Is the DFT matrix the unique* unitary matrix with all entries of same magnitude?
(*up to some trivial transformations)
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$\begingroup$ what about the identity matrix? $\endgroup$– SuvritCommented Mar 26, 2013 at 17:20
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1$\begingroup$ The identity matrix has lots of zeros. $\endgroup$– Steve HuntsmanCommented Mar 26, 2013 at 17:25
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$\begingroup$ 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. $\endgroup$– Steve HuntsmanCommented Mar 26, 2013 at 17:55
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$\begingroup$ I should have said "$N$th roots of unity" above. $\endgroup$– Steve HuntsmanCommented Mar 26, 2013 at 17:57
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$\begingroup$ ;-) duh! I somehow interpreted it to mean magnitude of the nonzeros! $\endgroup$– SuvritCommented Mar 26, 2013 at 21:59
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2 Answers
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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.)
answered Mar 26, 2013 at 23:26
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$\begingroup$ 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... $\endgroup$ Commented Mar 27, 2013 at 0:12
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No. For example, there are Hadamard matrices (after rescaling).
answered Mar 26, 2013 at 17:43