Weak majorization of higher-power coefficients in the exponential product formula

From papers

Let $A,B\in\mathbb{H}_n$ be Hermitian matrices, and let $H$ and $R_k$ denote the operators defined in the paper from the coefficients of $e^{(A+B)t}$ and $e^{At}e^{Bt}$ , respectively. Write $\lambda(\cdot)$ for the vector of eigenvalues in decreasing order, and let $\prec_w$ denote weak majorization. Higher-power weak-majorization conjecture. For every integer $k\ge 5$ ,

\lambda(H^k)\prec_w\lambda(R_k).

The preceding results establish this type of inequality for the cubic and quartic coefficients, while the displayed assertion proposes it for all higher powers; its status is not resolved in the supplied context.

Progress summary

Open

The conjecture remains open: results cover the cubic and quartic cases, but no verified progress covers higher powers.

The conjecture, formulated as Conjecture 5.5 in a January 2026 paper, asks whether the eigenvalues of the higher powers of the sum are weakly majorized by those of the corresponding symmetrized product coefficients for every $k\ge 5$ . The paper does not claim a proof or counterexample.

Known results

The cubic and quartic cases are proved: $\lambda(H^k)\prec_w\lambda(R_k)$ for $k=3,4$ .
The paper proves the related singular-value comparison $\lambda(H^k)\prec_w\sigma(Q_k)$ for $k=3,4$ .
For $k\ge 5$ , the authors give an explicit $D_5=R_5-H^5$ expansion but cannot organize it into the positivity decomposition required for a general proof.

Current status (as of August 2026): The cases $k=3,4$ are settled, while the higher-power conjecture for every $k\ge 5$ remains unresolved.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Teng Zhang, “Weak majorization inequalities for the cubic and quartic coefficients of e^(A+B)t versus e^Ate^Bt”, arXiv:2601.07286 (2026).

Additional references

2 papers in this index state this conjecture (2023–2026). The statement above is taken from the most recent of them; the others are arXiv:2301.07934.

1 week ago

Counterexample, including every odd order at least five.

Consider the higher-order weak-majorization conjecture stated as Conjecture 5.5 in arXiv:2601.07286. For Hermitian matrices $A,B$ , write

H=A+B,\qquad Q_k=\sum_{j=0}^k\binom{k}{j}A^jB^{k-j},\qquad R_k=\frac{Q_k+Q_k^*}{2}.

The conjecture asserts $\lambda(H^k)\prec_w\lambda(R_k)$ for every $k\ge5$ .

Take the real symmetric matrices

A=\begin{pmatrix}0&0\\0&-2\end{pmatrix}, \qquad B=\begin{pmatrix}0&1\\1&0\end{pmatrix}.

Direct exact computation gives

H^5=\begin{pmatrix}-12&29\\29&-70\end{pmatrix},\qquad Q_5=\begin{pmatrix}0&1\\121&-122\end{pmatrix}, \qquad R_5=\begin{pmatrix}0&61\\61&-122\end{pmatrix}.

Weak majorization at full rank would require

\operatorname{tr}(H^5)\le\operatorname{tr}(R_5).

Instead,

\operatorname{tr}(H^5)=-82>-122=\operatorname{tr}(R_5).

Thus the conjecture already fails for two-by-two real symmetric matrices at its first proposed order.

More strongly, the same matrices disprove every odd order $k\ge5$ . Set

Z=\operatorname{diag}(1,-1),\qquad S=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad A_c=cI+Z,\quad B=S.

Since $Z^2=S^2=I$ and $ZS+SZ=0$ , the exact exponential generating function for the trace difference is

\sum_{k\ge0}\frac{t^k}{k!} \left(\operatorname{tr}R_k-\operatorname{tr}(A_c+B)^k\right) =e^{ct}\left(1+\cosh(2t)-2\cosh(\sqrt2\,t)\right).

Coefficient extraction yields

\operatorname{tr}R_k-\operatorname{tr}(A_c+B)^k = \sum_{j=2}^{\lfloor k/2\rfloor} \binom{k}{2j}c^{\,k-2j}\bigl(4^j-2^{j+1}\bigr).

Whenever $c<0$ and $k\ge5$ is odd, every summand is strictly negative. Choosing $c=-1$ recovers the displayed integer matrices and proves failure simultaneously at every odd order $k\ge5$ .

This refutes precisely Conjecture 5.5 concerning the eigenvalues of $R_k$ . The separate singular-value question $\lambda(H^k)\prec_w\sigma(Q_k)$ in Problem 1.1 is not refuted by this argument.

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Shivam Patel · 15 hours ago