Question 8.8.2: Find the eigenvalues and eigenvectors of A = (1 2 1 6 -1 0 -......
Find the eigenvalues and eigenvectors of
\mathrm{A}=\left(\begin{array}{rrr} 1 & 2 & 1 \\ 6 & -1 & 0 \\ -1 & -2 & -1 \end{array}\right) . (5)
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
\begin{aligned} (\mathrm{A}-0 \mathrm{I} \mid \mathrm{0})= & \left(\begin{array}{rrr|r} 1 & 2 & 1 & 0 \\ 6 & -1 & 0 & 0 \\ -1 & -2 & -1 & 0 \end{array}\right) \stackrel{-6 R_{1}+R_{2}}{\stackrel{R_{1}+R_{3}}{\Rightarrow}}\left(\begin{array}{rrr|r} 1 & 2 & 1 & 0 \\ 0 & -13 & -6 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \\ & \stackrel{-\frac{1}{13} R_{2}}{\Rightarrow}\left(\begin{array}{rrr|r} 1 & 2 & 1 & 0 \\ 0 & 1 & \frac{6}{13} & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \stackrel{-2 R_{2}+R_{1}}{\Rightarrow}\left(\begin{array}{lll|l} 1 & 0 & \frac{1}{13} & 0 \\ 0 & 1 & \frac{6}{13} & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) . \end{aligned}
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,
(\mathrm{A}+4 \mathrm{I} \mid \mathrm{0})=\left(\begin{array}{rrr|r} 5 & 2 & 1 & 0 \\ 6 & 3 & 0 & 0 \\ -1 & -2 & 3 & 0 \end{array}\right) \stackrel{\substack{-R_{3} \\ R_{1} \leftrightarrow R_{3}}}{\Rightarrow}\left(\begin{array}{rrr|r} 1 & 2 & -3 & 0 \\ 6 & 3 & 0 & 0 \\ 5 & 2 & 1 & 0 \end{array}\right)
\begin{aligned} & \stackrel{\substack{-6 R_{1}+R_{2} \\ -5 R_{1}+R_{3}}}{\Rightarrow}\left(\begin{array}{rrr|r} 1 & 2 & -3 & 0 \\ 0 & -9 & 18 & 0 \\ 0 & -8 & 16 & 0 \end{array}\right) \stackrel{\substack{-\frac{1}{9} R_{2} \\ -\frac{1}{8} R_{3}}}{\Rightarrow}\left(\begin{array}{lll|l} 1 & 2 & -3 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 1 & -2 & 0 \end{array}\right) \\ & \underset{\Rightarrow}{\stackrel{-2 R_{2}+R_{1}}{-R_{2}+R_{3}}}\left(\begin{array}{rrr|r} 1 & 0 & 1 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \end{aligned}
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).