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Feb 17, 2022 at 17:09 vote accept Carlo Beenakker
Feb 17, 2022 at 16:30 answer added Noam D. Elkies timeline score: 3
Feb 14, 2022 at 18:34 comment added Carlo Beenakker I presume you mean $2k-N$ rather than $N-2k$? You might want to post this as an answer, so that I can accept it.
Feb 14, 2022 at 17:31 comment added Noam D. Elkies (For general $T,V$ I suppose we shouldn't say that $\pi T \pi$ is "unitary on $V \cap TV$" because $V \cap TV$ need not be stable under $\pi T \pi$, though it's still true that $\| \pi T \pi(v) \| = \| v \|$ for all $v \in V \cap TV$.)
Feb 14, 2022 at 17:26 comment added Noam D. Elkies In any case it's not hard to show $N-2k$ is a lower bound. Let $T$ be any unitary operator on ${\bf C}^N$, and $\pi$ the orthogonal projection to any $k$-dimensional subspace $V \subset {\bf C}^N$. Then $\pi T \pi$ is a "sub-unitary" operator on $V$ that is unitary on $V \cap TV$, so has $1$ as a singular value of multiplicity at least $N-2k$. In our setting, $T$ is the discrete Fourier transform, and $V$ is a coordinate subspace. I guess that special properties of this $T$ (such as invertibility of any submatrix) will show that the multiplicity is exactly $\max(N-2k,0)$.
Feb 14, 2022 at 13:01 history edited Carlo Beenakker CC BY-SA 4.0
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Feb 14, 2022 at 12:55 history asked Carlo Beenakker CC BY-SA 4.0