If omega neq 1 is a cube root of unity and H | vedclass

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(C) Given $H = \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix}$.We calculate the powers of $H$:$H^2 = \begin{bmatrix} \omega & 0 \ 0 & \omega

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$H$

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(C) Given $H = \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix}$.We calculate the powers of $H$:$H^2 = \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix} \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix} = \begin{bmatrix} \omega^2 & 0 \ 0 & \omega^2 \end{bmatrix}$.By induction,$H^n = \begin{bmatrix} \omega^n & 0 \ 0 & \omega^n \end{bmatrix}$.For $n = 70$,$H^{70} = \begin{bmatrix} \omega^{70} & 0 \ 0 & \omega^{70} \end{bmatrix}$.Since $\omega^3 = 1$,we have $\omega^{70} = (\omega^3)^{23} \cdot \omega = (1)^{23} \cdot \omega = \omega$.Therefore,$H^{70} = \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix} = H$.

If $\omega \neq 1$ is a cube root of unity and $H = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$,then $H^{70}$ is equal to:

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