# Proof of Jordan-Chevalley decomposition

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.

## 2 Answers

"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*}

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.