$M$ and $S$ have the same characteristic polynomial?

33 Views Asked by At

I have read a linear algebra theorem called Jordan–Chevalley decomposition, which says that:

If $M\in\mathcal{M}_n(\mathbb{C})$ then there exists $S,N\in M\in\mathcal{M}_n(\mathbb{C})$ such that $M=S+N$, where $S$ is diagonalizable, $N$ is nilpotent, and $SN=NS$.

But I don't understand why $M$ and $S$ have the same characteristic polynomial?

an idea plaese?