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.
Best Answer
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.