I am currently working on Jean-Pierre Serre's "Linear representations" german translation in chapter 6 and I do not understand the last part of the proof of the following theorem.
a) Let $\rho_i : G \longrightarrow GL(V_i)$ for $i = 1,...,n$ be distinct irreducible representations of the group G. The map $\rho: \mathbb{C}[G] \longrightarrow \prod_{i}^{n} End(V_i)$, defined by the $\rho_i$, is surjective.
Proof:
Suppose the Image of $\rho$ is a proper subspace of $\prod_{i}^{n} End(V_i)$. Then there exists a non zero linear form $\phi : \prod_{i}^{n} End(V_i) \longrightarrow \mathbb{C}$ that vanishes on the Image of $\rho$, i.e. $\phi (\sum_{i}^{n} \rho_i(s)) = 0 $ for all $s \in G$.
This will give us a relation of the form $\sum_{k} \lambda_k m_k(s) = 0$ for all $s \in G$. Here, the $m_k(s)$ are the coefficients in the Matrix of the representation $\rho_i(s)$.
So far the proof makes sense to me, but I do not understand the following part at all. We want to show that all $\lambda_k$ are zero, because then we have found our contradiction to the non zeroness of $\phi$. Serre only writes:
Due to the orthogonality relations of the coefficients we can conclude that $\lambda_k = 0$ for all $k$.
The orthogonality relations from chapter 2.2 in the english version are meant here, but I just cannot see how I have to apply them to get that all $\lambda_k = 0$.
I have found another thread to the same theorem, but the proof is different at the end and I hope someone can help me with my problem.
Help would be highly appreciated.
Pick an $i \in \{1, \dots, n\}$. We multiply both sides of the equation $\sum_{k} \lambda_k m_k(s)= 0$ by $\bar m_i(s)$, and we take a sum over all $s \in G$, giving $$ \sum_k \left( \lambda_k \sum_{s \in G} m_k(s) \bar m_i(s) \right)= 0$$ But by orthogonality, $$\sum_{s \in G} m_k(s) \bar m_i(s) = \begin{cases} |G| & {\rm if \ } k = i \\ 0 &{\rm if \ } k \neq i \end{cases}$$ So $$ \sum_k \left(\lambda_k \sum_{s \in G} m_k(s) \bar m_i(s) \right) = \lambda_i |G|.$$ Thus we have shown that $\lambda_i|G| = 0$, i.e. $\lambda_i = 0$. Since the choice of $i$ was arbitrary, this completes the proof.