Crouzeix’s Conjecture: How OpenAI Proved the Complete Matrix Inequality with the Sharp Constant 2

A matrix can have only zero eigenvalues and still amplify a vector by a factor of two. That small surprise explains why Crouzeix’s conjecture matters: eigenvalues do not tell the whole story of nonnormal matrices. The right geometric object is their numerical range.

OpenAI’s September 26, 2026 direct proof establishes the sharp constant 2 for polynomial evaluation with arbitrary finite-dimensional matrix coefficients. Earlier scalar proofs had already appeared. The advance here is the complete inequality, which remains valid as both matrix dimensions and polynomial degree vary independently.

Its mechanism connects convex geometry, Fourier coefficients, and a positive block matrix. Following that connection reveals what was proved, why multiplication order matters, and where the exact constant comes from.

1. What Is Crouzeix’s Conjecture? The Original Numerical-Range Problem

The numerical range of a matrix collects its quadratic-form values on unit vectors:

Here the star denotes conjugate transpose, and vector norms are Euclidean. Matrix norms are induced operator norms, measuring maximum vector amplification. The numerical range is compact, convex, and contains every eigenvalue. The scalar conjecture asks for

For a normal matrix, which commutes with its adjoint, the spectral theorem already controls polynomial evaluation using eigenvalues. Contrast the nonnormal matrix

Its square vanishes, so both eigenvalues are zero. Yet its norm is 2. With $p(z)=z$, the maximum over eigenvalues is zero, making an eigenvalue-only bound impossible. Its numerical range is instead the closed unit disk.

2. Why the Sharp Constant 2 Was So Difficult to Prove

Crouzeix posed the scalar question in 2004 and established a universal complete bound of 11.08 in 2007. The Crouzeix–Palencia theorem reduced it to $1+\sqrt{2}$ in 2017. That was already dimension independent, but still above the conjectured optimum.

Scalar proofs appeared in 2026, including Shanmu Jin’s preprint and Emiel Lorist and Felix Schwenninger’s argument. Thus “Crouzeix conjecture solved” needs a scope qualification: OpenAI’s direct paper tackles the complete extension. Scalar functional-calculus arguments can use commutation identities that fail when coefficients become matrices.

Complete constant-two results also existed for restricted base dimensions. The direct paper cites the two-dimensional case and a 2026 result for dimensions at most three. Neither restriction settles the theorem for arbitrary base matrices. The issue was retaining the optimal bound while removing both dimension restrictions.

As in our Naimark problem coverage, identifying the contribution requires separating an earlier resolution from a new proof or stronger statement.

3. What OpenAI’s Complete Crouzeix Theorem Actually Proves

Theorem 1.1 allows any positive base dimension $n$, any positive coefficient dimension $m$, and any nonnegative degree $d$. Write $P(z)=\sum_{k=0}^{d}B_kz^k$, with $B_k\in M_m(\mathbb{C})$. Its tensor evaluation is

The tensor product places the base matrix first and acts on a space of dimension $nm$. For matrix-valued polynomials, the complete Crouzeix conjecture becomes

Setting $m=1$ recovers the scalar statement. Completeness requires exactly the same constant for every coefficient size. In spectral-set terminology, this is the complete 2-spectral set bound, expressed here for polynomials.

For developers, it supplies a worst-case certificate for polynomial matrix transformations. A scalar polynomial approximation error can be bounded over the numerical range before evaluating the matrix. It does not demonstrate faster training or a numerical-software speedup, a distinction also central to our log-concave sampling explainer.

For example, a residual represented by a scalar polynomial $r$ satisfies $\|r(A)b\|\leq2\max_{z\in W(A)}|r(z)|\,\|b\|$. This follows immediately from the theorem and the operator-norm definition. Finding a small maximum remains an approximation problem; the theorem does not choose the polynomial for you.

OpenAI’s release announcement attributes the collection to an internal frontier model. The family 325 scope documentation reports formalization covering the finite-dimensional bound, sharpness, and broader operator results. Those broader results exceed this paper’s stated theorem. This article does not independently rebuild Lean, run its comparator, or establish journal peer review. Our Fermat formalization guide explains why statement scope and verification remain distinct.

4. Mapping the Numerical Range Onto a Smooth Convex Domain

First enclose the numerical range strictly inside a bounded convex domain $\Omega$ with regular real-analytic boundary. This temporary smoothness permits convergent boundary expansions. An exterior conformal map $h$ sends the exterior unit disk onto the exterior of the domain and continues across its boundary.

