I Linear Algebra - Inner Product problem

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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

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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?
 
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

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##(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

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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 ...
 

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