Question 9.5.2: Unitarily diagonalize A = [2 1 + i 1 - i 3].

The characteristic polynomial of A is

C(\lambda ) = \left | \begin{matrix} 2 – \lambda & 1 + i \\ 1 – i & 3 – \lambda \end{matrix} \right | = \lambda ^2 − 5\lambda + 4 = (\lambda − 4)(\lambda − 1)

Hence, the eigenvalues of A are \lambda _1 = 4 and \lambda _2 = 1.

For \lambda _1 = 4,

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

Thus, a corresponding eigenvector is {\bf{z}}_1 = \left [ \begin{matrix} 1 + i \\ 2 \end{matrix} \right ]. For \lambda _2 = 1,

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

Thus, a corresponding eigenvector is {\bf{z}}_2 = \left [ \begin{matrix} 1 + i \\ -1 \end{matrix} \right ].

To unitarily diagonalize A, we need an orthonormal basis for \mathbb{C} ^{2} of eigenvectors of A.

Hence, we normalize {\bf{z}}_1 and {\bf{z}}_2 and take

U = \left [ \begin{matrix} (1 + i) / \sqrt{6} & (1 + i) / \sqrt{3} \\ 2 / \sqrt{6} & -1 / \sqrt{3} \end{matrix} \right ]

to get

U^{∗}AU = diag(4, 1)