Prove that $ND = DN$ where $D$ is a diagonalizable and $N$ is a nilpotent matrix.

Let $A$ be an $n \times n$ complex matrix. Prove that there exist a diagonalizable matrix $D$ and a nilpotent matrix $N$ such that
a. A = D + N
b. DN = ND
and show that these matrices are uniquely determined.

I think I've solved the part a but don't have an idea to continue. Here is what I've tried:

Let $D = \begin{bmatrix} a_{11} & 0 \\ * & a_{nn} \\ \end{bmatrix}$ and $N = \begin{bmatrix} 0 & * \\ 0 & 0 \end{bmatrix}$. Since D is a lower triangular matrix, its determinant is equal to product of its diagonal entries and hence its characteristic polynomial is $(x-a_{11})...(x-a_{nn}).$ Then D is diagonalizable. Similarly, characteristic polynomial of $N$ is $x^n$ then by Cayley-Hamilton $N^n = 0$.
Update: My assumption for D to be diagonalizable was wrong, eigenvalues need not to be distinct. Update after answers: Thank you all for your help. It was just a question asked in the end of the chapter "Canonical Forms". So I just know Jordan form, Rational form, etc. I don't know anything about the Lie Algebra, Semi simple matrices, Representation theory, Perfect field mentioned in the answers. Honestly, answers didn't help me to understand the solution but they seem useful so maybe they help other people. Thanks.

• Can we assume that eery matrix can be put into Jordan canonical form? – Omnomnomnom May 10 '15 at 2:16
• Do you know Jordan decomposition? – Michael Burr May 10 '15 at 2:16
• $D$ of given form need not be diagonalizable; consider $$D = \begin{pmatrix}0&0\\1&0\end{pmatrix}.$$ But what if you apply such an argument to the Jordan Canonical Form? – Travis May 10 '15 at 2:17
• I know about the Jordan form but why do we assume it can be put into Jordan form? – rackne May 10 '15 at 2:20
• The field of complex numbers is algebraically closed. Hence any $n \times n$ matrix has $n$ (not necessrly distinct ) eigenvalues! – mich95 May 10 '15 at 2:51

It is the Jordan-Chevalley decomposition. cf. http://en.wikipedia.org/wiki/Jordan%E2%80%93Chevalley_decomposition

It is valid when $A\in M_n(K)$ where $K$ is a perfect field. Then $D$ (a semi-simple matrix) and $N$ (a nilpotent matrix) are polynomials in $A$ with coefficients in $K$.

Moreover, there is a Newton-like method that permits to calculate, with a polynomial complexity, the previous polynomials. In particular, we stay in $K$ and don't go in its algebraic closure.

EDIT. @ rackne , you are not an eagle. Indeed Tlön Uqbar Orbis Tertius did the job up to 80 %. Now I write the end of the required proof.

Orbis proved that there is an invertible $P$ s.t. $P^{-1}AP=diag(a_1I_{\alpha_1}+N_1,\cdots,a_kI_{\alpha_k}+N_k)$ where the $(a_i)$ are distinct complex numbers, $\sum_i\alpha_i=n$ and $N_i$ is nilpotent. Then put $D=Pdiag(a_1I_{\alpha_1},\cdots,a_kI_{\alpha_k})P^{-1}$ and $N=Pdiag(N_1,\cdots,N_k)P^{-1}$. Clearly $DN=ND$, $D$ is diagonalizable and $N$ is nilpotent.

Hint: Let $P$ be the minimal polynomial for $A$. It can be written as a product of factors having the form $(X-a)^n$, for example $P = \prod_{i=1}^n (X-a_1)^{n_i}$. Then, the so-called "kernel lemma" states that $$\ker P(A) = \bigoplus_{i=1}^n \ker (A - a_i \mathrm{Id})^{n_i}$$ and each of the subspaces $E_i = \ker (A - a_i \mathrm{Id})^{n_i}$ is stable.

Now note that on $E_i$, you have $A = a_i \mathrm{Id} + (A - a_i \mathrm{Id})$ and the second term is nilpotent.

You could use this to have a general decomposition of the form $D+N$.

To prove that $DN=ND$ and $D,N$ are unique, you could try to prove that $D$ and $N$ are indeed polynomials in $A$.

You can see Prop. 4.2, p.17 in Humphreys, James E. (1972), Introduction to Lie Algebras and Representation Theory. This is exactly what you need.