Linear Algebra - Inner Product problem

[RikaWolf]

TL;DR Summary

Prove/Disprove - Inner Product topic

I need help to know if I'm on the right track:
Prove/Disprove the following:
Let u ∈ V . If (u, v) = 0 for every v ∈ V such that v ≠ u, then u = 0.
(V is a vector-space)
I think I need to disprove by using v = 0, however I'm not sure.

 

[fresh_42]

Summary: Prove/Disprove - Inner Product topic

I need help to know if I'm on the right track:
Prove/Disprove the following:
Let u ∈ V . If (u, v) = 0 for every v ∈ V such that v ≠ u, then u = 0.
(V is a vector-space)
I think I need to disprove by using v = 0, however I'm not sure.

I assume you have a real, finite dimensional vector space here.

You are right, $(u,0)=0$ is always true for any $u$, which makes me think there is a typo. The condition is probably $(u,v)=0$ for every $v\neq 0$, then $u=0$.

What does it mean, that $(u,v)=0$ for all $v\neq 0$ and what does it mean if you chose a basis and express $u$ according to this basis?

 

[RikaWolf]

I assume you have a real, finite dimensional vector space here.

You are right, $(u,0)=0$ is always true for any $u$, which makes me think there is a typo. The condition is probably $(u,v)=0$ for every $v\neq 0$, then $u=0$.

What does it mean, that $(u,v)=0$ for all $v\neq 0$ and what does it mean if you chose a basis and express $u$ according to this basis?

Well then, assuming v= {b1...bn}, u={a1...an} there could still be a (u,v)=0 when v= {-an...-a} and u isn't 0

 

[fresh_42]

$(u,v)=0$ means that $u$ and $v$ are perpendicular to each other. Now how can a vector $u$ be perpendicular to all the rest? Have you had the Gram Schmidt orthogonalization algorithm in your course, yet?

 

[PeroK]

Summary: Prove/Disprove - Inner Product topic

I need help to know if I'm on the right track:
Prove/Disprove the following:
Let u ∈ V . If (u, v) = 0 for every v ∈ V such that v ≠ u, then u = 0.
(V is a vector-space)
I think I need to disprove by using v = 0, however I'm not sure.

Let me add to the above by explaining how I thought about this:

First, why are we not given that $(u, u) = 0$? Well, one of the inner product axioms is that $(u, u) = 0$ iff $u =0$.

Now, that gives me an idea of how to prove that $u = 0$. If I could show that $(u, u) = 0$ then that would be enough.

So, how to show that $(u, u) = 0$?

First, I can see an easy way that is a bit of a cheat. So, if I don't use that, what else can I do?

Well, there's the linearity of the inner product: $(u, v+w) = (u, v) + (u, w)$. Perhaps I could use that ...