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?
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:
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!