Uniqueness of the Jordan decomposition I have seen it said that a matrix $M$ (over $\mathbb{C}$, say) has a unique decomposition $M = D + N$ where $D$ is diagonal and $N$ is nilpotent. I'm having trouble seeing this, since the Jordan form of a matrix is only unique up to ordering of the blocks. Wouldn't a different ordering give you different matrices?
 A: Sami Ben Romdhane's answer shows that the decomposition must be unique. However, I think your question still applies, whether how this can be, if the Jordan decomposition is not unique, namely only up to reordering of the blocks. 
The Jordan decomposition of $M$ looks like
$$M=PJP^{-1}=P(J_D+J_N)P^{-1}$$
where $J_D$ is the diagonal of $J$ and $J_N$ contains the offdiagonal entries. Now, the matrices in your decomposition are clearly $D=PJ_DP^{-1}$ and $N=PJ_NP^{-1}$. If you look at another Jordan decomposition which leads to a reordering of the blocks you will have
$$M=P'J'P'^{-1}=P'(J'_D+J'_N)P'^{-1}$$
The matrices $J'_D$ and $J'_N$ will of course be different from $J_D$ and $J_N$, but the matrix $P'$ "contains" this block reordering, such that $D=PJ_DP^{-1}=P'J'_DP'^{-1}=D'$ and $N=PJ_NP^{-1}=P'J'_NP'^{-1}=N'$. 
If it's not yet evident, you can also write $J'=RJR^T$, where $R$ is a permutation matrix that leads to this block reordering (note $RR^T=I$). Then $P'=PR^T$ and the result follows.
A: This is called Jordan–Chevalley decomposition: If $M$ is a matrix with  coefficients in a field $\mathbb{F}$  such that it's caracteristic polynomial splits over  $\mathbb{F}$ then there's a unique $(D,N)$ where $D$ is diagonalizable matrix and $n$ is nilpotent and $M=D+N$ and $DN=ND$.
For the unicity of this decomposition, suppose there's $(D,N)$ and $(D',N')$ such that
$$M=D+N=D'+N',$$
then $D-D'=N'-N=0$ since the zero matrix is the unique matrix diagonalizable and nilpotent.
