Question

Suppose h#; #i1 and h#; #i2 are inner products on v such that hv;wi1 d 0 if and only if hv;wi2 d 0. prove that there is a positive number c such that hv;wi1 d chv;wi2 for every v;w 2 v.

Answer 1

Answer:

We take v in the vector space V different from 0. Since v is not 0, then

$< v,v>_1 \, >0 \, ;, <v,v>_2 \, >0$

Lets take k>0 such that

$<v,v>_2 \, = k*\, <v,v>_1$

Now, we take any vector w. We want to show that $<v,w>_2 = k*<v,w>_1$ .

Since v is any non-zero vector, then this will prove that $<a,b>_2 = k*<a,b>_1$ for any vectors a,b. The reason is that for any vector different from 0, lets name it x, there will exist a constant  $k_x$ such that $<x,y>_2 = k_x *<x,y>_1$ for any y (this is for the same reason a constant exists for v). Since y can be anything, then it can be v. But that means that $k_x = k$, because v also has its constant k.

Now, lets show that $<v,w>_2 = k*<v,w>_1$ .  Lets take a constant c such that $c*<v,v>_2 = <v,w>_2$ . We have that

$0 = <v,w>_2 - c* <v,v>_2 = <v, w-cv>_2$

Thus

$<v,w-cv>_1 = 0$

Which means that

$0 = <v,w-cv>_1 = <v,w>_1 - c<v,v>_1$

Which means that $<v,w>_1 = c<v,v>_1$ . As a consequence

$<v,w>_2 = c <v,v>_2 = ck*<v,v>_1 = k * <v,w>_1$

Which proves what we were looking for.