Are the matrices U = [1 -i i 1] and V = [1/√3(1 + i) 1/√6(1 + i) - 1/√3i 2/√6i] unitary?

Question

Are the matrices [latex]U = \left [ \begin{matrix} 1 & -i \ i & 1 \end{matrix} ight ][/latex] and [latex]V = \left [ \begin{matrix} \frac{1}{\sqrt{3} } (1 + i) & \frac{1}{\sqrt{6} } (1 + i) \ - \frac{1}{\sqrt{3} } i & \frac{2}{\sqrt{6} } i \end{matrix} ight ][/latex] unitary?

Accepted Answer

Observe that

[latex] \left\langle \left [ \begin{matrix} 1 \ i \end{matrix} ight ] , \left [ \begin{matrix} 1 \ i \end{matrix} ight ] ight angle = \overline{\left [ \begin{matrix} 1 \ i \end{matrix} ight ]} \cdot \left [ \begin{matrix} 1 \ i \end{matrix} ight ] = \left [ \begin{matrix} 1 \ -i \end{matrix} ight ] \cdot \left [ \begin{matrix} 1 \ i \end{matrix} ight ] = 1(1) + (−i)i = 2[/latex]

Hence, [latex]\left [ \begin{matrix} 1 \ i \end{matrix} ight ][/latex] is not a unit vector. Thus, U is not unitary.

For V, we have [latex] V^{∗} = \left [ \begin{matrix} \frac{1}{\sqrt{3} } (1 - i) & \frac{1}{\sqrt{3} } i \ \frac{1}{\sqrt{6} } (1 - i) & - \frac{2}{\sqrt{6} } i \end{matrix} ight ][/latex] , so

[latex] V^{∗}V = \left [ \begin{matrix} \frac{1} {3} (2 + 1) & \frac{1} {3 \sqrt{2} } (2 - 2) \ \frac{1} {3 \sqrt{2} } (2 - 2) & \frac{1}{6} (2 + 4) \end{matrix} ight ] = \left [ \begin{matrix} 1 & 0 \ 0 & 1 \end{matrix} ight ][/latex]

Thus, V is unitary.

Question 9.3.8: Are the matrices U = [1 -i i 1] and V = [1/√3(1 + i) 1/√6(1 ...

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