TITLE: Let $J$ be a $k \times k$ jordan block, prove that any matrix which commutes with $J$ is a polynomial in $J$
QUESTION [8 upvotes]: Let $J$ be a $k \times k$ jordan block, prove that any matrix which commutes with $J$ is a polynomial in $J$.
I appreciate your hints, Thanks

REPLY [9 votes]: One direction of the equivalence is easy. For any polynomial $p$ and any square matrix $A$, $p(A)A=Ap(A)$. 
For the other direction, we can assume that the $k\times k$ Jordan block $J_k$ has zeros on the diagonal. Indeed, any other Jordan block can be written in the form $\beta I + J_k$ and it is easy to see that $A$ commutes with $\beta I+J_k$ if and only if it commutes with $J_k$. 
Let $A=(a_{ij})$ be a $k\times k$ matrix and assume that $AJ_k-J_kA=0$. 
We first show that the this implies that $A$ is an upper triangular (UT) and Toeplitz (T) matrix
and proceed by induction on $k$. So assume that if a $(k-1)\times (k-1)$ matrix commutes with a $(k-1)\times (k-1)$ Jordan block, it is UT&T.
Let's write $J_k$ and $A$ in the partitioned form
$$
A=\begin{bmatrix}A_{11}&a_{12}\\a_{21}^*&\alpha_{22}\end{bmatrix},
\quad
J_k=\begin{bmatrix}J_{k-1}&e_{k-1}\\0&0\end{bmatrix},
$$
where $A_{11}$ and $J_{k-1}$ are $(k-1)\times(k-1)$. Let's have a look on the commutator:
$$
\begin{split}
AJ_k-J_kA&=\begin{bmatrix}A_{11}&a_{12}\\a_{21}^*&\alpha_{22}\end{bmatrix}\begin{bmatrix}J_{k-1}&e_{k-1}\\0&0\end{bmatrix}-\begin{bmatrix}J_{k-1}&e_{k-1}\\0&0\end{bmatrix}\begin{bmatrix}A_{11}&a_{12}\\a_{21}^*&\alpha_{22}\end{bmatrix}\\
&=\begin{bmatrix}A_{11}J_{k-1}-J_{k-1}A_{11}-e_{k-1}a_{21}^*&A_{11}e_{k-1}-J_{k-1}a_{12}-\alpha_{22}e_{k-1}\\a_{21}^*J_{k-1}&a_{21}^*e_{k-1}
\end{bmatrix}\\
&=\begin{bmatrix}A_{11}J_{k-1}-J_{k-1}A_{11}-e_{k-1}a_{21}^*&(A_{11}-\alpha_{22}I)e_{k-1}-J_{k-1}a_{12}\\a_{21}^*J_{k-1}&a_{21}^*e_{k-1}
\end{bmatrix}.
\end{split}
$$
Since $AJ_k-J_kA=0$ by assumption, the last two block rows imply that 
$$
a_{21}^*e_{k-1}=0, \quad a_{21}^*J_{k-1}=0.
$$
The first equation implies that the last entry of $a_{21}=0$ is zero, while the other says that the first $k-2$ entries of $a_{21}$ are zero. Therefore, $a_{21}=0$ and indeed the matrix $A$ is UT (since by the induction assumption, $A_{11}$ is UT).
The first block of $AJ-JA=0$ implies that
$$
A_{11}J_{k-1}-J_{k-1}A_{11}-e_{k-1}a_{21}^*=A_{11}J_{k-1}-J_{k-1}A_{11}=0
$$
because we already showed that $a_{21}=0$ and hence from $A_{11}J_{k-1}-J_{k-1}A_{11}=0$ we have that $A_{11}$ is UT&T (by the induction assumption).
It remains to show that if $A_{11}$ is T then from $(A_{11}-\alpha_{22}I)e_{k-1}-J_{k-1}a_{12}=0$ we have that $A$ is T. We have
$$
(A_{11}-\alpha_{22}I)e_{k-1}
=
\begin{bmatrix}
a_{1,k-1}\\
\vdots\\
a_{k-2,k-1}\\
a_{k-1,k-1}-\alpha_{22}
\end{bmatrix},
\quad
J_{k-1}a_{12}=\begin{bmatrix}
a_{2,k}\\
\vdots\\
a_{k-1,k}\\
0
\end{bmatrix}.
$$
Since both vectors are equal, it gives $a_{i,k-1}=a_{i+1,k}$ for $i=1,\ldots,k-2$ and $\alpha_22=a_{k-1,k-1}$. Since $A_{11}$ is UT&T, the diagonal of $A_{11}$ is constant and thus the diagonal of $A$ is constant as well and the last $k-2$ entries of $a_{12}$ are "copies" of the first $k-2$ entries of the last column of $A_{11}$. Therefore, $A$ is T.
Summarizing, $A$ and $J_k$ commute implies that $A$ is a UT&T matrix and consequently it can be written in the form
$$
A=\begin{bmatrix}
a_1 & a_2 & \ldots & a_k \\
    & a_1 & \ldots & a_{k-1} \\
    &     & \ddots & \vdots \\
    &     &        & a_1
\end{bmatrix}.
$$
By realizing how the powers of $J_k$ look like, it is easy to see that $A$ can be written as
$$
A=a_1 J_k^0 + a_2 J_k^1 + \ldots + a_k J_k^{k-1} =: p(J_k),
\quad p(t)=a_1+a_2t +\ldots+a_k t^{k-1}.
$$
Q.E.D.