[Math] Irreducibility of a character if and only if inner product equals 1

charactersrepresentation-theory

I am working through the following representation theory notes:
Notes

I am having some trouble understanding why the Corollary to Theorem 15 is true.

Theorem 15
Suppose $G$ is a finite group with irreducible characters $\chi_{1},\dots \chi_{k}$ over $\mathbb{C}$ If $\chi$ is any character, expressible as a sum of irreducible characters by $\chi = \Sigma m_{i} \chi_{i} $ then:

  1. For each i , $ m_{i}=\langle \chi_{i} | \chi_{i} \rangle $
  2. $\langle \chi | \chi \rangle = \Sigma m_{i}^{2} $

Corollary A character $\chi$ is irreducible if and only if $\langle \chi | \chi \rangle=1$

I understand that if $\chi$ is irreducible then it must be one of $\chi_{1},\dots \chi_{k}$ and then $\langle \chi | \chi \rangle=1$

I am having trouble understanding why the converse must be true.
For example if I had $\chi =\frac{3}{5}\chi_{1}+\frac{4}{5}\chi_{2} $ then $\langle \chi | \chi \rangle = \Sigma m_{i}^{2}=1 $ but this is reducible.

I would appreciate some clarification on what is going on and a 'proof' of the corollary. Thank you.

Best Answer

You are not allowed to have rational numbers in the linear combination of the characters. The $m_i$'s are natural numbers. Theorem 10 in your notes says that 

Every character is a sum of irreducible characters.

So you can't have half of a character.

So then, if $\chi = m_1\chi_1 + \dots + m_k\chi_k$, you see that $\chi$ is irreducible if and only if exactly one $m_i =1$ and the rest zero.

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