Explanation of a theorem 
Theorem. Let $V$ be a linear representation of $G$, with character $\phi$ and suppose $V$ decomposes into a direct sum of irreducible representations:
  $$
V= W_1 \oplus \cdots \oplus W_k
$$
  Then if $W$ is an irreducible representation with character $\chi$, the number of $W_i$ isomorphic to $W$ is equals $(\phi|\chi)$.

Can someone give me a concrete but simple example of this theorem?
 A: Consider the cyclic group $G = \langle s \mid s^{2} \rangle$, and consider the three representations 
$$
\rho_1 : G \to \operatorname{GL}(\mathbb{C}) = \mathbb{C}^{\times} \ : \ s \mapsto 1,
$$
$$
\rho_2 : G \to \operatorname{GL}(\mathbb{C}) = \mathbb{C}^{\times} \ : \ s \mapsto -1,
$$
$$
\rho_3 : G \to \operatorname{GL}(\mathbb{C}^{3}) \ : \ s \mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}
$$
Then the characters $\chi_i$ of $\rho_i$ are given by
$$
\begin{array}{c |  c c c}
G & 1 & s \\
\hline
\chi_1 & 1 & 1 \\
\chi_2 & 1 & -1 \\
\chi_3 & 3 & -1 \\
\end{array}
$$
We calculate that 
\begin{align*}
& \langle \chi_1, \chi_1 \rangle = \frac{1 + 1}{2} = 1, & \langle \chi_2, \chi_2 \rangle = \frac{1 + (-1)^{2}}{2} = 1, & & \langle \chi_3, \chi_3 \rangle = \frac{3^{2} + (-1)^{2}}{2} = 5, \\
& \langle \chi_1, \chi_2 \rangle = \frac{1 - 1}{2} = 0, & \langle \chi_1, \chi_3 \rangle = \frac{3 - 1}{2} = 1, & & \langle \chi_2, \chi_3 \rangle = \frac{3 + 1}{2} = 2. \\
\end{align*}
Now notice that $\rho_1, \rho_2$ are irreducible (dimension one), and $\rho_3$ is reducible since $\langle \chi_3, \chi_3 \rangle = 5$. We have two possibilities for $\rho_3$, either $\rho_3$ is the direct sum of a single two dimensional representation and a single one dimensional, or is the direct sum of one one-dimensional representation, and two copies of the a different one dimensional representation. Moreover, since $\langle \chi_1, \chi_3 \rangle, \langle \chi_2, \chi_3 \rangle \neq 0$, with $\rho_1, \rho_2$ irreducible, we must have that $\rho_1, \rho_2$ are subrepresentations of $\rho_3$. It follows that $\rho_3$ is equivalent to either $\rho_1 \oplus \rho_1 \oplus \rho_2$, or $\rho_1 \oplus \rho_2 \oplus \rho_2$. Since $\langle \chi_2, \chi_3 \rangle = 2$, it follows that $\rho_3$ is equivalent $\rho_1 \oplus \rho_2 \oplus \rho_2$. Now in this contrived example it is easy to see this directly. But the key take home is that if $\rho$ is a representation of a finite group $G$ over an algebraically closed field $\mathsf{k}$ satisfying $\rho = \bigoplus_i \rho_i^{\oplus n_i}$ for $\rho_i, \rho_j$ pairwise non-isomorphic irreducible representations, then 
\begin{align*}
\langle \chi, \chi_k \rangle = \operatorname{dim}_{\mathsf{k}}\operatorname{Hom}_{\mathsf{k}\left[G\right]}(\rho, \rho_k) & = \operatorname{dim}_{\mathsf{k}}\left( \bigoplus_{i} n_i \operatorname{Hom}_{\mathsf{k}\left[G\right]}(\rho_i, \rho_k) \right) \\
& = \sum_i n_i \operatorname{dim}_{\mathsf{k}}\left(\operatorname{Hom}_{\mathsf{k}\left[G\right]}(\rho_i, \rho_k) \right) \\
& = \sum_i n_i \langle \chi_i, \chi_k \rangle \\
& = \sum_i n_i \delta_{i,k} = n_k,
\end{align*}
where the penultimate line is due to Schur's lemma. 
