Prove that the following are not inner products.

Question

Show that each of the following (a) <u, v> = u₁ v₁ + u₂ v₂; and (b) <u, v> = u₁ v₂ u₃ + v₁ u₂ v₃

is not an inner product on $ℝ^3$, where u = (u₁, u₂, u₃) and v = (v₁, v₂, v₃).

A function is an inner product if it satisfies the axioms:

1. <u,v> = <v,u>
2. <u,v+w> = <u,w> + <u,w>
3. c<u,v>=<cu,v>
4. <v,v>=0 and <v,v>=0 ⇔ v=0

(a) <u,v> = u₁v₁ + u₂v₂

Solution:

<u,v> = u₁v₁ + u₂v₂

<u,v> = v₁u₁ + v₂u₂ (Commutativity)

<u,v> = <v,u> (Axiom 1 satisfied).

Let w=(w₁, w₂) <u,v+w> = <(u₁, u₂), (v₁ + w₁, v₂ + w₂)>

<u,v+w> = u₁ (v₁ + w₁)+ u₂ (v₂ + w₂)

<u,v+w> = (u₁v₁ + u₂v₂) + (u₁w₁ + u₂w₂)

<u,v+w> = <u,v> + <u,w> (Axiom 2 satisfied).

Let c∈ℝ, then

c<u,v> = c(u₁v₁ + u₂v₂)

c<u,v> = (cu₁)v₁ + cu₂(v₂)

c<u,v> = <cu,v> (Axiom 3 satisfied).

<v,v> = v₁² + v₂² ≥ 0

Moreover, <v,v> = 0

⇔ v₁² + v₂² = 0

⇔ v₁ = v₂ = 0 since v₁² ≥ 0, v₂² ≥ 0

⇔ v = (0, 0) (Axiom 4 satisfied).

Therefore, <u,v> = = u₁v₁ + u₂v₂ is an inner product on $ℝ^3$.

Am I doing the right thing in (a)? I'm not sure about it. However, I am stuck in (b). Am I going to do the same process for (b)?

Answer 1

For a) use the comment by lulu.

For b) take $u=(1,1,-1)$ and note that $ \langle u, u \rangle <0$.