Prove: y=ξz if [x,y]=0 & [x,z]=0

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Discussion Overview

The discussion revolves around proving the relationship between two linear functionals, y and z, under the condition that their pairings with a vector x are zero whenever the pairing of x with z is zero. The participants explore the implications of this condition and seek a proof without relying on the dual basis, while considering both finite and infinite dimensional vector spaces.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a proof strategy involving the dual basis but seeks an alternative approach due to the context of the problem.
  • Another participant questions the meaning of the notation [x,y] and suggests that it represents the pairing between a vector x and the linear functional y.
  • A third participant introduces the rank-nullity theorem to discuss the dimensions of the kernels of linear functionals, suggesting that if both functionals are non-zero, they must have the same kernels.
  • There is a suggestion that if [x_0,z] is non-zero, then any vector in the space can be expressed as a combination of x_0 and an element from the kernel of y, which could be used to further the proof.
  • A later reply asserts that the argument can be adapted for infinite dimensional spaces, maintaining that the relationship holds under these conditions.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of certain mathematical concepts, particularly regarding finite versus infinite dimensional spaces. There is no consensus on a definitive proof method, and the discussion remains unresolved.

Contextual Notes

Some participants note the potential limitations of the proof strategies discussed, particularly concerning assumptions about dimensionality and the nature of the linear functionals involved.

Matthollyw00d
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Q: Prove that if y and z are linear functionals (on the same vector space) such that [x,y]=0 whenever [x,z]=0, then there exists a scalar ξ such that y=ξz.
(Hint: if [x0,z]≠0, write ξ=[x0,y]/[x0,z].)

I'm fairly certain there's an obvious proof using the dual basis, but this is in the section before that, so I'm trying to do it without that, and can't seemed to get the proper result. Any help would be great, thanks!
 
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What does [x,y] mean? Is x a vector and [x,y] the pairing between x and the linear functional y?

If so, here's a hint: Let n be the dimension of the vector space. The dimensions of the kernels of linear functionals are either n (if it is the 0 functional) or n-1. This follows from the rank nullity theorem. Assuming that the functionals are non-zero (if one is zero the problem is trivial), this means that they have the same kernels. If [x_0,z]\ne 0 then anything in the vector space can be written as x_0 + k where k is in the kernel. Now use the hint.
 
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[x,y]=y(x)

Your hint would be adequate if we can assume finite dimensionality, but I don't think we can in this problem.
 
I think you can still modify the argument to show it works in the infinite dimensional case: you can still show that everything in V can be written as a multiple of x_0 + something in the kernel of y.
 

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