Question

If $\omega \ne 1$ is the complex cube root of unity and matrix $ H =\begin{bmatrix}\omega&0\\ 0&\omega\end{bmatrix}$, then $H^{70}$ is equal to -

• A. 0
• B. -H
• C. $H^2$
• D. H

Answer 1

Answer: (D)

$H^{2} = \begin{bmatrix}\omega&0\\ 0&\omega\end{bmatrix}\begin{bmatrix}\omega &0\\ 0&\omega \end{bmatrix} = \begin{bmatrix}\omega^{2} &0\\ 0&\omega^{2} \end{bmatrix}$
If $H^{k} = \begin{bmatrix}\omega^{k} &0\\ 0&\omega^{k} \end{bmatrix}$, then $H^{k+1} = \begin{bmatrix}\omega^{k+1} &0\\ 0&\omega^{k+1} \end{bmatrix}$
So by mathematical induction,
$ H^{70} = \begin{bmatrix}\omega ^{70} &0\\ 0&\omega ^{70} \end{bmatrix} = \begin{bmatrix}\omega &0\\ 0&\omega \end{bmatrix} = H$