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

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SUMMARY

The expression (u,v) = -u2v2 does not qualify as an inner product in R2 due to its violation of the positive definiteness axiom. Specifically, for vectors u = (1,2) and v = (2,2), the calculation yields (u,v) = -4, which is less than zero, contradicting the requirement that (v,v) must be greater than or equal to zero. This discussion highlights the importance of adhering to the axioms of inner products, particularly axiom 4, which mandates that the inner product of a vector with itself must be non-negative.

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  • Familiarity with vector notation in R2
  • Knowledge of mathematical axioms related to inner products
  • Basic algebraic manipulation skills
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  • Explore examples of valid inner products in R2
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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
 
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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?
 
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
 

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