Why Does (u,v) = -u2v2 Not Qualify as an Inner Product in R2?

[ephemeral1]

Homework Statement

State why (u,v) is not an inner product for u=(u1,u2) and v=(v1,v2) in R2
(u,v)=-u2v2

Homework Equations

(u,v)=(v,u)
c(u,v)=(cu,v)
(v,v)=>0 and (v,v)=0 if only if v=0

The Attempt at a Solution

I am having trouble understanding this problem and how to start it. Please help. Thank you

 

[ephemeral1]

I would think that axiom 4 is not satisfied.
if u=(1,2) and v=(2,2), then -u2v2=-4, which is less than 0, which violates axiom 4 that states (v,v) greater than or equal to 0. Is this right?

 

[lanedance]

I think the axiom you refer to only guarantees positive definiteness for the inner product of a vector with itself (v,v), though it shouldn't be hard to alter you argument to work in that case

note the inner product of 2 arbitrary vectors can be negative