Is this an inner product on R3?

[jacko_20]

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 I've wrongly proved all the axioms :S I am 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

 

[mjsd]

your last step is wrong

proving u2 = u3=0 leaves u1 arbitrary and so the vector u = (u1, 0, 0) which is non-zero violates axiom 4 b).

 

[jacko_20]

Thanks mjsd! So it is correct that none of the other axioms apart from 4.b) depend on u1 or v1?