Proving $g(u,v)≠0$ with Linear Independence

In summary, in order to prove that there always exists a vector w whose contraction with a lightlike vector u is always different from zero, we can use the fact that both vectors are linearly independent and the non-degeneracy condition on the pseudo-Riemannian metric. This can be done in a coordinate system or in a free coordinate way. By considering the common plane for the two vectors and using the properties of bilinearity of the metric, we can arrive at the conclusion that u is not lightlike. Therefore, there always exists a vector w whose contraction with u is non-zero.
  • #1
isaacdl
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TL;DR Summary
Contraction of a lightlike vector with another vector in a free coordinate scheme.
I'm trying to prove that there exist always a vector w whose contraction with a lightlike vector u (g(u,u)=0) it's always different from zero:
$g(u,v)≠0$I know how to do this with coordinates, but in a free cordinate scheme I'm totally lost.

Any help?

PD: Both vectors are linearly independent.
 
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  • #2
What have you tried so far?
How did you do it in a coordinate system?
Can you generalize that to work without coordinates?
 
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  • #3
In a coordinates it's straight forward. If u=(1,1,0,0) and w=(1,0,0,0) timelike or same to ligthlike, it's not difficult to prove they are colinear. But I don't know how to traduce it in a free coordinate way.
 
  • #4
pseudo-Riemannian metrics are non-degenerate; can you argue by contradiction?
 
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  • #5
Mmm, I think I have to use the fact w and u are linearly independent. My first idea was to use the properties of bilinearity of the metric. So by contradiction you mean to consider g(u,v)=0 and get to u is not lightlike?
 
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  • #6
As in, if there were no ##\mathbf{w}## giving ##g(\mathbf{u},\mathbf{w}) \neq 0## that’s the same as saying ##g(\mathbf{u},\mathbf{w}) = 0## for all ##\mathbf{w} \in \mathbf{R}^4##. How does that mesh with the non-degeneracy condition on ##g##?
 
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  • #7
Consider the common plane for the two vectors…. Given the lightlike vector, can you write the “other vector” using it?
 
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  • #8
ergospherical said:
As in, if there were no ##\mathbf{w}## giving ##g(\mathbf{u},\mathbf{w}) \neq 0## that’s the same as saying ##g(\mathbf{u},\mathbf{w}) = 0## for all ##\mathbf{w} \in \mathbf{R}^4##. How does that mesh with the non-degeneracy condition on ##g##?
Ok, I was using proof by contradiction wrong. Ok, if ##g(\mathbf{u},\mathbf{w}) = 0## for all ##\mathbf{w} \in \mathbf{R}^4##, by non-degenacy condition if S is non-degenerated u is a null vector u=0. But I'm not sure if the space S or equiv, g | S is non-degenerate. The only info I have is that ##S=(v_1,...,u,..._w,...,v_m)## is a linearly independent set.

What I have to find now is that ##u## is not lightlike, right?PD: I don't know why LATEX is not showing properly.
 
  • #9
robphy said:
Consider the common plane for the two vectors…. Given the lightlike vector, can you write the “other vector” using it?
Yes, I can see it graphically, but my problem is to show it in a free cordinate way.
 
  • #10
@isaacdl if ##g(\mathbf{u}, \mathbf{w}) = 0## for all possible ##\mathbf{w}## then the non-degeneracy of ##g## would imply that ##\mathbf{u} = 0##, but this is not true by assumption (##\mathbf{u}## is a non-zero lightlike vector).
 
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  • #11
ergospherical said:
@isaacdl if ##g(\mathbf{u}, \mathbf{w}) = 0## for all possible ##\mathbf{w}## then the non-degeneracy of ##g## would imply that ##\mathbf{u} = 0##, but this is not true by assumption (##\mathbf{u}## is a non-zero lightlike vector)
Yeah totally understood that part, but what I said in the previous post is that I don't know if g in S is non-degenerated. It's not given in the exercise.

PD: Wait, u are totally right, we can use it, it's given in a previous part of the exercise! Thanks a lot.
 

1. How do you prove that g(u,v)≠0 using linear independence?

To prove that g(u,v)≠0 using linear independence, we need to show that the vectors u and v are linearly independent. This means that there is no non-zero linear combination of u and v that equals zero. We can do this by setting up a system of equations and showing that the only solution is when both u and v are equal to zero.

2. Why is linear independence important in proving g(u,v)≠0?

Linear independence is important because it ensures that the vectors u and v are not dependent on each other. If they were dependent, it would be possible to find a non-zero linear combination of u and v that equals zero, which would contradict the statement that g(u,v)≠0. Therefore, proving linear independence is crucial in proving g(u,v)≠0.

3. Can you provide an example of a proof for g(u,v)≠0 using linear independence?

Yes, for example, let u = (1, 2) and v = (3, 4). To prove that g(u,v)≠0, we need to show that the vectors u and v are linearly independent. We can set up the following system of equations: a(1, 2) + b(3, 4) = (0, 0). Solving this system, we get a = 0 and b = 0, which means that the only solution is when both u and v are equal to zero. Therefore, u and v are linearly independent, and g(u,v)≠0.

4. Are there any other methods for proving g(u,v)≠0 besides linear independence?

Yes, there are other methods such as using the properties of determinants or using the properties of inner products. However, linear independence is one of the most commonly used methods for proving g(u,v)≠0.

5. How does proving g(u,v)≠0 using linear independence relate to real-world applications?

Proving g(u,v)≠0 using linear independence is relevant in many real-world applications, especially in fields such as physics and engineering. For example, in physics, linear independence is used to show that forces acting on an object are not dependent on each other. In engineering, it is used to show that different components of a system are not dependent on each other, which is crucial in designing efficient and reliable systems.

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