On the unit circle, write $q(\lambda)=\lambda h^{\prime}(\lambda)$. This nonzero vector points outward, although it is not generally a unit normal. Convexity supplies the supporting-line inequality

Every interior point lies on the inward side of the tangent line. The conjugation turns complex multiplication into the appropriate real inner product. Our Brennan conjecture explainer introduces related conformal-map language, although the estimate here concerns matrix evaluation.

5. How Faber Polynomials Turn Boundary Geometry Into Coefficients

This map parametrizes the exterior, rather than sending the domain itself into a disk. That choice produces Laurent expansions at infinity. The analytically continued collar around the unit circle later ensures genuine convergence, instead of leaving the proof dependent on formal series manipulations.

Theorem 2.1 defines its polynomial basis through an expansion at infinity:

If $h(\lambda)=c\lambda+c_0+O(\lambda^{-1})$, then $b_0=1$, and $b_k$ has degree $k$ with leading coefficient $c^{-k}$. These are Faber polynomials for that exterior coordinate.

For positive k, their crucial feature appears after returning to the boundary:

A nonconstant basis polynomial has one positive Fourier mode and only negative modes besides it. The mixed coefficients $s_{kj}$ describe that correction. They come from subtracting the circle’s Cauchy kernel from the mapped kernel. Analytic continuation makes their double series absolutely and uniformly convergent on the boundary, justifying the integrations that follow.

The correction in the next section has expansion $s(\lambda,\mu)=\sum_{k,j\geq1}s_{kj}\lambda^{-k}\mu^{-j}$. The first index labels the input Faber polynomial and the second labels its negative output frequency. The argument does not assume that swapping those indices leaves the coefficients unchanged.

6. The Positive Kernel That Controls the Negative Fourier Modes

Define the mixed correction and its real kernel by

For distinct unit-circle points, the real part of the circle term contributes one after doubling. Consequently $K=2\operatorname{Re}(q(\lambda)/(h(\lambda)-h(\mu)))\geq0$ by the supporting-line inequality. The apparent diagonal singularity cancels in the correction, so continuity preserves positivity there.

Use normalized angular measure $d\tau=dt/(2\pi)$. Every mixed term has nonzero frequency in each variable, giving

These two identities make the associated integral operator an $L^2$ contraction: Jensen’s inequality controls each averaged value, and the other marginal controls its integral. Both normalizations matter. A one-sided averaging argument would leave the total energy uncontrolled.

More explicitly, $(Tu)(\mu)=\int_{\mathbb T}K(\lambda,\mu)u(\lambda)\,d\tau(\lambda)$. Applied to $u(\lambda)=\sum_{k=1}^{N}C_k\lambda^k$, integration cancels the constant and conjugate correction terms. The surviving mixed term produces precisely the negative-frequency part required by the Faber expansion.

7. The Coefficient Norm Estimate: Where the Factor 2 First Appears

Write $G=\sum_{k=0}^{N}b_kC_k$. The Hilbert–Schmidt norm $|C|_2=(\operatorname{tr}(C^{*}C))^{1/2}$ is the square root of the sum of squared entry magnitudes. Unlike the operator norm, it gives a Hilbert-space coefficient geometry.

The boundary expansion separates into a constant, positive frequencies, and negative frequencies. Parseval’s identity makes their squared energies add. The contraction bounds negative energy by positive energy, proving Proposition 2.2:

Here the boundary norm integrates squared Hilbert–Schmidt norms against normalized angular measure. Orthogonality prevents cancellation between different frequencies, the same basic principle used in Fourier analysis of wave estimates.

The lower bound simply retains the constant and positive energies, which are already nonnegative. For the upper bound, the negative part contributes at most one additional copy of the positive energy. No pointwise comparison between individual Fourier coefficients is required.

Only nonconstant coefficients can generate negative-frequency energy. Their weight doubles; the constant’s does not. Keeping that asymmetry is essential to the final comparison.

8. The Resolvent and the Positivity Behind OpenAI’s Proof

Now reintroduce the matrix through its resolvent:

The scalar expansion becomes $R(\lambda)=\sum_{k\geq0}b_k(A)\lambda^{-k}$. Strict spectral separation gives geometric coefficient decay and uniform convergence. Its average is the identity.

More decisively, $R+R^{*}\succeq0$, where the symbol means positive semidefinite. To see why, set $B=h(\lambda)I-A$. Direct multiplication gives

Testing the right side against a unit vector produces twice the supporting-line expression with $z=v^{*}Av$. That point belongs to the numerical range, so the result is nonnegative. Invertibility of the resolvent denominator transfers positivity back. The boundary geometry has become a matrix inequality.

