Unitarily diagonalize A = [4i 1 + 3i -1 + 3i i].

Question

Unitarily diagonalize [latex]A = \left [ \begin{matrix} 4i & 1 + 3i \ -1 + 3i & i \end{matrix} ight ][/latex].

Accepted Answer

We have [latex]C(\lambda ) = \lambda ^2 − 5i \lambda + 6[/latex]. Using the quadratic formula, we get eigenvalues [latex]\lambda _1 = 6i[/latex] and [latex]\lambda _2 = −i[/latex]. We find that

[latex]A − 6iI = \left [ \begin{matrix} -2i & 1 + 3i \ -1 + 3i & -5i \end{matrix} ight ] \Rightarrow {\bf{z}}_1 = \left [ \begin{matrix} 3 - i \ 2 \end{matrix} ight ] [/latex]

[latex]A + iI = \left [ \begin{matrix} 5i & 1 + 3i \ -1 + 3i & 2i \end{matrix} ight ] \Rightarrow {\bf{z}}_2 = \left [ \begin{matrix} 1 + 3i \ -5i \end{matrix} ight ] [/latex]

Hence, taking

[latex]U = \left [ \begin{matrix} (3 - i) / \sqrt{14} & (1 + 3i) / \sqrt{35} \ 2 / \sqrt{14} & −5i / \sqrt{35} \end{matrix} ight ][/latex]

gives

[latex]U^{∗}AU = \left [ \begin{matrix} 6i & 0 \ 0 & -i \end{matrix} ight ] [/latex]

Question 9.5.4: Unitarily diagonalize A = [4i 1 + 3i -1 + 3i i].

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