# Signature of quadratic form associated to an integral circulant matrix with only real eigenvalues

I am stil stuck with the following:

Let $C$ be a symmetric circulant matrix with integer coefficients of order $n=4k$ (e.g., $C=circ(-1,1,1,1)).$

Assume that $C^{-1}$ is a polynomial (with rational coefficients) in $C;$ say $C^{-1} = P(C).$

Question: Can we get the signature of the quadratic form $Q(x,y) = x^{T}Cy$ from this data ?

More precisely: How it may vary the signature depending on the polynomial $P,$ (if it happens that these signature depends at all on $P$).

In the case $C=circ(-1,1,1,1))$ the signature is $(1,3)$ that means: $1$ positive square and $3$ negative squares; moreover, we have $$C^{-1} = C/4.$$

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I'm not sure why this is formulated in terms of matrices. An $n\times n$ circulant matrix is of the form $f(S)$ where $f$ is a polynomial of degree $\leq n-1$, and $S$ is the cyclic shift matrix, whose eigenvalues are the $n$th roots of unity. So can't your question (once one puts in all the properties you impose on $C$) just be stated in terms of the values of a certain self-adjoint trigonometric polynomial on roots of unity? Perhaps I have misunderstood. – Yemon Choi Jan 26 '11 at 22:58
I feel your idea is great. But I do not see well how to build such trigonometric polynomial. What happens e.g., in the special case when we take $C=circ(-1,1,1,1)$ ? (forgetting (say) some of the technical conditions imposed on $C$ to simplify). Just to take a look to the trigonometric polynomial, to see if I can manage him. – Luis H Gallardo Jan 26 '11 at 23:53
Luis, I am stuck today on a computer which does not display the mathematics on this site properly, but I will try to reply properly later. In the case where the top row of your matrix is -1, 1, 1, 1 then I think the matrix is equal to f(S) where f is the Laurent polynomial $f(z) = -1 + z + z^{-1} + z^2/2 + z^{-2}/2$ and S is the cylic shift of order 4. Because f is not unique, it might be that some choices are better than others; I will think about this. – Yemon Choi Jan 27 '11 at 17:31