For every n∈N and every Hermitian matrix-valued Laurent polynomial P(z,w)=∑_(j,k)∈ FA_j,kz^jw^k∈ M_n!(C[z^pm1,w^pm1]) satisfying P(e^it,e^is)succeq 0 for all (t,s)∈R^2, does there exist an integer m≥ 1 and a matrix polynomial Q(z,w)∈ M_m,n!(C[z,w]) such that P(e^it,e^is)=Q(e^it,e^is)^Q(e^it,e^is) for all (t,s)∈R^2?

Mehta–Slofstra–Zhao and Savchuk–Schmüdgen factorization questions

For every $n\in\mathbb{N}$ and every Hermitian matrix-valued Laurent polynomial $P(z,w)=\sum_{(j,k)\in F}A_{j,k}z^jw^k\in M_n\!\left(\mathbb{C}[z^{\pm1},w^{\pm1}]\right)$ satisfying $P(e^{it},e^{is})\succeq 0$ for all $(t,s)\in\mathbb{R}^2$ , does there exist an integer $m\geq 1$ and a matrix polynomial $Q(z,w)\in M_{m,n}\!\left(\mathbb{C}[z,w]\right)$ such that $P(e^{it},e^{is})=Q(e^{it},e^{is})^*Q(e^{it},e^{is})$ for all $(t,s)\in\mathbb{R}^2$ ?

References

Primary source

arXiv

Additional references

Addendum to "Factoring non-negative operator valued trigonometric polynomials in two variables" — arXiv

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

An August 2026 addendum claims to settle the factorization questions and repair a flaw in an earlier proof, but independent verification is absent.

These questions concern when nonnegative operator-valued trigonometric polynomials in two variables admit the desired matrix Fejér–Riesz factorization. The latest addendum claims the theorem holds and pinpoints the earlier difficulty in Theorem $3.7$ .

August 2026 addendum

The addendum to Factoring non-negative operator valued trigonometric polynomials in two variables establishes the stated factorization theorem and corrects a consequence drawn from an earlier proof. It therefore claims a complete resolution of the tracked questions, but the result remains unrefereed and independently unassessed.

Current status (as of August 2026): The addendum claims the factorization questions are settled and the earlier proof corrected, but this claim remains unverified.

Sources

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