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Assume $M$ is an invertible positive matrix of rank $N$. The Von Neumann entropy $H$ of a matrix $M$ with eigenvalues $\{ \lambda_n \}$ is

$H[M] = -\sum_{n=1}^N \lambda_n \ln \lambda_n$.

In principle, the eigenvalues are encoded in the characteristic polynomial

$\phi_t (M) = \mathrm{det}(tI-M) = \prod_n(t-\lambda_n) = t^{N} + a_{N-1} t^{N-1} \cdots + a_1 t +a_0 $.

The trace $\mathrm{Tr}\, M$ is given by the coefficient $a_{N-1}$ in the characteristic polynomial:

$\lim\limits_{t \to \infty} \dfrac{\phi_t (M)-t^N}{t^{N-1}} = a_{N-1} = (-1)^{N-1} \, \mathrm{Tr}\, M = (-1)^{N-1} \sum_n \lambda_n$.

Is there a similar relationship between the entropy and the characteristic polynomial?

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To define $H$ you also need that the eigenvalues are positive, right? – Qiaochu Yuan Jul 18 '12 at 18:10
Sorry, yes, I will fix that. Thanks. – Jess Riedel Jul 18 '12 at 18:28

I'm not sure if this qualifies as simple (or if this is helpful at all), but we have $$ \frac{\phi'_M(t)}{\phi_M(t)}=\sum_n\frac{1}{t-\lambda_n} $$

Using the residue theorem, we can write $$ H[M]=\frac{-1}{2\pi i}\oint\frac{\phi'_M(z)}{\phi_M(z)}z\log(z)\,dz $$ where the integral is taken over a closed contour containing all of the eigenvalues of $M$ (I guess we're either working on the Riemann surface of $\log(z)$, or we chose a branch of $\log(z)$).

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Yes. The entropy is a power series in the coefficients of the characteristic polynomial. (If the eigenvalues all lie on the same side of $0$.)

Expand $\lambda_n \ln \lambda_n$ as a power series in $\lambda_n$. This allows you to write $H[M]$ as an infinite sum of symmetric polynomials in the $\lambda_n$. A symmetric polynomial in the $\lambda_n$ can be written as a polynomial of the elementary symmetric polynomials, which are $a_0,...,a_{N-1}$. So you can write $H[M]$ as an infinite sum of polynomials in the $\lambda_n$.

Is that a simple enough relation?

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$f(x) = x \ln x$ is not differentiable at $x=0$, so I don't think I can do the power series, right? – Jess Riedel Jul 18 '12 at 18:35
You do a Taylor series around some nonzero number, depending on your eigenvalue spread. If, as I understand, they're all positive real, you take it around some number that is at at least half the largest eigenvalue. – Will Sawin Jul 18 '12 at 19:27

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