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A counterexample to Fisk’s Question 10

Abstract

A positive upper bivariate polynomial has a non-Hurwitz coefficient-Toeplitz minor.

Definition 1.1 (Upper bivariate polynomials).

Formalization. D5/S3/Zeros/FiskToeplitzRefutation.Upper (✓ std3).

Citation. Steve Fisk (2008). Questions about determinants and polynomials. URL: https://arxiv.org/abs/0808.1850v1.

Commentary.

PDF p. 1: “All polynomials f(x₁,…,x_d) with real coefficients such that f(σ₁,…,σ_d)≠0 for all σ₁,…,σ_d in the open upper half plane.” Here d=2. The carrier is Polynomial (Polynomial ℝ), with x in the coefficient ring and y outermost. Polynomial.eval₂ embeds the real coefficients into ℂ and evaluates at x and y.

Definition 1.2 (Positive coefficients on the support).

Formalization. D5/S3/Zeros/FiskToeplitzRefutation.PositiveUpper (✓ std3).

Citation. Steve Fisk (2008). Questions about determinants and polynomials. URL: https://arxiv.org/abs/0808.1850v1.

Commentary.

PDF p. 1: “P⁺_d = all polynomials in U_d with all positive coefficients.” Every nonzero coefficient (f.coeff j).coeff i is strictly positive; absent monomials have coefficient zero. The condition Upper excludes the zero input.

Definition 1.3 (Closed left half-plane roots).

Formalization. D5/S3/Zeros/FiskToeplitzRefutation.HurwitzStable (✓ std3).

Citation. Steve Fisk (2008). Questions about determinants and polynomials. URL: https://arxiv.org/abs/0808.1850v1.

Commentary.

PDF p. 1: “H₁ = all polynomials with all roots in the closed left half plane.” The real coefficients are mapped to ℂ before taking Polynomial.roots. The zero polynomial has the empty root multiset, so structural zero minors are included. The counterexample minor is nonzero.

Definition 1.4 (The coefficient matrix (5)).

Formalization. D5/S3/Zeros/FiskToeplitzRefutation.toeplitz (✓ std3).

Citation. Steve Fisk (2008). Questions about determinants and polynomials. URL: https://arxiv.org/abs/0808.1850v1.

Commentary.

Question 10, PDF p. 4: “Suppose f∈P⁺₂, and write f = ∑ f_i(x)y^i. Form the infinite matrix”. Matrix (5) has f₀,f₁,f₂,… in its first row, with each next row shifted one column to the right and zeros below the diagonal. Rows and columns are indexed by ℕ from zero. Polynomial.coeff is zero beyond the degree.

Definition 1.5 (Ordered finite minors).

Formalization. D5/S3/Zeros/FiskToeplitzRefutation.minor (✓ std3).

Citation. Steve Fisk (2008). Questions about determinants and polynomials. URL: https://arxiv.org/abs/0808.1850v1.

Commentary.

Question 10, PDF p. 4: “Show that the resulting matrix is totally stable. That is, every minor is a stable polynomial.” The polynomial is the determinant of the selected submatrix. StrictMono row and column maps in claim preserve the source’s order and determinant sign. The size-zero determinant is one.

Definition 1.6 (Question 10).

Formalization. D5/S3/Zeros/FiskToeplitzRefutation.claim (✓ std3).

Citation. Steve Fisk (2008). Questions about determinants and polynomials. URL: https://arxiv.org/abs/0808.1850v1.

Commentary.

Question 10, PDF p. 4: “Suppose f∈P⁺₂, and write f = ∑ f_i(x)y^i. Form the infinite matrix”. Matrix (5) is toeplitz. “Show that the resulting matrix is totally stable. That is, every minor is a stable polynomial.” The quantifiers range over every real bivariate polynomial, every natural size and every two strictly increasing maps Fin n → ℕ. The roots must have nonpositive real part.

Theorem 1.7 (Question 10 is false).

Proof. Machine-checked in Lean as D5/S3/Zeros/FiskToeplitzRefutation.result (✓ std3). ∎

Resolves. Problems/fisk10-toeplitz (refuted) by D5/S3/Zeros/FiskToeplitzRefutation.result.

Source. Repository-derived.

Acknowledgement. Steve Fisk (2008). Questions about determinants and polynomials. URL: https://arxiv.org/abs/0808.1850v1.

Commentary.

Take f=(4x+y+1)(x+y+4). Each affine factor has positive imaginary part when x and y do, and all six nonzero coefficients are positive. Rows 0,…,4 and columns 1,…,5 give a tridiagonal minor with determinant 5(x+1)(21x²+33x+21)(13x²−x+13). The complex number (1+15i√3)/26 is a root with real part 1/26. Thus this nonzero minor fails HurwitzStable. The five-by-five determinant factors as b(b²−a)(b²−3a), where a=4x²+17x+4 and b=5x+5.

References