I got the answer:
According to the Cauchy Schwartz inequality:
|<Tv,v>| <= ||Tv||.||v||
T is unitary so:
||Tv||=||v||
So:
|<Tv,v>| <= ||Tv||.||v||=||v||.||v||=<v,v>
QED
Let V be an inner product space of finite dimension over C (complex numbers). Let T be a unitary linear operator from V to V. Prove that:
|<Tv,v>| is less than or equal to <v,v>
My working so far:
Since T is unitary <Tv,Tv>=<v,v>, but I'm not sure if this helps at all since I don't know how to show that |<Tv,v>| <= <Tv,Tv>.
Thanks