Let \(J=J_n(\lambda)\) be a single \(n\times n\) Jordan block over a field \(F\). Determine all matrices \(X\in M_n(F)\) satisfying \[ XJ=JX. \]
Solution
Write \[ J=\lambda I+N, \] where \[ N=J_n(0) \] is the nilpotent Jordan block with \(1\)s on the superdiagonal. Since \(\lambda I\) commutes with every matrix, \[ XJ=JX \iff XN=NX. \]
Let \(X=(x_{ij})\). Comparing entries in \(XN=NX\) gives \[ x_{i,j-1}=x_{i+1,j} \] whenever both sides are defined, together with zeros below the main diagonal. Hence \(X\) is upper triangular and constant along each superdiagonal: \[ X= \begin{pmatrix} a_0&a_1&a_2&\cdots&a_{n-1}\\ 0&a_0&a_1&\cdots&a_{n-2}\\ 0&0&a_0&\cdots&a_{n-3}\\ \vdots&\vdots&\ddots&\ddots&\vdots\\ 0&0&\cdots&0&a_0 \end{pmatrix}. \] Such a matrix is exactly \[ a_0I+a_1N+\cdots+a_{n-1}N^{n-1}. \] Therefore \[ C_{M_n(F)}(J) =F[J] =\{f(J):f\in F[x],\ \deg f<n\}. \] In particular, the centralizer is an \(n\)-dimensional commutative \(F\)-algebra.