# Commutant of Jordan Block

Matrices $$B \in \mathbf{C}^{n\times n}$$ commuting with a given Jordan block $$A$$ are known to be upper triangular Toeplitz matrices. I have seen convincing proofs, but I wanted to derive this fact by my own method (in Dirac notation) which does not seem to work:

1. Consider the generalized eigenbasis (Jordan basis) for A and its adjoint $$A^\dagger$$ belonging to eigenvalue $$\lambda$$: $$|R_i\rangle \in \ker(A-\lambda)^i- \ker(A-\lambda)^{i-1}, |L_j\rangle \in \ker(A^\dagger-\lambda^*)^i- \ker(A^\dagger-\lambda^*)^{i-1}$$ giving rise to Jordan chains: $$|R_{i-1}\rangle = \ker(A-\lambda)|R_i\rangle, |L_{i-1}\rangle = \ker(A^\dagger-\lambda^*)|L_i\rangle$$

2. Express $$B$$ in this (right and left) basis: $$B = \sum_{ij} B_{ij}|R_i\rangle \langle L_j|$$

3. Apply $$A$$ on left resp. right:

$$AB=\sum_{ij} B_{ij}A|R_i\rangle \langle L_j|= \sum_{ij} B_{ij}\lambda|R_i\rangle \langle L_j|+\sum_{ij} B_{ij}|R_{i-1}\rangle\langle L_j|$$

$$BA=\sum_{ij} B_{ij}|R_i\rangle \langle L_j|A= \sum_{ij} B_{ij}\lambda|R_i\rangle \langle L_j|+\sum_{ij} B_{ij}|R_i\rangle\langle L_{j-1}|$$

1. So we have the equality $$\sum_{ij} B_{ij}|R_{i-1}\rangle\langle L_j|=\sum_{ij} B_{ij}|R_i\rangle\langle L_{j-1}|$$

2. Redefining summation indices via $$k:=i-1$$ and $$l:=j-1$$ yields

$$\sum_{kj} B_{k+1,j}|R_k\rangle\langle L_j|=\sum_{il} B_{i,l+1}|R_i\rangle\langle L_l|$$

1. Since $$|R_i\rangle \langle L_j|$$ constitutes a basis for the space of matrices, this equality holds for the respective coefficients separately: $$B_{i+1,j}=B_{i,j+1}$$

But Toeplitz matrices are characterized by $$B_{i+1,j}=B_{i,j-1}$$! What is wrong here?

Your error is actually fairly straightforward. It's that you've expressed $$B$$ with your right and left bases, but when we say that the matrices that commute with a Jordan block are Toeplitz matrices, we express those matrices inputs and outputs both in the right basis.
The key to fixing your argument is to switch everything to the right basis at the end. It's obvious from the matrix for a Jordan block (and then taking its conjugate transpose), that $$|L_j\rangle = |R_{n-j}\rangle$$. Using this equality, your equality of matrix entries becomes $$B_{i+1,n-j}=B_{i,n-(j+1)}=B_{i,n-j-1}$$, and reindexing with $$j=n-j$$, we get $$B_{i+1,j}=B_{i,j-1}$$ as desired.