2
$\begingroup$

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 $p\le 10$): \begin{equation} \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!} \end{equation} 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 $x^p$.

$\endgroup$
0
$\begingroup$

Clearly we have: \begin{equation} \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] } \end{equation} Now by expanding the exponential in a Taylor series we have: \begin{equation} 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} \end{equation} 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.