Question 9.4.1: Let A = [5 -6 3 -1]. Find its eigenvectors and diagonalize o...

We have

C(\lambda ) = det(A − \lambda I) = \left | \begin{matrix} 5 – \lambda & -6 \\ 3 & -1 – \lambda \end{matrix} \right | = \lambda ^2 − 4\lambda + 13

So, by the quadratic formula, we get that the eigenvalues of A are \lambda _1 = 2 + 3i and \lambda _2 = 2 − 3i.

For \lambda _1 = 2 + 3i,

A − \lambda _{1}I = \left [ \begin{matrix} 3 – 3i & -6 \\ 3 & -3 – 3i \end{matrix} \right ] \sim \left [ \begin{matrix} 1 & -(1 + i) \\ 0 & 0 \end{matrix} \right ]

Hence, an eigenvector corresponding to \lambda _1 = 2 + 3i is z_1 = \left [ \begin{matrix} 1 + i \\ 1 \end{matrix} \right ].

For \lambda _2 = 2 − 3i,

A − \lambda _{2}I = \left [ \begin{matrix} 3 + 3i & -6 \\ 3 & -3 + 3i \end{matrix} \right ] \sim \left [ \begin{matrix} 1 & -(1 – i) \\ 0 & 0 \end{matrix} \right ]

Thus, an eigenvector corresponding to \lambda _2 = 2 − 3i is z_2 = \left [ \begin{matrix} 1 – i \\ 1 \end{matrix} \right ].

It follows that A is diagonalized to \left [ \begin{matrix} 2 + 3i & 0 \\ 0 & 2 – 3i \end{matrix} \right ] by P = \left [ \begin{matrix} 1 + i & 1 – i \\ 1 & 1 \end{matrix} \right ].