The complete Crouzeix conjecture

The explanation

Take the 2-by-2 matrix A with top row (0, 2) and bottom row (0, 0). Its only eigenvalue is 0. So looking at eigenvalues alone suggests A is tiny. But A stretches the vector (0, 1) to (2, 0), so its size, or norm, is 2. A matrix whose eigenvalues miss its true size like this is called nonnormal. Even a nonzero nilpotent matrix, one with a power equal to zero, has spectrum {0}Source 1.

A better guide is the numerical range W(A). It collects every value u*Au, over unit vectors u. For this A, the numerical range is the disc of radius 1 around 0. Now take the polynomial p(z) = z. Its largest size on W(A) is 1. But p(A) = A has norm 2. So the ratio here is exactly 2.

Crouzeix conjectured in 2004 that 2 is always enough. For any polynomial p, the norm of p(A) should be at most twice the largest value of |p| on W(A). He proved a universal complete bound of 11.08 in 2007Source 2. Crouzeix and Palencia later lowered it to 1 + √2, about 2.414Source 3. Badea, Crouzeix and Delyon proved the complete constant two for all 2-by-2 matricesSource 4. Recent preprints give scalar proofs with constant twoSource 5. Åhag, Czyż and Virtanen prove the complete form for matrices of size at most 3Source 6.

The word complete matters. Here the polynomial's coefficients are themselves m-by-m matrices. The question asks for the factor 2 for every m, every matrix size, and every degree. The obstacle is that a scalar proof need not survive this step. Coefficient matrices need not commute, so their order of multiplication mattersSource 7.

The first paper gives a direct proof for matrices of every size and every coefficient size. This resolves the complete Crouzeix conjecture in its matrix form, including numerical ranges that are points or line segmentsSource 8. The constant two cannot be loweredSource 9. The example above shows why. The proof places the numerical range inside a smooth convex region. It then controls the polynomial through coefficient estimates along the region's boundary.

The second paper proves the same sharp bound for every bounded operator on a complex Hilbert spaceSource 10. A Hilbert space is a space with lengths and angles, possibly of infinite dimension. No separability assumption is neededSource 11. It also proves a structural result for finite matrices. A change of inner product, distorting lengths by a factor of at most two, turns a related map into a contractionSource 12. A contraction is a map that never increases lengths. One fixed boundary density then represents every matrix-valued test function. The same change and density work for every coefficient size and every functionSource 13.

How sure is this? Lean, a computer proof checker, proves the complete inequality with constant 2 for every matrix and every coefficient size. It proves that no smaller constant works. It also covers every bounded operator on any complex Hilbert space, and the structural result for finite matrices. No specialist review has been published, but the main statements have Lean proofs.

What it doesn't show. This does not improve the factor 2, which is the best possible, and the structural result is stated only for finite matrices.