Etingof problem 2.15.1 Representations of sl(2) I'm studying from Etingof's Introduction to Representation Theory. This is problem 2.15.1, part a. I feel I'm close to the solution. Here's what I have.
Problem: A representation of sl(2) is a vector space V with a triple of operators E, F, H such that (I) $HE–EH = 2E$, (II) $HF–FH = -2F$, and (III) $EF-FE = H$. Let $V$ be a finite dimensional representation of sl$(2)$ (the ground field is $\mathbb C$).
Part (a) Take eigenvalues of H and pick one with the largest real part. Call it $\lambda$. Let $\overline V(\lambda)$ be the $\textit{generalized}$ eigenspace corresponding to $\lambda$. Show that E, restricted to $\overline V(\lambda)$, is $0$.
Partial Solution: Let $v$ be a generalized eigenvector of $H$, i.e., there is a nonnegative integer $k$ such that $(H - \lambda I)^kv = 0$ (let $k$ be minimal in this regard). Expanding $(H - \lambda I)^k$ we get
\begin{align*}
(H - \lambda I)^kv &= 0\\
\sum_l^k \binom{k}{l}H^{k-l}(-\lambda)^lv & = 0\\
\sum_l^{k-1}\binom{k}{l} H^{k-l}(-\lambda)^lv & = -(-\lambda)^kv\quad\text{factor one }H\\
H\left(\sum_l^{k-1} \binom{k}{l}H^{k-1-l}(-\lambda)^lv\right) & = (-1)^{k+1}\lambda^kv
\end{align*}
Let $v^\prime = \sum_l^{k-1} \binom{k}{l}H^{k-1-l}(-\lambda)^lv$. We see that (this is an error) $v^\prime$ is a (proper) eigenvector of $H$ with eigenvalue $(-1)^{k+1}\lambda^k$. Recall that by assumption, the real part of $(-1)^{k+1}\lambda^k$ is at most Re($\lambda$).
Now apply relation (I). We compute
\begin{align*}
2Ev^\prime &= HEv^\prime - EHv^\prime\\
HEv^\prime &=  2Ev^\prime + EHv^\prime\\
H(Ev^\prime) &= \big(2 + (-1)^{k+1}\lambda^k  \big)(Ev^\prime)
\end{align*}
Observe $Ev^\prime$ is an eigenvector of $H$ with eigenvalue $2 + (-1)^{k+1}\lambda^k$, or else $Ev^\prime = 0$.
Ok. If it is somehow true that Re$(2 + (-1)^{k+1}\lambda^k)$ is greater than Re$(\lambda)$, the solution is complete...
 A: Ok, I figured it out.
First consider $v$ in $V(\lambda)$, that is a (proper) eigenvector of $H$. Compute $HEv = 2Ev + EHv =  (2+\lambda)Ev$, which is a contradiction with the assumption on $\lambda$, unless $Ev = 0$. Thus $E$ is $0$ when restricted to the eigenspace of $\lambda$.
Here I would like to note that ``generalized eigenvalues" and eigenvalues are the same thing. Suppose $(H-\gamma I)^kw = 0$ and $k>1$ is minimal. Then $(H-\gamma I)\left[ (H-\gamma I)^{k-1} w\right] = 0$ and $ (H-\gamma I)^{k-1} w$ is nonzero. This shows $w^\prime =  (H-\gamma I)^{k-1} w$ is an eigenvector of $H$ with eigenvalue $\gamma$.
Now let $w$ be in $\overline V(\lambda)\setminus V(\lambda)$ and let $k$ be the minimal nonnegative integer such that $(H - \lambda I)^kw = 0$, so that $(H - \lambda I)^{k-1}w$ is a nonzero $\lambda$-eigenvector of $H$. So we know $0 = E(H-\lambda I)^{k-1}w$. We compute the permutation
\begin{align*}
E(H-\lambda I) &= EH - \lambda E = (HE - 2E) - \lambda E = [H-(\lambda + 2)I]E,
\end{align*}
So we have $[H - (\lambda + 2)]^{k-1}Ew = 0$, i.e., $Ew$ is a generalized eigenvector of $H$ with eigenvalue $\lambda + 2$ (a contradiction) or else $Ew = 0$, as desired.
