# Inner products

• Oct 4th 2007, 03:16 AM
Jacko
Inner products
We define: <u,v>=u2v2+u3v3
For vectors u=(u1,u2,u3) and v=(v1,v2,v3) in R3.
Explain the reasons why this is not an inner product on R3.

I have completed the 4 axioms as below:
1. <u,v>= u2v3 + u3v3
=v2u2 + v3u3
=<v,u>

2.<cu,v> = cu1v2+cu2v2
= c(u2v2+u3v3)
= c<u,v>
3.<u,v+w>=u2(v2+w2)+u3(v3,w3)
= u2v2+u2w2+u3v3+u3w3
= <u,v>+<u,w>
4.a) <u,u>=u22 + u32
greater than or equal to zero as u2^2 greater than or equal to zero and u3^2 is greater than or equal to zero
b) <u,u>=0 then u2=0 and u3=0.

Somehow ive wrongly proved all the axioms :S im assuming this does not define an inner product as it does not include u1 and v1 in the inner product, therefore cannot be an inner product in R^3. I would greatly appreciate anyone looking over my work to help me! Thanks
• Oct 4th 2007, 04:59 AM
Plato
If u=(1,0,0) then what is <u,u>?
• Oct 4th 2007, 08:36 PM
Jacko
Even if u=(1,0,0) <u,u> does not depend on u1 as stated in the inner product equation. <u,u>=u2^2 + u3^3 which is equal to zero in this case. I understand now that 4b) is not satisfied as u does not equal the zero vector, but doesnt axiom 4a still hold?