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Mark44
#5
Mar30-12, 02:30 PM
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Quote Quote by bugatti79 View Post
Oh I see,

Axiom 1 <x,x> >=0 since we have that x_n for n=1,2,3 are in R
You're waving your arms here. Expand <x, x> using the definition of this inner product and it should be obvious why <x, x> >= 0.
Quote Quote by bugatti79 View Post




Yes, the bar on top of them.
Anyway,

<y,x>=y1x1+y2x2+y3x3
=x1y1+x2y2+x3y3
=<x,y>

Is axiom 4 ok?



Thanks
Yes, #4 is fine.