Proof of Jordan-Chevalley decomposition

Question

Let $A$ be a square matrix over $\mathbb{C}$. Prove there are matrices $D$ and $N$ such that $A = D + N$ such that $D$ is diagonalizable, $N$ is nilpotent and $DN = ND$.

I can see that any nilpotent matrix has to satisfy $N^l=0$ for some $l$. I'm not sure how to go about proving that all these conditions hold for any square matrix A.

Answer 1

"Intuition": We construct $D$ and $N$ in the following fashion: We want to show, that for any invertible matrix $S \in \text{GL}(n,K)$ we have $$ A = S^{-1} \tilde{D} S + S^{-1} ( J(A) - \tilde{D}) S, $$ where $J(A)$ is the Jordan decomposition of $A$.

This is true because from Jordan decomposition we that the exists an invertible matrix $S \in \text{GL}(n,K)$, so that \begin{align*} A = S^{-1} J(A) S = S^{-1} (\tilde{D} + J(A) - \tilde{D}) S = S^{-1} \tilde{D} S + S^{-1} ( J(A) - \tilde{D}) S. \end{align*}

Existence: Let $\tilde{D}$ be the diagonal matrix whose entries are the eigenvalues of $A$. Because every diagonalisable matrix is similar to a diagonal matrix, we know, that $S^{-1} \tilde{D} S$ is diagonalisable und let $D := S^{-1} \tilde{D} S$.

Now $J(A) - \tilde{D}$ is a upper triangular matrix, whose diagonal only contains zeros and therefore nilpotent. Also, $N := S^{-1}( J(A) - \tilde{D}) S$ is nilpotent and we have $A = D + N$.

Commutativity:

Because $\tilde{D}$ is a diagonal matrix we have \begin{align*} DN & = S^{-1} \tilde{D} S S^{-1} ( J(A) - \tilde{D}) S = S^{-1} \tilde{D} ( J(A) - \tilde{D}) S = S^{-1} ( J(A) - \tilde{D}) \tilde{D} S \\ & = S^{-1} ( J(A) - \tilde{D}) S S^{-1} \tilde{D} S = ND \end{align*}

Answer 2

A good thing to remember is that complex numbers form an algebraically closed field. So A is similar to a triangular matrix. So I think you can narrow your proof by only proving the result for a triangular matrix and concluding by using similarity.