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I have seen the proof of this in GTM9. But I have a problem with the statement "Since ${\rm ad}(x)(L)\subseteq L$, it follows from proposition 4.2(c) that ${\rm ad}(s)(L)\subseteq L$ and ${\rm ad}(n)(L)\subseteq L$." I understand that there is a polynomial $p(T)$ such that $s=p(x)$, but ${\rm ad}(x)$ is not $p({\rm ad}(s))$, right? The ${\rm ad}$ operation can preserve the usual operation $*$ in ${\rm End}(V)$?

Glare
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Tate Yang
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  • I can't really parse the question as you've written it unfortunately, but the statement you quoted is a result of the Lemma A immediately after Proposition 4.2(c) (page 18 in the 1972 edition): "if $x\in{\rm End}(V)$ has Jordan decomposition $x=s+n$, then ${\rm ad}(x) = {\rm ad}(s)+{\rm ad}(n)$ is the Jordan decomposition of ${\rm ad}(x)$." Proposition 4.2(c) now implies the subspace inclusions in your quote. – Glare Apr 10 '22 at 05:09
  • Related: https://math.stackexchange.com/q/487384/96384, https://math.stackexchange.com/q/2308807/96384. – Torsten Schoeneberg Apr 10 '22 at 06:44
  • I got it. Thanks. – Tate Yang Apr 10 '22 at 07:46

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