endomorphism as sum of two endomorphisms (nilpotent and diagonalizable)

$V$ is a field over $\mathbb{C}$. Show that $\phi: V \to V$ can be written as $\phi = \psi + \sigma$ where $\psi$ is diagonalizable and $\sigma$ is nilpotent.

I managed to show this first part (you can transform so that $\phi$ is on jordan form, and then split this matrix in a diagonal and a nilpotent ...).

But the next part doesn't work for me:

Show then, that $\psi \circ \sigma = \sigma \circ \psi$.

I tried to do it with the matrices $B$ of $\psi$ and $C$ of $\sigma$, which I know can be transformed so that $B$ is diagonal and $C$ is nilpotent. But the multiplication of these isn't commutative e. g.:

$\begin{pmatrix} 1 & 0 \\ 0 & 2 \\ \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} \neq \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 2 \\ \end{pmatrix}$

Where my mistake?

Alright the example had a mistake since $\left(\begin{smallmatrix} 1 & 2 \\ 0 & 1 \end{smallmatrix}\right)$ isn't jordan form.

But how to proove this now for $B$ the diagonal part and $C$ the other entries of a matrix in jordan form. How to proove for these matrices that $BC =CB$?

Solution 1:

Hint: You can make use of the fact that a Jordan form matrix is a block matrix.

Just show the claim for a single Jordan block, and then argue via block matrices that it holds for the whole thing.

It may be helpful to note that the diagonal part of a Jordan block is a scalar multiple of the unit matrix.