This question is essentially exercise 8.4 from the book "Introduction to Lie Algebras" by Erdmann and Wildon:
"Suppose that $V$ is a finite-dimensional $\mathfrak{sl}(2,\mathbb{C})$-module. Show that $V$ is determined, up to isomorphism, by the eigenvalues of the action $h$ (where $ h = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$)."
I know about Weyl's theorem and the classification of $\mathfrak{sl}(2,\mathbb{C})$-modules, and I've done a few low-dimensional examples to convince myself that the above statement is true, but I'm having difficulty articulating a proof of this.
My attempts at a proof have involved decomposing $V$ into the direct sum of irreducible submodules, looking at the largest eigenvalue $\lambda$ of $h$ on $V$, and then noting that by the classification results, we must also have eigenvalues $\lambda - 2, \lambda - 4, \dots 2-\lambda, - \lambda$. Thus we have a submodule isomorphic to $V_\lambda$ (using the notation from the book). Then repeating this argument gives the result.
However, I'm not very satisfied by this "proof", and haven't been able to make it more precise. What's the best way to go about this?
Let $V$ be an $\mathfrak{sl}(2,\mathbb{C})$-module.
Following the notation of the book, write $W_r = \{w \in V: h \cdot w = rw\}$ and denote the $d+1$-dimensional irreducible $\mathfrak{sl}(2,\mathbb{C})$-module by $V_d$.
Consider some direct sum decomposition of $V$ into irreducible submodules. The summands are invariant under $h$, so an eigenspace $W_r$ is the direct sum of the eigenspaces of these summands. In any irreducible module $V_d$, each eigenspace is one-dimensional and we have eigenvectors of $h$ for eigenvalues $d, d-2, d-4, \ldots, -d+2, -d$.
Therefore for $d \in \mathbb{Z}_+$, the dimension $\dim W_d$ is exactly the number of summands isomorphic to one of the following: $V_d, V_{d+2}, V_{d+4}, \ldots$
Hence if $n_d$ denotes the number of summands that are isomorphic to $V_d$, then $n_d = \dim W_d - \dim W_{d+2}$. These dimensions are determined by the action of $h$.