I'm working through these notes on representation theory: http://math.mit.edu/~etingof/replect.pdf.
Currently I'm looking at exercise 1.55 part j. I've done all of the previous parts and most of the rest of the question but I really can't work out how to do the first part of j:
'Deduce from (i) that the eigenspace $V(\lambda)$ of $H$ is $n + 1$-dimensional'.
In part (i) we've shown that $V$ has a subrepresentation $W = V_{\lambda}$ such that $V/W = nV_{\lambda}$ for some $n$. Here $V$ is a representation of $\mathfrak{sl}_2 \mathbb{C}$ and $\lambda$ is the maximal eigenvalue for the operator representing of the element $$ h = \begin{pmatrix} 1 && 0 \\ 0 && -1 \end{pmatrix} $$ of $\mathfrak{sl}_2 \mathbb{C}$.
For this part of the problem we are assuming (towards a contradiction) that $V$ is the smallest (minimal dimension) representation that cannot be decomposed into a direct sum of irreducible subrepresentations.
I think I might be missing something obvious. Does anyone have any suggestions for how to approach this part? Thanks for any help.
Presumably you understand from the rest of the question that the $H$-eigenspace decomposition of $W=V_{\lambda}$ is $W=\bigoplus_{n=0}^\lambda W(\lambda-2n)$ with each eigenspace, $W(\lambda-2n)$, being 1-dimensional.
By part (i) of the question together with the decomposition above, you know that the $\lambda$-eigenspace of $V/W\cong nV_\lambda$ is $n$-dimensional. Since the $\lambda$-eigenspace of $W$ is $1$-dimensional, it follows that the $\lambda$-eigenspace of $V$ is $(n+1)$-dimensional.
Edit: We explain how to get a basis of $V(\lambda)$. We first note that it is clear that $\dim V(\lambda)\leq n+1$. We now prove equality below.
Note that if $v+W\in V/W$ is a $\lambda$-eigenvector, then $Hv=\lambda v+w$ for some $w\in W$. Write $w=\sum_{n=0}^{\lambda}w_{\lambda-2n}$, where $w_{\lambda-2n}\in W(\lambda-2n)$, and set $$v'=v-\sum_{n=1}^\lambda\frac{1}{2n}w_{\lambda-2n}.$$ Then, $Hv'=\lambda v'+w_\lambda$ and $v+W=v'+W$. Therefore, we may assume $Hv=\lambda v+w$ with $w\in W(\lambda)$. In particular, $v\in \overline{V}(\lambda)$ is a generalized $\lambda$-eigenvector. But, $H$ acts diagonally on $V(\lambda)$, so $w=0$.
It now follows that we may choose a basis $\{v_1+W,\ldots,v_n+W\}$ for the $\lambda$-eigenspace of $V/W$ so that $v_1,\ldots,v_n\in V(\lambda)$. Now, choosing any nonzero $w\in W(\lambda)$ yields a linearly independent set $\{v_1,\ldots,v_n,w\}\subset V(\lambda)$, which must be a basis.