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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
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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
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2 Answers

up vote 5 down vote accepted

Call a unitary matrix flat if all its entries have the same absolute value. In operator theory these arise as a class of type-II matrices, which were used by Vaughan Jones in his work on link invariants. Currently they are also of interest in physics, because of their connection with "mutually unbiased bases". In this context they are known as generalized Hadamard matrices (which is a good enough name, but has a different meaning among design theorists).

The basic examples are Hadamard matrices and character tables of abelian groups, as noted by Mark and Steve. The class of flat unitary matrices is closed under Kronecker product, and this gives us examples which are neither Hadamard nor character tables. A survey of the subject by physicists appears as arXiv:quant-ph/0512154. (And if you search on quant-ph for articles with "Hadamard" in the title, you'll find many more papers on the subject.)

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Thank you for this quick answer! –  user32545 Mar 26 '13 at 19:55
    
I didn't know that I noted the character tables, but thanks for spelling it out! Now it's clear to me what direction I was headed... –  Steve Huntsman Mar 27 '13 at 0:12
    
Thanks for the reference! –  Olivier Leveque Mar 27 '13 at 14:12
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No. For example, there are Hadamard matrices (after rescaling).

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