Commutivity of Endomorphism with Jordan Decomposition I asked this question earlier, but am still a bit stuck on the following question, and would appreciate any help.
Let $\mathbb{F}$ be algebraically closed. If $T=D+N$ is a Jordan decomposition of a linear operator, ($D$ is diagonal and $N$ is nilpotent), prove for $S\in \text{End}(V)$, that $T$ commutes with $S$ if and only if $D$ and $N$ commute with $S$. That is, show $TS=ST\iff NS=SN$ and $DS=SD$.
I got one part already, that is, suppose $N$ and $D$ commute with $S$. Then clearly:
$$TS=(N+D)S=NS+DS=SN+SD=S(N+D)=ST$$
But I am a bit stuck on the other way. I was thinking I could maybe make use of the fact that because $\mathbb{F}$ is algebraically closed, we know every polynomial $f(x)\in \mathbb{F}[x]$ factors as $f(x)=(x-c_1)^{e_1}\cdot \ldots \cdot (x-c_n)^{e_n}$, where $e_i\geq 1$ and $c_i$ are distinct roots. I know given two polynomials $f(x),g(x)$, clearly they commute, that is $f(x)g(x)=g(x)f(x)$. Thus $f(T)$ and $g(T)$ commute. I can thus represent $D$ and $N$ as polynomials in $T$, so $DN=ND$. But how can I show $DS=SD$ and $NS=SN$ given $TS=ST$? Any help would be much appreciated.

Hint In fact, use - as you say - that $D$ and $N$ are polynomials in $T$. There is a lemma :

Lemma : The commutant of $S$, 
  $$
C(S):=\{S\mid ST=TS\}
$$
  is an algebra (i.e. closed by products and linear combinations). 

Hence, if $S$ commutes with $T$, it commutes with all powers of $T$ and polynomials $P(T)$. 
