Jordan Normal Form matrix decomposition into the sum of parts that commute. I'm learning about Jordan Normal Form matrices, and I've read that we can decompose a Jordan Normal Form matrix into the sum of two parts $$J=N_J+D_J$$where $D_J$ is the diagonal part (i.e. the eigenvalues) and $N_J$ is the superdiagonal part (hence is nilpotent). It is then stated that $D_J$ and $N_J$ commute ($N_JD_J=D_JN_J$). 
Why do $D_J$ and $N_J$ always commute? Initially I thought this would relate to the eigenvalues of the parts, but $N_J$ is nilpotent so has eigenvalues of zero only, whereas $D_J$ does not. Can someone point me in the right direction?

This is simply because if $D=\begin{bmatrix}d_1&0&\dots&0\\0&d_2&\dots&0\\\dots&\dots&\dots&\dots\\0&0&\dots &d_n\end{bmatrix}$, left multiplying any matrix $A$ by $D$ results in multiplying row no $k$ of $A$ by $d_k$ ($\;k=1,\dots, n$), and right multiplying results in multiplying column no $k$ by $d_k$, and that for the nilpotent part of a Jordan matrix, it has the same effect, as there is only $1$ non zero element per row and per column.
