Symmetric matrix power inequality for row sums

About 1 year old · traced to 2025

Let $M$ be any $n\times n$ symmetric matrix with entries satisfying $0\leq M_{ij}\leq1$ , and let $\mathbf{1}$ be the $n$ -dimensional all-ones vector. Matrix power inequality conjecture. For every $\alpha\geq0$ ,

\mathbf{1}^{\mathsf T}M^\alpha\mathbf{1}\leq\mathbf{1}^{\mathsf T}(M\mathbf{1})^\alpha.

The paper proposes this inequality as a sufficient route to proving the preceding trace and row-sum bounds when $q>1$ . Its validity is not established in the source.

References

Primary source

Phuc Nguyen, Josiah Couch, Rahul Bansal, Alexandra Morgan, Chris Tam, Miao Li, Rima Arnaout and Ramy Arnaout, “Which Similarity-Sensitive Entropy (Sentropy)?”, arXiv:2511.03849 (2026).

matrix-inequalities spectral-theory

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader supplied a concrete counterexample in two dimensions, but it has not been independently verified.

The conjecture asks whether the stated power inequality holds for every allowed symmetric matrix and every nonnegative exponent. The source presents it only as a proposed route to stronger bounds and does not establish its validity.

Posted attempt

A reader claims a complete disproof using a strictly positive-definite $2\times2$ matrix with entries in $[0,1]$ and exponent $\alpha=3/2$ , asserting that the two sides are $9/\sqrt{10}$ and $1+3\sqrt{6}/4$ , respectively, with the left side larger. This attempt has not been independently verified.

Current status (as of August 2026): The conjecture has an explicit unverified counterexample claim, so it is not settled; absent verification, the original question remains mathematically open.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

The conjecture fails even for a strictly positive definite $2\times2$ matrix.

Let

M= \begin{pmatrix} \frac12&\frac12\\[2pt] \frac12&1 \end{pmatrix}, \qquad \alpha=\frac32, \qquad \mathbf1=\binom11.

Every entry belongs to $[0,1]$ , and

\det M=\frac14>0,\qquad \operatorname{tr}M=\frac32,

so $M$ is positive definite and its real fractional power is unambiguously defined. The positive square root is

M^{1/2} = \frac{M+\frac12I}{\sqrt{5/2}},

as follows immediately from the Cayley–Hamilton identity

M^2-\frac32M+\frac14I=0.

Since

M\mathbf1=\binom1{3/2},\qquad \mathbf1^{\mathsf T}M\mathbf1=\frac52,\qquad \mathbf1^{\mathsf T}M^2\mathbf1=\frac{13}{4},

we obtain

\mathbf1^{\mathsf T}M^{3/2}\mathbf1 = \frac{\frac{13}{4}+\frac12\cdot\frac52}{\sqrt{5/2}} = \frac9{\sqrt{10}}.

On the other hand,

\mathbf1^{\mathsf T}(M\mathbf1)^{3/2} = 1+\left(\frac32\right)^{3/2} = 1+\frac{3\sqrt6}{4}.

These quantities satisfy the strict reverse inequality

\frac9{\sqrt{10}} > 1+\frac{3\sqrt6}{4}.

Indeed, after multiplying by positive quantities and squaring, this inequality reduces to

149>60\sqrt6,

which holds because

149^2-(60\sqrt6)^2=22201-21600=601>0.

Therefore

\mathbf1^{\mathsf T}M^\alpha\mathbf1 > \mathbf1^{\mathsf T}(M\mathbf1)^\alpha

for the admissible positive-definite matrix $M$ and exponent $\alpha=3/2$ .