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This is an important and well-known lemma used in proving the Lie and Engel theorem. But the proof I've written is much shorter and simpler than the usual one on this result, which involves extending the shared eigenspace of h to its completion.

This makes me worried that I may have missed an important step here. Please critique me and point out any flaws you can see in this argument.

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Steven Xu
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$v^t (hx) v=v^t (xh)v$ is simply false statement. take $x=E_{1,2}$, $h=E_{3,1}$ $v=e_1+e_2+e_3$. \begin{split} v^T (hx) v &= v^T \delta_{1,1} E_{2,3} v\\ &= 1 \end{split} but \begin{split} v^T (hx) v &= v^T \delta_{2,3} E_{1,1} v\\ &= 0 \end{split} or, did I missed anything?

ankai liu
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  • Excellent answer, thank you very much! What I failed to acknowledge in my previous answer, was that the assumption that all elements of the representation are Hermitian, in which case the vt(hx)v=vt(xh)v identity holds (just take the hermitian adjoint of both sides).

    The proof presented is a minor simplification of the standard argument - but as you have astutely identified, it is applicable only in the Hermitian case.

     – Steven Xu Dec 19 '17 at 06:16
  • @StevenXu If this answer was useful for you – and it strongly appears as if – then you should consider to vote it up, shouldn't you? – Hanno Jan 02 '19 at 13:54
  • @Hanno Indeed I should, and I did indeed upvote at the time - but I don't have sufficient reputation for my "upvote" to be publically displayed. – Steven Xu Jan 04 '19 at 03:06
  • @StevenXu Ah, I see. And I see that I'm not so familiar with the rules in this realm $\stackrel{..}{-}:$ At least some push towards the rep frontier is possible $\ddot\smile$ – Hanno Jan 04 '19 at 07:04
  • In the formula line after "but": $h$ and $x$ should be exchanged, shouldn't they? – Hanno Jan 05 '19 at 18:52