0
$\begingroup$

Question

Let $V$ be a vector space over the complex field $C$ , and let $T$ be an linear map $V\rightarrow V$. Show that exists a pair of linear maps $D$ and $N$ such that $D+N=A$, $D$ is diagonalizable, $N$ is nilpotent, and DN=ND.

I found this endomorphism as sum of two endomorphisms (nilpotent and diagonalizable) where he said "you can transform so that ϕ is on jordan form, and then split this matrix in a diagonal and a nilpotent"

But I didn't get his explanation. How can I decompose the jordan marix of T? any why does it proves the commute and sum of the linear maps?

Any help would be appreciated, Thanks!

$\endgroup$
  • $\begingroup$ Have you ever heard of Dunford decomposition? $\endgroup$ – Gabriel Romon Apr 28 '14 at 16:55
  • $\begingroup$ No, not at all. Not this and not semi-simple transformations. I guess I suppose to solve it using Jordan Form because that's what we focus on right now.. $\endgroup$ – Splash Apr 28 '14 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.