Exercise on representations of $\mathfrak{sl}_2 \mathbb{C}$ (Etingof 1.55.j) 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.
 A: 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.
