# Homework Help: Determine whether it's an inner product on R^3

1. Dec 15, 2011

### sam0617

1. The problem statement, all variables and given/known data
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.

2. Relevant 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

3. 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.

2. Dec 15, 2011

### HallsofIvy

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

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

3. Dec 15, 2011

### sam0617

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.