Fully reducible rep-invariant subspace

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How could I show that a representation D is fully reducible if and only if for every invariant subspace V_{1} \in V then also V_{1}^{T} (meaning orthogonal to V1) is also invariant?
http://www.crystallography.fr/mathcryst/pdf/nancy2010/Souvignier_irrep_syllabus.pdf
(Lemma 1.6.4)
in fact I don't understand the inaffected scalar product it mentions...
 
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It's OK I think I came into the answer...
if we take v \in V_{1} and w \in V_{1}^{T}
then <v,w>=0
but v doesn't change subspace under the action of the repr
<D(g)v, w>=0= <v,D^{t}(g) w>=<v,D(g^{-1}) w>=<v,D(g) w>
However this needs the unitarity for the g^{-1} step
 

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