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Jan 21, 2011 at 15:38 comment added Luis H Gallardo @unknown mannekin pisse: Assume that $C$ is also a Hadamard matrix. I am afraid not to yet understand why when the circulant matrix $C$ is symmetric or close to be symmetric, the associated quadratic form $Q$ transforming $x \to x^T Cx$ (or something alike when $C$ is not symmetric) must have a very special signature, of the type $(1,*)$ or $(-1,*)$, (that forces $n=4$) while normally (for more non-circulant Hadamard's) the signature should be of the type $ ((n+\sqrt{n})/2,(n-\sqrt{n})/2)$.
Jan 6, 2011 at 10:13 comment added Luis H Gallardo Let $C=circ(r_0,r_1, \ldots ,r_{n-1})$ be the circulant matrix with first row $r_0, \ldots ,r_{n-1}$. The $R(\omega)$ are the eigenvalues of $C$. We may consider the special case when $C$ is orthogonal in order to see different signs taken by the eigenvalues.
Dec 30, 2010 at 22:18 vote accept Luis H Gallardo
Dec 30, 2010 at 22:16 history edited user631 CC BY-SA 2.5
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Dec 30, 2010 at 21:51 history answered user631 CC BY-SA 2.5