Negative inertia of entrywise powers

For $\alpha>0$, let $M_n(\alpha)$ be the largest possible number ofnegative eigenvalues of $A^{\circ\alpha}$, where $A$ ranges over the realpositive definite matrices of order $n$ with strictly positive entries.For $n\geq3$, we prove\[ M_n(\alpha)=n-2 \qquad\left(0<\alpha<\frac1{n-2}\right),\]and, for every fixed $\alpha>0$,\[ \lim_{n\to\infty}\frac{M_n(\alpha)}n = \begin{cases} 0,&\alpha\in\mathbb N,\\ 1,&\alpha\notin\mathbb N. \end{cases}\]For a positive integer $m$, put\[ p_m(n)=\min\left\{p\geq1:n\leq\binom{p+m-1}{m}\right\}.\]For $m\geq2$, we prove\[ M_n(\alpha)\geq n-p_m(n) \qquad\left(\frac1m\leq\alpha<\frac1{m-1}\right),\]and hence $n-M_n(\alpha)=O_\alpha(n^{1/m})$ on this interval.At reciprocal exponents we further prove that, for $m\geq2$, $p\geq3$,and $1\leq r\leq\lfloor p/3\rfloor$,\[ M_{\binom{p+m-1}{m}+r}(1/m) \geq \binom{p+m-1}{m}+r-p.\]Finally, we characterize the equality $M_{m+2}(1/m)=m-1$ by a reversetensor inequality.

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Publication Details

Journal
Zenodo (CERN European Organization for Nuclear Research)
Published
2026-09-09
DOI
https://doi.org/10.5281/zenodo.22670490
Primary Topic
Matrix Theory and Algorithms
Type
preprint
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preprint

Negative inertia of entrywise powers

Wei Xie
Zenodo (CERN European Organization for Nuclear Research)
Matrix Theory and Algorithms
preprint

Negative inertia of entrywise powers

Wei Xie
preprint en

Abstract

For $\alpha>0$, let $M_n(\alpha)$ be the largest possible number ofnegative eigenvalues of $A^{\circ\alpha}$, where $A$ ranges over the realpositive definite matrices of order $n$ with strictly positive entries.For $n\geq3$, we prove\[ M_n(\alpha)=n-2 \qquad\left(0<\alpha<\frac1{n-2}\right),\]and, for every fixed $\alpha>0$,\[ \lim_{n\to\infty}\frac{M_n(\alpha)}n = \begin{cases} 0,&\alpha\in\mathbb N,\\ 1,&\alpha\notin\mathbb N. \end{cases}\]For a positive integer $m$, put\[ p_m(n)=\min\left\{p\geq1:n\leq\binom{p+m-1}{m}\right\}.\]For $m\geq2$, we prove\[ M_n(\alpha)\geq n-p_m(n) \qquad\left(\frac1m\leq\alpha<\frac1{m-1}\right),\]and hence $n-M_n(\alpha)=O_\alpha(n^{1/m})$ on this interval.At reciprocal exponents we further prove that, for $m\geq2$, $p\geq3$,and $1\leq r\leq\lfloor p/3\rfloor$,\[ M_{\binom{p+m-1}{m}+r}(1/m) \geq \binom{p+m-1}{m}+r-p.\]Finally, we characterize the equality $M_{m+2}(1/m)=m-1$ by a reversetensor inequality.

Zenodo (CERN European Organization for Nuclear Research)
Matrix Theory and Algorithms
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