Verify that \mathrm{K}=\left(\begin{array}{r}1 \\ -1 \\ 1\end{array}\right) is an eigenvectorof the 3 \times 3 matrix
\mathrm{A}=\left(\begin{array}{rrr} 0 & -1 & -3 \\ 2 & 3 & 3 \\ -2 & 1 & 1 \end{array}\right) .
By carrying out the multiplication AK we see
\mathrm{A K}=\left(\begin{array}{rrr} 0 & -1 & -3 \\ 2 & 3 & 3 \\ -2 & 1 & 1 \end{array}\right)\left(\begin{array}{r} 1 \\ -1 \\ 1 \end{array}\right)=\left(\begin{array}{r} -2 \\ 2 \\ -2 \end{array}\right)=(-2)\left(\begin{array}{r} 1 \\ -1 \\ 1 \end{array}\right)=\left.\begin{array}{c} \text{eigenvalue}\\\downarrow \quad \\ (-2) \mathrm{K}. \end{array}\right.
We see from the preceding line and Definition 8.8.1 that \lambda=-2 is an eigenvalue of \mathrm{A}.