When is the orbit of a vector spanning the whole space? Let $\rho:G\to\mathrm{GL}(V)$ be a linear representation of a finite group $G$ over a complex vector space $V$. One may ask whether there is a point $v\in V$ so that the orbit $\rho(G)v$ spans all of $V$. Apprently there is an answer in terms of characters:

If $\chi$ is the character of $\rho$ and $\psi_1,...,\psi_m$ is an enumeration of the irreducible characters of $G$, then such a point exists if and only if $\langle \chi,\psi_i\rangle \le \psi_i(1)$ for all $i\in\{1,...,m\}$.

I thought about this and found an involved topological argument. But this seems like a basic and well-known fact. So I wonder, what is the easiest proof that you know of? And do we need to assume some machinery or does it quickly follow from first principles?
 A: It is helpful to work here with the group algebra $\mathbb{C}G$, and consider the representation as a linear map $\rho \colon \mathbb{C}G \to \operatorname{End}_\mathbb{C}(V)$ of algebras. For any vector $v \in V$, let $\varphi_v \colon \mathbb{C}G \to V$ be the linear map $x \mapsto \rho(x)v$, then the linear span of the orbit $\{\rho(g)v \mid g \in G\}$ is precisely the image of $\varphi_v$. So we can restate the question: when does $V$ admit a vector $v$ such that $\operatorname{im} \varphi_v = V$? There are some necessary conditions coming from dimensions: for example we have $\dim \operatorname{im} \varphi_v \leq \dim \mathbb{C}G$.
Let $\psi_1, \ldots, \psi_m$ be the irreducible characters of $G$. For any representation $V$, denote by $V[i] \subseteq V$ the $\psi_i$-isotypic component of $V$: this is the sum of all subrepresentations with character $\psi_i$. The map $\varphi_v$ is a homomorphism of representations and hence it breaks up along isotypic components, so it is a direct sum of the maps $\varphi_v[i] \colon \mathbb{C}G[i] \to V[i]$. Applying the same dimension observation as above to each component gives that $\dim V[i] \leq \dim \mathbb{C}G[i]$, which is the inequality $\langle \chi, \psi_i \rangle \psi_i(1) \leq \psi_i(1)^2$ where $\chi$ is the character of $\operatorname{im} \varphi_v$, so we find that the condition $\langle \chi_V, \psi_i \rangle \leq \psi_i(1)$ is necessary. In more straightforward terms, this is saying that the irreducible $W_i$ can only appear in a homomorphic image of $\mathbb{C}G$ as many times as it appears in $\mathbb{C}G$ itself.
To show that this dimension condition is sufficient, we appeal to the Jacobson density theorem:

(Density theorem): Let $\rho_W \colon \mathbb{C}G \to \operatorname{End}_\mathbb{C}(W)$ be an irreducible $\mathbb{C}G$-module, then the image of $\rho_W$ is the whole of $\operatorname{End}_\mathbb{C}(W)$.

Now, given an isotypic representation $W^{\oplus n}$, provided that $n \leq \dim W$ we can pick $n$ linearly independent vectors $w_1, \ldots, w_n$, then I claim that $w := (w_1, \ldots, w_n) \in W^{\oplus n}$ is a vector whose $\mathbb{C}G$-orbit is the whole of $W^{\oplus n}$. For any other point $(x_1, \ldots, x_n) \in W^{\oplus n}$, by independence of the $w_i$ there exists some linear endomorphism $A: W \to W$ with $Aw_i = x_i$, and by the density theorem this operator $A$ is in the image of $\mathbb{C}G$, so there is some $a = \sum_{g \in G} a_g g \in \mathbb{C}G$ such that $a \cdot (w_1, \ldots, w_n) = (x_1, \ldots, x_n)$.
Let me know if this is clear!
