0; v_1=v_2+v_3,~v_2=a,~v_3=b\]
\[E=\left[ \begin{array}{ccc} v_1 \\ v_2 \\v_3 \end{array}\right]=a\left[ \begin{array}{ccc} 1 \\ 1 \\0 \end{array}\right]+b\left[ \begin{array}{ccc} 1 \\ 0 \\1 \end{array}\right]\]