I have come across the following statement in various sources without any proof. Apparently they say the proof is trivial. However, I don't see the triviality in this case:
$\rho_1,\cdots , \rho_r$ are isomorphism classes of irreducible repreentations of finite group G and $\rho$ is any representation of G. We also have $\chi_i$ and $\chi$ as the characters of $\rho_i$ and $\rho$ and $n_1=<\chi, \chi_i>$
Theorem: Two representations $\rho$ and $\rho'$ of a finite group are equivalent if and only if their characters are equal.
I was able to prove (not rigorously) that the $\chi = n_1\chi_1+\cdots n_r\chi_r$ but I dont see how this helps in our proof. Maybe the forward direction is trivial. That is if two representations are equivalent then their matrix representation will be equivalent too (sa,e matricies), so they will have the same trace. I don't know how to prove the reverse direction.
Best Answer
It follows from the orthogonality relations for characters. Note that if $\chi = \sum n_i \chi_i,$ then $n_i = \langle \chi, \chi_i \rangle.$ So, if two reps $\rho, \sigma$ have the same characters, they have the same set of $n_i.$ If that is true, you are saying that both $\rho$ and $\sigma$ are equivalent to the same representation. Thus they are eqiuivalent.