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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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    Would [math.se] be a better home for this question? – 2017-02-03
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    It would also be useful to use [MathJax for math notation](http://physics.stackexchange.com/help/notation), rather than relying on images – 2017-02-03
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    True, I have uploaded it to Math Stack Exchange as well. But I know that physicists also deal with lie groups and lie algebras a lot. – 2017-02-05
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    I encourage you to rewrite your question using MathJax as we can better interact (and with details :). – 2018-02-25

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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?

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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. – 2017-12-19