Originally Posted by
Deveno sure, just add the 0-vector [0,0,0].
the point being, if we have an orthogonal set in R^3 that is all non-zero vectors, say v1,v2,v3. then consider any linear combination:
u = av1 + bv2 + cv3. if this is 0, then <v1,u> = 0, since <v1,0> = 0 (here <x,y> is just the dot product in R^3).
but <v1,u> = <v1,av1> + <v1,bv2> + <v1,cv3> = a<v1,v1> + b<v1,v2> + c<v1,v3> = a<v1,v1> (by orthogonality).
but since <v1,v1> is not 0 (because v1 is not 0), we must have a = 0.
now show that <v2,u> = 0 implies b = 0, and that <v3,u> = 0 implies c = 0, showing {v1,v2,v3} is linearly independent.
the only assumption we make is that none of the v's are 0-vectors, so any orthogonal set of non-zero vectors in R^3 is linearly independent.