Let $G$ be a finite group, $\chi: G \longrightarrow \mathbb{C}$ an irreducible character. Then the following two maps are also characters: $$\chi_S : g \longrightarrow \frac{1}{2}(\chi(g)^2 + \chi(g^2));$$ $$\chi_A : g \longrightarrow \frac{1}{2}(\chi(g)^2 - \chi(g^2)).$$ This is very useful when calculating character tables, and I realized that in almost all examples I treated, it turns out that either $\chi_A$ is irreducuble, or $\chi$ has multiplicity 1 in $\chi_A$, and the same goes for $\chi_S$. I wondered wether this is just luck, or if there is some reason why this is true, maybe some sufficient condition for this to be true...
I know it might not be a very clear question, and I apologize for that.
If you tried a linear character $\lambda \colon G \to \mathbb C$, then you must have $\lambda_A = 0$. Hence it is neither irreducible nor $\langle \lambda_A, \lambda \rangle = 1$ in general.