0

Here is a problem about additive Jordan Decomposition, that is, V is a finite dimension complex vector space, for a linear map T∈End(V), we have T=D+N, and DN=ND, where D is diagonalizable and N is nilpotent. The decomposition is unique.

Suppose $V$ is a finite dimensional complex vector space and $u\in End(V)$. Let $ad(u):End(V)\rightarrow End(V)$ be the map defined by $$ad(u)(v)=u\circ v-v\circ u$$ for $v\in End (V)$. It's easy to check that $ad(u)$ is a linear map. Suppose $u=u_s+u_n$ is the decomposition, it's easily to check $ad(u)=ad(u_s)+ad(u_n)$, how could a show it is exactly the decomposition for $ad(u)$, i.e., how could I prove $ad(u_s)$ is still diagonalizable and $ad(u_n)$ is still nilpotent.

Here is the link $A$ is some fixed matrix. Let $U(B)=AB-BA$. If $A$ is diagonalizable then so is $U$? showing that $ad(u_s)$ is diagonalizable.

So my question is why $ad(u_n)(v)=u_n\circ v-v\circ u_n$ is nilpotent when $u_n$ is nilpotent. If we check

$$(u_n\circ v-v\circ u_n)^{2n}=u_n\circ v\circ u_n\circ v\circ\cdots\circ u_n\circ v+\Sigma,$$

I can't see that $(u_n\circ v)^{2n}=u_n\circ v\circ u_n\circ v\circ\cdots\circ u_n\circ v=0$, so how could we know that?

I notice there is a similar problem here, Jordan Decomposition, but in fact it did not show why ad preserve nilpotent property.

Zijie Tang
  • 265
  • 1
    You have mistaken $\left(\operatorname{ad}(u)(v)\right)^{2n}$ for $\left(\operatorname{ad}(u)\right)^{2n}(v)$. The word $u\circ v\circ u\circ v$ does not appear in the expansion of the latter. – user1551 Jan 10 '25 at 12:20

0 Answers0