# 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 $BC =CB$?

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Note that $\left( \begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array} \right)$ is not in Jordan form... –  Johannes Kloos May 15 '12 at 16:10
@JohannesKloos: You're right. Ok. But I don't know how to show this thent. I tried to do it by calculating the sum for each entry of the product (row times column) but it doesn't really work.. –  steltjen May 15 '12 at 16:17
I'm assuming that you want $V$ to be a vector space? And $\phi$ is linear? –  Martin Argerami May 15 '12 at 17:02
Your mistake is that your original matrix should be $\psi$ and $\sigma$ should be the zero matrix (and those commute). The point is that your original matrix is already diagonalizable. The Jordan decomposition is not related to how a matrix happens to look in standard coordinates, as you found out when your guess didn't commute. –  KCd May 16 '12 at 0:52