# Inner product in Maschke's Theorem

I am working through Maschke's Theorem on page 16 in Bruce Sagan's The Symmetric Group:

In order to prove the theorem the author constructs an inner product $\langle v, w \rangle' = \sum_{g \in G} \langle gv , gw \rangle$.

However, I fail to see why $\langle v, w \rangle'$ satisfies the definiteness property of the inner product. In other words, I do not know how to prove $\langle v,v \rangle = 0$ if and only if $v = 0$.

The "if"-implication is obvious, but I cannot see why $v = 0$ if $\langle v, w \rangle' = 0$.

Thank you very much for your help!

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You have $\langle gv,gv\rangle \geqslant 0$ for all $g\in G$ and $v\in V$, with equality only for $v = 0$, since $G$ acts as a group of automorphisms of $V$. – Daniel Fischer Jul 5 '14 at 15:00

If $\langle v, v \rangle' = \sum_{g \in G} \langle gv , gv \rangle = 0$ but each of the terms are non-negative so they must all be zero... $\langle gv , gv \rangle = 0$ implies that $v = 0$.
As $g.V=V$ for all $g \in G$ (because this is an action of $G$ on $V$), and each of these is a linear map, we have that $g:V \rightarrow V$ is a linear isomorphism. So for all $g \in G$, $g.v=0$ if and only if $v=0$. So in order to show $\langle v,v \rangle'=0 \Rightarrow v=0$, observe that $\langle v, v \rangle' = \sum_{g \in G} \langle gv , gv \rangle = 0$ if and only if $v =0$ by the remark above and by positive-definiteness of $\langle .,.\rangle$.