Determine whether it's an inner product on R^3

[sam0617]

Homework Statement

Let u = (u₁, u₂, u₃)
and v = (v₁, v₂, v₃)
Determine if it's an inner product on R³.
If it's not, list the axiom that do not hold.

Homework Equations

the 4 axioms to determine if it's an inner product are
(all letters representing vectors)

1. <u,v> = <v,u>
2. <u+v, w> = <u, w> + <v,w>
3. <ku, v> = k<u,v>
4. <v,v> ≥ 0 and < v,v> = 0 if and only if v = 0

The Attempt at a Solution

So <u, v> is defined as
u₁v₁ + u₃v₃

I'll skip the ones that did work and show axiom 4 which did not hold but I'm confused as to why this doesn't hold. I have a guess but have to make sure that I'm thinking correctly.

Axiom 4 does not hold:
<v, v > = v₁v₁ + v₃v₃
= v₁² + v₃² ≥ 0
and
<0, 0> = (0)(0) + (0)(0) = 0

now to check the other way:
if <v, v > = 0
implies that since
v₁² = 0 => v₁ = 0
v₃² = 0 => v₃ = 0

then it goes to say it's not an inner product on R³. Am I correct to say it's not an inner product on R³ because there are only 2 components for axiom 4? and not 3? (i.e. no v₂ showing anywhere)

Thank you for any help. Will be much appreciated.

 

[HallsofIvy]

Homework Statement

Let u = (u₁, u₂, u₃)
and v = (v₁, v₂, v₃)
Determine if it's an inner product on R³.
If it's not, list the axiom that do not hold.

Homework Equations

the 4 axioms to determine if it's an inner product are
(all letters representing vectors)

1. <u,v> = <v,u>
2. <u+v, w> = <u, w> + <v,w>
3. <ku, v> = k<u,v>
4. <v,v> ≥ 0 and < v,v> = 0 if and only if v = 0

The Attempt at a Solution

So <u, v> is defined as
u₁v₁ + u₃v₃

Okay, I was wondering if you were going to mention that!

I'll skip the ones that did work and show axiom 4 which did not hold but I'm confused as to why this doesn't hold. I have a guess but have to make sure that I'm thinking correctly.

Axiom 4 does not hold:
<v, v > = v₁v₁ + v₃v₃
= v₁² + v₃² ≥ 0
and
<0, 0> = (0)(0) + (0)(0) = 0

now to check the other way:
if <v, v > = 0
implies that since
v₁² = 0 => v₁ = 0
v₃² = 0 => v₃ = 0

then it goes to say it's not an inner product on R³. Am I correct to say it's not an inner product on R³ because there are only 2 components for axiom 4? and not 3? (i.e. no v₂ showing anywhere)

Thank you for any help. Will be much appreciated.

Specifically, because there is no "v₂", If u= (0, 1, 0) u is not 0 but <u, u>= 0, contradicting that last law.

 

[sam0617]

Okay, I was wondering if you were going to mention that!


Specifically, because there is no "v₂", If u= (0, 1, 0) u is not 0 but <u, u>= 0, contradicting that last law.

Oh okay, that makes sense.
So just to clarify, because if <v, v > = 0 if and only if v = the zero vector
but we don't know v₂ due to how <u, v > is defined so that means v₂ could be a non-zero number.

Correct me if my logic is wrong.

Thank you again, HallsofIvy.