In Lie algebras we study $\mathfrak{sl}(2)$ (the complex span of the usual matrices $X,Y,H$ where $X$ and $Y$ are the raising and lowering operators respectively). The defining commutator relations are $[X,Y]=H,[H,X]=2X,[H,Y]=-2Y$.
Let $V$ be a finite dimensional representation of $\mathfrak{sl}(2)$ and suppose $0\neq v\in V$ has positive $H$ eigenvalue $\lambda$. I want to show that lowering it by $Y$ also gives me something non-zero, i.e. $Yv\neq 0 $.
Thoughts: Assume that $Yv=0$ then $0=XYv=[X,Y]v+YXv=Hv+YXv=\lambda v+YXv$. So done if we had assumed $Xv=0$. But did not make significant progress beyond this point.
Background: This is a step used in a proof that all root spaces of a complex finite dimensional semisimple Lie algebra are 1-D.
We have a minimal $n\geq 1 $ such that $X^{n-1}v\neq0,X^{n}v=0$ as $V$ finite dimensional.
Now $Yv = 0$ implies $0=X^iYv=(\lambda +(\lambda+2)+...(\lambda+2(i-1)))X^{i-1}v+YX^iv=0$. Putting $i=n$ gives a contradiction.