Question 8.8.4: Find the eigenvalues and eigenvectors of A = (9 1 1 1 9 1 1 ......

The characteristic equation

\operatorname{det}(\mathrm{A}-\lambda \mathrm{I})=\left|\begin{array}{ccc} 9-\lambda & 1 & 1 \\ 1 & 9-\lambda & 1 \\ 1 & 1 & 9-\lambda \end{array}\right|=-(\lambda-11)(\lambda-8)^{2}=0

shows that \lambda_{1}=11 and that \lambda_{2}=\lambda_{3}=8 is an eigenvalue of multiplicity 2 .

For \lambda_{1}=11, Gauss-Jordan elimination gives

(\mathrm{A}-11 \mathrm{I} \mid \mathrm{0})=\left(\begin{array}{rrr|r} -2 & 1 & 1 & 0 \\ 1 & -2 & 1 & 0 \\ 1 & 1 & -2 & 0 \end{array}\right) \stackrel{\substack{\text { row } \\ \text { operations }}}{\Rightarrow}\left(\begin{array}{rrr|r} 1 & 0 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right).

Hence k_{1}=k_{3} and k_{2}=k_{3}. If k_{3}=1, then

\mathrm{K}_{1}=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right).

Now for \lambda_{2}=8 we have

(\mathrm{A}-8 \mathrm{I} \mid \mathrm{0})=\left(\begin{array}{lll|l} 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \end{array}\right) \stackrel{\substack{\text { row } \\ \text { operations }}}{\Rightarrow}\left(\begin{array}{lll|l} 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right).

In the equation k_{1}+k_{2}+k_{3}=0 we are free to select two of the variables arbitrarily. Choosing, on the one hand, k_{2}=1, k_{3}=0, and on the other, k_{2}=0, k_{3}=1, we obtain two linearly independent eigenvectors:

\mathrm{K}_{2}=\left(\begin{array}{r} -1 \\ 1 \\ 0 \end{array}\right) \quad \text { and } \quad \mathrm{K}_{3}=\left(\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right)

corresponding to a single eigenvalue.