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Let $\mathfrak{g}$ be a Lie Algebra with basis $X,Y$ and a Lie bracket characterised by: $[X,Y]=Y$

I‘m supposed to show that for every $n$-dim representation, i.e. $\rho: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathbb{C}^n)$, $\rho([X,Y])=[\rho( X), \rho( Y)]$, it must hold that $\exists n\in \mathbb{N}: \rho(Y)^n=0$.

This is possible by considering the eigenvectors of $\rho(X)$. Because:

$$ \rho(X)v=\lambda v \wedge [\rho(X),\rho(Y)]v=\rho(Y)v \\ \Rightarrow \rho(X)\rho(Y)v=(\lambda +1)\rho(Y)v $$

Which means that $\rho(Y)v$ is also an eigenvector, but since eigenvectors with different eigenvalues are linearily independent this can’t go on forever and $\rho(Y)^nv=0$. The desired result would follow if the eigenvectors of $\rho(X)$ would span the space, but I dont see why that would hold. What am I missing?

Henry T.
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  • Or have I just forgotten a linear algebra fact, that just states exactly that? – Henry T. Aug 15 '22 at 19:37
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    Yes, let $A=\rho(Y)$ and $B=\rho(X)$ and use this linear algebra result from the duplicate. We have $AB-BA=[A,B]$ because $\rho$ is a representation. But note that we need characteristic zero. – Dietrich Burde Aug 16 '22 at 07:50

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