Common kernel of irreducible representations of finite groups Let $G$ be a finite group and $\rho_1,\rho_2,\cdots,\rho_m$ be pairwisely non-isomorphic irreducible representations of $G$. Then is it right that $\bigcap\limits_{i=1}^m\ker\rho_i=1$?
 A: This is true if you mean that the list above exhausts all irreducible representations. 
To see this, note that by extending a representation $\rho:G\to GL_n(\mathbb{C})$ linearly, we get a representation $\tilde{\rho}: \mathbb{C}G\to M_n(\mathbb{C})$. Since elements of $G$ form a basis for $\mathbb{C}G$, we see that $\ker(\rho)=\{1\}$ if, and only if, $\ker(\tilde{\rho})=\{0\}$. Therefore, we are left to check that $\bigcap_i \tilde{\rho_i}=\{0\}$. 
Well, there is an isomorphism
$$\mathbb{C}G\cong M_{n_1}(\mathbb{C})\oplus M_{n_2}(\mathbb{C})\oplus\cdots\oplus M_{n_m}(\mathbb{C})$$
and $\rho_i$ is the projection onto the $i$th factor under this isomorphism. Thus, $x\in\bigcap_i\ker(\tilde{\rho_i})$ if and only if $x\mapsto 0$ under the isomorphism above. That is, $x=0$.
A: Suppose $\chi$ is a character of $G$: $\chi=\sum_{i=1}^k n_i\chi_i$, where $\chi_1,...,\chi_k$ are all the irreducible characters of $G$.
We claim that $\ker\chi=\bigcap_{i=1}^k\{\ker\chi_i: n_i>0\}$.

Proof: $g\in\ker\chi\iff \sum_{i=1}^kn_i\chi_i(g)=\chi(g)=\chi(1)=\sum_{i=1}^kn_i\chi_i(1)$.
But $|\chi_i(g)|\le\chi_i(1)$ for all $i=1,...,k$ implies
$\chi_i(g)=\chi_i(1)\text{ when }n_i>0$ which is equivalent to $
> g\in\bigcap_{i=1}^k\{\ker\chi_i: n_i>0\}$. The reverse direction is
immediate.

Note that the regular character is $\rho=\sum_{i=1}^k\chi_i(1)\chi_i$.
Therefore $$\bigcap_{i=1}^k\ker\chi_i=\ker\rho=\{1\}.$$
