A short question about the DFT matrix

Is the DFT matrix the unique* unitary matrix with all entries of same magnitude?

(*up to some trivial transformations)

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what about the identity matrix? –  Suvrit Mar 26 '13 at 17:20
The identity matrix has lots of zeros. –  Steve Huntsman Mar 26 '13 at 17:25
Let $M$ a matrix of the form you describe, with $M_{jk} = C e^{i\omega_{jk}}$ and $C > 0$. Unitarity implies that $C^2 \sum_\ell e^{i(\omega_{\ell k} - \omega_{\ell j})} = \delta_{j k} = C^2 \sum_\ell e^{i(\omega_{j \ell} - \omega_{k \ell})}$. Taking $j = k$ gives that $C = N^{-1/2}$, where $N = \dim M$. For $j \ne k$, $\sum_\ell e^{i(\omega_{j \ell} - \omega_{k \ell})} = 0$. The only way this can happen is if the angles $\omega_{j \ell} - \omega_{k \ell}$ balance out''. Roots of unity are a particularly nice way for this to happen, but as Mark Meckes points out, not the only one. –  Steve Huntsman Mar 26 '13 at 17:55
I should have said "$N$th roots of unity" above. –  Steve Huntsman Mar 26 '13 at 17:57
;-) duh! I somehow interpreted it to mean magnitude of the nonzeros! –  Suvrit Mar 26 '13 at 21:59