# Characteristic polynomial type of identity.

Let $N$ be a positive integer, $x$ be a real number and let $Q$ be a real $N$-dimensional matrix. The following identity holds: \begin{eqnarray} &&\det\left({\mathbb 1} - x Q\right) = e^{-x \text{Tr}[Q]}\cdot \\&& \left(1 -\frac{x^2}{2} Tr[Q^2] - \frac{x^3}{3} Tr[Q^3] - \frac{x^4}{4} (-\frac{1}{2} Tr[Q^2]^2 + Tr[Q^4]) + x^5(\frac{1}{6} Tr[Q^2] Tr[Q^3] - \frac{1}{5} Tr[Q^5]) + x^6 (-\frac{Tr[Q^2]^3}{48} + \frac{Tr[Q^3]^2}{18} + \frac{Tr[Q^2] Tr[Q^4]}{8} - \frac{Tr[Q^6]}{6}) + O(x^7)\right) \end{eqnarray} As a matter of fact it seems that the coefficient at $x^p$ has the following form(we have checked this identity for all $p\le 10$): $$\left( coeff @ x^p \right) = \sum\limits_{l=1}^{\lfloor p/2 \rfloor} (-1)^l \sum\limits_{\stackrel{2 \le p_1 \le p_2 \le \dots \le p_l \le p-2 (l-1)}{p_1+\dots+p_l=p}} \prod\limits_{\xi=1}^l \frac{Tr[Q^{p_\xi}]^{d_\xi}}{p_\xi d_\xi!}$$ where $d_\xi$ is the multiplicity of the number $p_\xi$ for $\xi=1,\dots,l$. We have shown this identity by expanding the determinant in question in a Taylor series and then applying formulae given in Calculate a multiple sum of inverse integers. to the expansion coefficients and finally by resuming the resulting series. The question is to prove or disprove the formula for the coefficient at $x^p$.

Clearly we have: $$\det\left({\mathbb 1} - x Q\right) = e^{Tr \log\left({\mathbb 1} - x Q\right)} = e^{-x Tr(Q)} \cdot e^{- \sum\limits_{j=2}^\infty \frac{x^j}{j} Tr[Q^j] }$$ Now by expanding the exponential in a Taylor series we have: $$e^{- \sum\limits_{j=2}^\infty \frac{x^j}{j} Tr[Q^j] } = 1 + \sum\limits_{p=2}^\infty x^p \sum\limits_{l=1}^{\frac{p}{2}} \frac{(-1)^l}{l!} \sum\limits_{\stackrel{j_1+\dots+j_l = p}{j_1,\dots,j_l \ge 2}} \prod\limits_{\xi=1}^l \frac{Tr[Q^{j_\xi}]}{j_\xi}$$ Now let us write a sequence $\vec{j} := (j_1,\dots,j_l)$ as $\left(\underbrace{j_1,\dots,j_1}_{d_1},\underbrace{j_2,\dots,j_2}_{d_2},\dots,\underbrace{j_s,\dots,j_s}_{d_s}\right)$ where $j_1 < j_2 < \dots < j_s$ and $d_1+\dots+d_s=l$. Clearly there are $l!/(d_1! d_2! \cdot \dots \cdot d_s!)$ sequences $\vec{j}$ that all have the same decomposition in question. All those sequences will yield the same contribution to the coefficient. From this follows the result.