Question 8.8.2: Find the eigenvalues and eigenvectors of A = (1 2 1 6 -1 0 -......

To expand the determinant in the characteristic equation

$\operatorname{det}(\mathrm{A}-\lambda \mathrm{I})=\left|\begin{array}{ccc} 1-\lambda & 2 & 1 \\ 6 & -1-\lambda & 0 \\ -1 & -2 & -1-\lambda \end{array}\right|=0,$

we use the cofactors of the second row. It follows that the characteristic equation is

$-\lambda^{3}-\lambda^{2}+12 \lambda=0 \quad \text { or } \quad \lambda(\lambda+4)(\lambda-3)=0$

Hence the eigenvalues are $\lambda_{1}=0, \lambda_{2}=-4, \lambda_{3}=3$. To find the eigenvectors, we must now reduce $(\mathrm{A}-\lambda \mathrm{I} \mid \mathrm{0})$ three times corresponding to the three distinct eigenvalues.

For $\lambda_{1}=0$, we have

Thus we see that $k_{1}=-\frac{1}{13} k_{3}$ and $k_{2}=-\frac{6}{13} k_{3}$. Choosing $k_{3}=-13$ gives the eigenvector

$\mathrm{K}_{1}=\left(\begin{array}{r} 1 \\ 6 \\ -13 \end{array}\right).$

For $\lambda_{2}=-4$,

implies $k_{1}=-k_{3}$ and $k_{2}=2 k_{3}$. Choosing $k_{3}=1$ then yields a second eigenvector

$\mathrm{K}_{2}=\left(\begin{array}{r} -1 \\ 2 \\ 1 \end{array}\right).$

Finally, for $\lambda_{3}=3$, Gauss-Jordan elimination gives

$(\mathrm{A}-3 \mathrm{I} \mid \mathrm{0})=\left(\begin{array}{rrr|r} -2 & 2 & 1 & 0 \\ 6 & -4 & 0 & 0 \\ -1 & -2 & -4 & 0 \end{array}\right) \stackrel{\substack{\text { row } \\ \text { operations }}}{\Rightarrow}\left(\begin{array}{lll|l} 1 & 0 & 1 & 0 \\ 0 & 1 & \frac{3}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array}\right),$

and so $k_{1}=-k_{3}$ and $k_{2}=-\frac{3}{2} k_{3}$. The choice of $k_{3}=-2$ leads to a third eigenvector,

$\mathrm{K}_{3}=\left(\begin{array}{r} 2 \\ 3 \\ -2 \end{array}\right).$