# A Vandermonde like determinant with exponentials

Let $$n\geq m$$ be non negative integers, and consider a list of $$(n+m+1)$$ distinct numbers (complex or real). I am interested in getting a closed form formula for the following determinant: $$\det\left[\sum_{k=i}^{n+i}s_k^{m+1-j}e^{-s_k}\left[\prod_{l=i,l\neq k}^{n+i}(s_k-s_l)\right]^{-1}\right]_{i,j=1\ldots m+1}$$, i.e.

$$\det\left(\begin{array}[ccc] \,\sum_{k=1}^{n+1}s_k^me^{-s_k}\left[\prod_{l=1,l\neq k}^{n+1}(s_k-s_l)\right]^{-1}& \ldots &\sum_{k=1}^{n+1}e^{-s_k}\left[\prod_{l=1,l\neq k}^{n+1}(s_k-s_l)\right]^{-1} \\ \vdots & \ddots & \vdots \\ \sum_{k=m+1}^{n+m+1}s_k^me^{-s_k}\left[\prod_{l=m+1,l\neq k}^{n+m+1}(s_k-s_l)\right]^{-1}& \ldots &\sum_{k=m+1}^{n+m+1}e^{-s_k}\left[\prod_{l=m+1,l\neq k}^{n+m+1}(s_k-s_l)\right]^{-1} \end{array}\right)$$.

This determinant can be rewritten as the determinant of an alternant matrix $$\left(F_{j}(S_i)\right)_{i,j=1\ldots m+1}$$, for suitable multivariate functions $$F_j:\mathbb{C}^{n+1}\rightarrow\mathbb{C}$$.

If the exponential are not there, then this determinant should some king of Schur polynomial. I am wondering if this determinant can be interpreted in terms of representation theory.

• The matrix looks like a product of simpler matrices, can you rewrite it? The denominator product might be related to the derivative of a characteristic polynomial, see Q&A in mathoverflow.net/questions/236323. Where does the matrix come from? Jan 19 at 15:23
• I do not succeed in rewritting the marix as a product. I looked at the lonked reference, I am currently trying to see if one can adapt your preprint to my case. The problem comes from interpolation of quasi-polynomials $R(x)=q(x)+exp(x)p(x)$, where $p(x),q(x)$ are polynomials. Jan 20 at 14:42
• Any comments on my answer below? Feb 13 at 7:37

After slight renumbering $$(m,n)\mapsto (m,n)-1$$ and horizontal reflection $$j\mapsto m+1-j$$, let's call the $$m\times m$$ matrix in the OP's determinant $$A=\left[\sum_{k=i}^{i+n-1}s_k^{j-1}e^{-s_k}\prod_{l=i,l\neq k}^{i+n-1}(s_k-s_l)^{-1}\right]_{\,i,j=1}^{\,m},\tag{1}$$ and define $$r=m+n-1$$. Note that this answer also holds for $$n.
$$A$$ can be factorized similar to this posting using the $$r \times m$$ Vandermonde matrix $$V_{r,m}=\left[ s_\rho^{\mu-1}\right]_{\rho,\mu=1}^{r,m},\tag{2}$$ the diagonal matrix $$E_r = \mathrm{diag}\,[e^{-s_\rho}]_{\rho=1}^r\tag{3}$$ and the matrix of inverse derivatives $$D_{m,r}^{(o)}=\left[ \frac{1}{P_{\mu,o}'(s_\rho)} \begin{cases} 1 & \mu \leq \rho < m+\mu \\ 0 & \text{else} \end{cases} \right]_{\,\mu,\rho=1}^{\,m,r},\tag{4}$$ of the polynomials $$P_{l,o}(s)=\prod_{\rho=l}^{l+o-1}(s-s_\rho)\tag{5}$$ according to $$A = D_{m,r}^{(m)} E_{r} \, V_{r,m}.\tag{6}$$ Note that $$D$$ is not diagonal in this case, e.g., for $$m=3$$, $$n=3$$, where $$r=5$$: $$D_{3,5}^{(3)}= \left( \begin{array}{ccccc} \frac{1}{\left(s_1-s_2\right) \left(s_1-s_3\right)} & \frac{1}{\left(s_2-s_1\right) \left(s_2-s_3\right)} & \frac{1}{\left(s_3-s_1\right) \left(s_3-s_2\right)} & 0 & 0 \\ 0 & \frac{1}{\left(s_2-s_3\right) \left(s_2-s_4\right)} & \frac{1}{\left(s_3-s_2\right) \left(s_3-s_4\right)} & \frac{1}{\left(s_4-s_2\right) \left(s_4-s_3\right)} & 0 \\ 0 & 0 & \frac{1}{\left(s_3-s_4\right) \left(s_3-s_5\right)} & \frac{1}{\left(s_4-s_3\right) \left(s_4-s_5\right)} & \frac{1}{\left(s_5-s_3\right) \left(s_5-s_4\right)} \\ \end{array} \right).\tag{7}$$
As the matrices $$D$$ and $$V$$ are not square, the determinant of $$A$$ does not factorize in a simple way. However, one could try to enlarge both matrices to $$r \times r$$ by inserting appropriate matrix elements $$O(\epsilon)$$, factorize and let $$\epsilon\to 0$$. Alternatively, one can naturally enlarge $$D$$ and $$V$$ and define the $$r \times r$$ matrix $$B=D_{r,r}^{(m)} E_{r} \, V_{r,r},\tag{8}$$ where $$\det B$$ factorizes. The matrix $$A$$ then is the upper left $$m \times m$$ submatrix of $$B$$.