Discussion
If
\[ A = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^{2} \\ 1 & \omega^{2} & \omega^{4} \end{pmatrix}, \]
where \(\omega = e^{2\pi i/3}\), then \(A^{2}\) =
Answer: C
\(A^{2}\) is the permutation matrix \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\).
Index rows and columns by \(j, k = 0, 1, 2\). Then \(A_{jk} = \omega^{jk}/\sqrt{3}\), and since \(\omega^{3} = 1\):
\[ \begin{aligned} (A^{2})_{jk} &= \frac{1}{3}\sum_{m=0}^{2} \omega^{jm}\omega^{mk} \\ &= \frac{1}{3}\sum_{m=0}^{2} \omega^{m(j + k)} \end{aligned} \]
The sum \(1 + \omega^{r} + \omega^{2r}\) is 3 when \(r\) is a multiple of 3 and 0 otherwise (because \(1 + \omega + \omega^{2} = 0\)). So \((A^{2})_{jk} = 1\) exactly when \(j + k \equiv 0 \pmod 3\): the pairs (0, 0), (1, 2) and (2, 1). That gives a 1 in the top-left corner and the swap of the last two coordinates.
A is the discrete Fourier matrix, which is unitary but not an involution, so \(A^{2} \ne I\); \(A^{2}\) has real 0/1 entries, so it is not A; and the matrix swapping the first two coordinates puts its 1s in the wrong places.
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