 Solution

Put $e=E_{+}$ and $f=E_{-}$. The supplied identities giveIf no $f^{k}v$ vanishes, this is the first alternative. Otherwise choose the least $N\ge 1$ with $f^{N}v=0$. Applying $e$ gives $N(\chi -N+1)f^{N-1}v=0$. Since $f^{N-1}v\ne 0$ and the field has characteristic zero, $\chi =N-1$. ThusThe endpoint must include zero: the trivial one-dimensional module has $\chi =0$ and $fv=0$, so the printed claim of strict positivity has that counterexample. All nonzero vectors in the displayed chain have distinct $H$-weights, and hence are linearly independent.