Crouzeix's Conjecture

A matrix is a table of numbers. A square matrix acts on a list of numbers, called a vector. Here is one, which sends the vector $(x, y)$ to $(100 y, 0)$ :

A = (\begin{matrix} 0 & 100 \\ 0 & 0 \end{matrix})

The norm of a matrix, written $‖ A ‖$ , is the largest factor by which it stretches the length of a vector. Here $‖ A ‖ = 100$ , because $(0, 1)$ goes to $(100, 0)$ .

A polynomial is a formula that only adds and multiplies numbers and variables. You can put a matrix into one. If $p (z) = z^{2} + 3 z + 2$ , then $p (A) = A^{2} + 3 A + 2 I$ . Here $A^{2}$ means "apply $A$ twice", and $I$ is the matrix that changes nothing. The question is how large $‖ p (A) ‖$ can be. For a large matrix, this is hard to calculate directly.

The first guess is to look at the eigenvalues. An eigenvalue is a number $λ$ , the Greek letter lambda, with $A v = λ v$ for some nonzero vector $v$ , so $A$ only rescales $v$ . But eigenvalues can miss the size of a matrix. The matrix above has only the eigenvalue 0, yet it stretches a vector by 100.

The numerical range of the matrix with rows

(1, 1.5)

and

(0, - 1)

. Each small dot is

x^{*} A x

for one vector

x

. The two large dots are the eigenvalues,

1

and

- 1

The numerical range does better, and it uses complex numbers. A complex number has the form $a + b i$ , where $i^{2} = - 1$ . It is a point on a flat map, with $a$ across and $b$ up.

Take a vector $x$ of length 1, apply $A$ , and measure how much of the result points along $x$ . This gives one complex number, written $x^{*} A x$ . The numerical range $W (A)$ is the set of these points for all such $x$ . It is a region with no dents, and it holds all the eigenvalues. For the matrix above, it is a disk with radius 50, so it shows the size that the eigenvalues miss.

Crouzeix's conjecture says that this region gives the size of $p (A)$ , up to a factor of 2. It says that this holds for every matrix and every polynomial:

‖ p (A) ‖ \leq 2 \max_{z \in W (A)} | p (z) |

Here $| p (z) |$ is the distance from 0 to $p (z)$ on the map, and $\max$ means the largest value over the region. So the right side uses $p$ as an ordinary formula. Take its largest size over the region, then double it. So a question about a large table becomes a question about one formula on one region.

The factor 2 cannot go lower. Use the matrix above with $p (z) = z$ . Then $‖ p (A) ‖ = 100$ , and the largest $| z |$ on the disk is 50. Earlier proofs reached a factor of 11.08, and then $1 + \sqrt{2}$ , which is about 2.414.

Alex Townsend and Anne Greenbaum report that Shanmu Jin posted a proof, whose key step came from a long run of GPT-5.6 Sol. They and Michel Crouzeix checked the proof and believe that it is correct. Emiel Lorist and Felix Schwenninger posted a separate proof of five pages, and they used the same model to explore strategies.