Exercise 4.1.3

With $w=e^{2\mathit{\pi i}∕3}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$, then $\omega =\overline{w}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$

$F_3=\left[\begin{array}{ccc}\hfill w^{0\times 0}\hfill & \hfill w^{1\times 0}\hfill & \hfill w^{2\times 0}\hfill \\ \hfill w^{0\times 1}\hfill & \hfill w^{1\times 1}\hfill & \hfill w^{2\times 1}\hfill \\ \hfill w^{0\times 2}\hfill & \hfill w^{1\times 2}\hfill & \hfill w^{2\times 2}\hfill \\ \hfill \hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill w\hfill & \hfill w^2\hfill \\ \hfill 1\hfill & \hfill w^2\hfill & \hfill w^4\hfill \end{array}\right]$

Then ${\mathrm{\Omega }}_3=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill \omega \hfill & \hfill {\omega }^2\hfill \\ \hfill 1\hfill & \hfill {\omega }^2\hfill & \hfill {\omega }^4\hfill \end{array}\right]$

The permutation matrix is then $P_3=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$

We have $F_3P_3 =

$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill w^2\hfill & \hfill w\hfill \\ \hfill 1\hfill & \hfill w^4\hfill & \hfill w^2\hfill \end{array}\right]$

$

Note, $w^2=e^{4\mathit{\pi i}∕3}=e^{(2\pi -2\pi ∕3)i}=e^{2\mathit{\pi i}}{e}^{-2\mathit{\pi i}∕3}=e^{-2\mathit{\pi i}∕3}=\omega$, and $w^4=w^2{w}^2={\omega }^2$, ${\omega }^2=e^{-4\mathit{\pi i}∕3}=e^{(-2\pi +2\pi ∕3)i}=e^{-2\mathit{\pi i}}{e}^{2\mathit{\pi i}∕3}=e^{2\mathit{\pi i}∕3}=w$

So we see that ${\mathrm{\Omega }}_3=F_3{P}_3$.

Similarly we can prove that $F_3={\mathrm{\Omega }}_3{P}_3$.