Question 8.8.5: Find the eigenvalues and eigenvectors of A = (6 -1 5 4)....
Find the eigenvalues and eigenvectors of \mathrm{A}=\left(\begin{array}{rr}6 & -1 \\ 5 & 4\end{array}\right).
The characteristic equation is
\operatorname{det}(\mathrm{A}-\lambda \mathrm{I})=\left|\begin{array}{cc} 6-\lambda & -1 \\ 5 & 4-\lambda \end{array}\right|=\lambda^{2}-10 \lambda+29=0.
From the quadratic formula, we find \lambda_{1}=5+2 i and \lambda_{2}=\bar{\lambda}_{1}=5-2 i.
Now for \lambda_{1}=5+2 i, we must solve
\begin{aligned} (1-2 i) k_{1}- \qquad k_{2} & =0 \\ 5 k_{1}-(1+2 i) k_{2} & =0 . \end{aligned}
Since k_{2}=(1-2 i) k_{1}, it follows, after choosing k_{1}=1, that one eigenvector is
\mathrm{K}_{1}=\left(\begin{array}{c} 1 \\ 1-2 i \end{array}\right)
From Theorem 8.8.1, we see that an eigenvector corresponding to \lambda_{2}=\bar{\lambda}_{1}=5-2 i is
\mathrm{K}_{2}=\overline{\mathrm{K}}_{1}=\left(\begin{array}{c} 1 \\ 1+2 i \end{array}\right).