9. Why Singular Vectors and Both Products FG and GF Are Essential

Normalize a polynomial $F$ so its boundary operator norm is at most one, and put $\gamma=\|F[A]\|$. The zero case is immediate. Otherwise choose unit singular vectors satisfying

Explicitly, $x=\sum_{i=1}^{m}Xe_i\otimes e_i$ and $y=\sum_{i=1}^{m}Ye_i\otimes e_i$, with $X,Y$ of size $n\times m$. The vectors $e_i$ are standard coefficient-space basis vectors. Unit normalization gives $|X|_2=|Y|_2=1$, while preserving all correlations between the two tensor factors.

Encode their tensor coordinates by matrices $X,Y$, and define $D=X^{*}RX$, $E=Y^{*}RY$, $M=Y^{*}RX$, and $L=X^{*}RY$. We use D and E for the paper’s diagonal fields P and Q. Subscripts denote their negative Fourier coefficients. Set $a^2=\sum_{k\geq0}|M_k|_2^2$.

The coefficient pairing gives $|y^{*}G[A]x|\leq a(\sum_k|C_k|_2^2)^{1/2}$. Taking $G=F$ shows $a>0$. Evaluation preserves multiplication in order, yielding two different tests:

Left and right multiplication by the boundary contraction F each decrease Hilbert–Schmidt norm. Applying both coefficient estimates therefore bounds both diagonal functionals. Duality reverses the coefficient weights:

A concrete warning is supplied by matrix units: $E_{ij}$ has one nonzero entry, equal to one, in row $i$ and column $j$. Then $E_{12}E_{21}=E_{11}$, whereas $E_{21}E_{12}=E_{22}$. Even constant matrix-valued polynomials can therefore give different ordered products. Completeness requires a proof that respects this elementary difference.

Concretely, test with entrywise conjugate coefficients, halving each nonconstant one, then pass through finite truncations. Interchanging FG and GF would erase one required estimate. Matrix coefficients do not grant that shortcut.

10. The Final Positive Block Argument: Deriving the Exact Bound 2

Compress the positive resolvent field with the combined matrix of singular-vector coordinates:

Let $E_D,E_E,E_C$ denote the integrated squared Hilbert–Schmidt norms of its two diagonal fields and its lower cross field. Parseval gives $E_D\leq4a^2/\gamma^2$ and $E_E\leq4a^2/\gamma^2$: the constant doubles before squaring, while each nonconstant mode appears with its adjoint.

For the cross field, $L_0^{*}=M_0$. Its energy is therefore

Nothing discards the constant coefficient. Finally set $J=\operatorname{diag}(I_m,-I_m)$. Both $\Delta$ and $J\Delta J$ are positive, so their product has nonnegative trace. Expanding that trace and integrating gives $2E_C\leq E_D+E_E$. Thus

The constant follows from matching coefficient weights with positive block geometry. It is an exact universal bound, rather than a numerical fit.

11. Extending the Proof to Every Numerical Range, and Showing 2 Is Optimal

Smoothness was scaffolding. Lemma 4.1 constructs analytic convex outer domains around every nonempty compact convex set, including a point or segment. Their maximum distance from the set tends to zero. Applying the domain bound and uniform continuity of the fixed polynomial makes the boundary maxima converge to the maximum on the numerical range.

The construction combines finitely many supporting directions using exponentials of affine functions. Including both coordinate axes ensures boundedness and a positive-definite Hessian. A regular level curve then supplies the analytic boundary. This explains why the approximation remains available even when the original set has no interior.

Sharpness returns to the displayed nilpotent example with $P(z)=z$. For a unit vector, $|u^{*}Au|=2|\overline{u_1}u_2|\leq1$, with equality when both coordinates equal $1/\sqrt{2}$. Every phase and radius up to one occurs. Meanwhile $Ae_2=2e_1$, so its norm is exactly 2. Consequently

Crouzeix’s conjecture reaches its precise endpoint: constant 2, arbitrary finite coefficient size, and no missing degenerate numerical ranges. For applications, the next task is obtaining useful numerical-range enclosures and approximation bounds. For readers, the reusable insight is how geometry becomes positivity, then a norm estimate. Continue with Binary Verse AI’s guide to evaluating AI mathematics and follow our research explainers for proofs unpacked with their assumptions intact.