Partition of a complex matrix Given $A\in \Bbb M(n,\Bbb C)$,I was asked to show that there exist two polynomial $g$ and $h$ with both constant term vanished,such that $g(A)$ is diagonalizable ,$h(A)$ is nilpotent and $A=g(A)+h(A)$
Well...I got totally confused,especially with requirement that constant term vanish.Why do they have to be so?Is $g$,$h$ related to the characteristic polynomial of $A$?
 A: Theorem (Jordan-Chevalley decomposition, décomposition de Dunford) Let $T$ be a linear operator on a finite-dimensional vector space $V\simeq F^m$. If the minimal polynomial of $T$ splits over $F$, then there exist two operators $D,N$ on $V$ such that
1) $T=D+N$
2) $DN=ND$
3) $D$ is diagonalizable and $N$ is nilpotent.
Moreover:
4) Such a pair $(D,N)$ is unique. 
5) There exist polynomials $d, n\in F[X]$ such that $D=d(T)$, $N=n(T)$, and $d(0)=n(0)=0$.
Remarks 


*

*The assumption that the minimal polynomial splits over $F$ is equivalent to assuming that $T$ be annihilated by some nonzero polynomial splitting over $F$. It is also equivalent to ask that the characteristic polynomial split. Of course, this is always satisfied over an algebraically closed field like in your case.

*The existence of a pair $(D,N)$ satisfying 1)2)3) follows immediately from the Jordan normal form of $T$. But we would still have to prove 4) and 5). Instead, I will give a proof that does not use the Jordan normal form. And you can actually use the theorem above to prove the existence of the Jordan normal form, as it restricts it to the case of nilpotent operators.
Sketch  We first prove that there exists a pair $(D,N)$ satisfying 1)2)3) and 5) using the Chinese Remainder Theorem on the PID $F[X]$. Then we use this particular pair to prove 4).
Proof  Let $m_T(X)=\prod_{j=1}^s (X-\lambda_j)^{m_j}$ be the minimal polynomial where the roots $\lambda_j$ are pairwise distinct. By the Chinese Remainder Theorem (distinguishing two cases: $\lambda_j\neq 0$ for every $j$ or there exists $j$ such that $\lambda_j=0$), there exists a polynomial $d \in F[X]$ such that
$$
d(X)\equiv \lambda_j \;\mathrm{mod}\; (X-\lambda_j)^{m_j}\quad \forall j\quad \mbox{and}\quad d(X)\equiv 0 \;\mathrm{mod}\; X
$$
Since $V=\bigoplus_{j=1}^s \ker (T-\lambda_j \mathrm{Id})^{m_j}$ and since the operator $d(T)$ coincides with $\lambda_j \mathrm{Id}_{N_j}$ on every invariant subspace $N_j=\ker (T-\lambda_j \mathrm{Id})^{m_j}$, we see that $d(T)$ is a diagonalizable operator (if you think of the Jordan normal form of $T$, then $d(T)$ is the operator we obtain by erasing all the off-diagonal entries). For $n$, we now set $n(X)=X-d(X)$. It is clear that $D:=d(T)$ and $N:=n(T)$ satisfy 1)2)5). Moreover, we have already seen that $D$ is diagonalizable. It is not much harder to check that $N$ is nilpotent by looking at its restrictions to every $N_j$.
It remains to establish uniqueness 4). Assume $(D',N')$ is another pair satisfying 1)2)3). Then $D-D'=N'-N$ and, since $D,N$ are polynomials in $T$, $D'$ and $N'$ both commute with $D$ and $N$. It follows that $S=D-D'=N'-N$ is diagonalizable and nilpotent. Thus $S=0$. $\Box$
Concrete procedure to compute the decomposition  Take any annihilating polynomial $p(X)=\prod_{j=1}^s (X-\lambda_j)^{m_j}$ for $T$. Do partial fraction decomposition
$$
\frac{1}{p(X)}=\sum_{j=1}^s \sum_{k=1}^{m_j}\frac{\alpha_{k,j}}{(X-\lambda_j)^j}
$$
For every $j$, set $u_j(X):=\sum_{k=1}^{m_j}\alpha_{k,j}(X-\lambda_j)^{m_j-j}$ and $q_j(X):=\prod_{i\neq j}(X-\lambda_i)^{m_i}$. Then
$$
D=\sum_{j=1}^s\lambda_j \cdot (u_jq_j)(T)
$$
That is: the spectral projection onto $N_j$ is given by $(u_jq_j)(T)$.
