Efficient $k$-Sign Consistency Verification of Hankel Matrices via Schur Polynomials

TL;DR

The paper addresses certifying (strict) $k$-sign consistency of Hankel matrices and operators, a problem with combinatorial growth in minors. It introduces a row-consecutive minor decomposition: every $k$-th order minor can be written as a nonnegative integer combination of row-consecutive $k$-minors, with coefficients governed by Kostka numbers and Littlewood–Richardson coefficients from Schur polynomial theory. For Hankel matrices, these decompositions yield that checking the $k$-sign consistency of a reshaped $k$-row Hankel matrix suffices, and for the Hankel operator this condition becomes necessary and sufficient. The approach extends to Toeplitz matrices and partly to circulant matrices, and leverages Pena bijection-based total positivity certificates to achieve practical computational complexity, notably $O(k^{2}(M+N))$ in the finite setting. Overall, the work bridges algebraic combinatorics with total positivity to deliver efficient, structure-aware certificates for $k$-sign consistency in structured matrix classes.

Abstract

We consider the problem of certifying (strict) $k$-sign consistency of a matrix, that is, whether all of its $k$-th order minors share the same (strict) sign. Although this problem is generally of combinatorial complexity, we show that for Hankel matrices it can be significantly simplified: our sufficient condition requires checking only the $k$-th order minors of a reshaped Hankel matrix with $k$ rows. Remarkably, when applied to the Hankel operator, this sufficient condition is also necessary. Comparable results were known only in the setting of (strictly) $k$-positive Hankel matrices and operators, in which all minors of order up to $k$ have the same (strict) sign. More concretely, we derive a formula expressing the $k$-th order minors of Hankel matrices as nonnegative integer linear combinations of $k$-th order minors with consecutive row indices. Our derivation uses Schur polynomial theory to show that the $k$-th order minors of any matrix are nonnegative integer linear combinations of row-consecutive $k$-th order minors, meaning minors formed from distinct columns whose consecutive row indices need not coincide across columns. For Hankel matrices, these minors coincide -- up to sign changes arising from column swaps -- with the usual $k$-th order minors with consecutive row indices. Our main result then follows by showing that the sum of certain signed nonnegative integer coefficients equals the corresponding Littlewood--Richardson coefficients. In our problem, the nonnegativity of these coefficients ensures that negatively signed column permutations are cancelled by positively signed ones. Our results also extend naturally to Toeplitz matrices and operators, and we present a partial analogue for circulant matrices.

Figures (2)

• Figure 1: Semi-standard Young tableaux for the Kostka number $K_{\lambda,\mu}$ with $\lambda = (2,1,0)$ and $\mu = (1,1,1)$. Each tableau has three rows of boxes, with the first row containing $\lambda_1 = 2$ boxes, the second row containing $\lambda_2 = 1$ boxes, and the third row containing $\lambda_3 = 0$ boxes, which builds the Young diagram of shape $\lambda$. Then, there are exactly two possibilities to fill in $\mu_1$-times the integer $1$, $\mu_2$-times the integer $2$, and $\mu_3$-times the integer $3$, such that the entries of the tableaux do not decrease along rows and strictly increase along columns.
• Figure 2: Skew semi-standard Young tableaux for the Littlewood-Richardson coefficient $c_{\lambda,\mu}^\gamma$ with $\lambda = (2,1,0)$, $\mu = (3,2,1)$ and $\gamma = (4,3,2)$. The Young diagram of shape $\lambda$ is as in \ref{['fig:Kostka']}, which is missing from the Young diagram of shape $\gamma$ to establish the skew Young diagram of shape $\gamma/\lambda$. Then, there are exactly two possibilities to fill in $\mu_1$-times the integer $1$, $\mu_2$-times the integer $2$ and $\mu_3$-times the integer $3$, such that the entries of the tableaux do not decrease along rows, strictly increase along columns, and reading the tableaux from right to left and top to bottom creates a lattice word. The lattice word property in the left tableau means that $(1)$, $(1,1)$, $(1,1, 2)$, $(1,1,2,1)$, $(1,1,2,1,3)$ and $(1,1,2,1,3,2)$ contain at least as many $1$s as $2$s and at least as many $2$s as $3$s, and similarly for the right tableau.

Theorems & Definitions (23)

• Definition 1
• Proposition 1
• Proposition 2
• Corollary 1
• Lemma 1
• Definition 2
• Corollary 2
• Lemma 2
• Theorem 1
• Remark 1