# Eigenvalues of generalized Vandermonde matrices

Given a strictly increasing sequence $0<x_1<x_2<\dots<x_n$ of $n$ strictly positive real numbers and a second strictly increasing sequence $e_1<\dots e_n$ of $n$ real numbers, the matrix with coefficients $x_i^{e_j}$ has experimentally always $n$ strictly positive real eigenvalues. This matrix is of course a Vandermonde matrix if $e_i=i-1$.

Are there counterexamples to this observation?

Is this known, at least for ordinary Vandermonde matrices?

It seems that positivity of $x_1,x_2,\dots$ cannot be dropped: chosing $e_j$ integral and the first few values of $x_i$ negative leads in general to complex eigenvalues.

Other remarks: (1) These matrices seem to be numerically highly instable, I have to compute with very high accuracy (or work with Sturm sequences over the integers). This limits the size of feasible cases somewhat.

(2) If all $e_j$ are positive integers, the characteristic polynomial has alternating coefficients by the definition of Schur-polynomials as can be seen by adapting David Speyer's answer to question 60938. Perhaps a clever continuity argument (at least in the case $e_1>0$) shows that the coefficients are always alternating?

If this observation is true, then there is perhaps a (more or less) naturally associated symmetric matrix lurking behind.

• This seems like a special case of the $e^{x_iy_j}$ matrix, whose invertibility was discussed on MO, I cannot locate that question right now (though I don't remember if positivity was discussed / proved). Jan 27, 2014 at 15:31

The kernel $x^y$ is strictly totally positive (STP), i.e., for any choices of $0 < x_1 < \cdots < x_n$ and $y_1 < y_2 < \cdots < y_n$ the matrix $K_{ij} = x_i^{y_j}$ is (strictly) totally positive, i.e., all minors are positive. Since it is STP, its eigenvalues are also positive---for a proof see Spectral Properties of Totally Positive Kernels and Matrices, in Total Positivity and its Applications, M. Gasca, C. A. Micchelli (eds.), pp. 477-511, Kluwer Academic Publishers, 1996
For a proof of the STP property of this kernel, notice that it is essentially a special case of the $e^{xy}$ kernel (since $x_i > 0$, we can write it as an exponent). The STP property of $e^{xy}$ is well-studied, and can be found for example in Pinkus's recent book (Chapter 4): Totally Positive Matrices.