Elements of Dual of Infinite_Dim. Space have Codimension One

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Discussion Overview

The discussion revolves around the properties of the kernel of linear maps in the context of infinite-dimensional normed spaces, specifically addressing whether the kernel of an element in the dual space has codimension one, similar to the finite-dimensional case.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant outlines a method for showing that the kernel of a linear map in a finite-dimensional space has codimension one, using the rank-nullity theorem.
  • Another participant proposes a strategy to demonstrate that a linear subspace W, which strictly contains the kernel of a linear map L, must equal the entire space V.
  • A subsequent reply questions the implications of the argument presented, suggesting that if the kernel were the whole space, then L would be the zero map, and asks how the infinite-dimensional nature of V is utilized in the argument.
  • A later reply indicates that the previous confusion has been resolved by the participant.

Areas of Agreement / Disagreement

The discussion includes some confusion regarding the implications of the arguments presented, particularly in relation to the properties of the kernel in infinite-dimensional spaces. There is no clear consensus on whether the kernel has codimension one in this context.

Contextual Notes

Limitations include the lack of clarity on how the infinite-dimensional aspect affects the properties of the kernel and whether the assumptions made in the finite-dimensional case apply directly to the infinite-dimensional scenario.

Bacle
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Hi:
In the case of a finite-dimensional normed space V, it is relatively-straightforward to
show that the kernel of any element of V* has 1 .

( Assume DimV=n):

We take a linear map L:V-->F ; F the base field. We choose a basis to represent L,
then we consider F as a vector space over itself; F is then 1-dimensional over
itself, then ,( by rank-nullity) , L has rank 1 , so it must have nullity n-1.

I don't see, though, how, in the case of V infinite-dimensional, how to show that, given
L in V* , that the kernel of L is a maximal (strict) subspace; I don't know if we still
say that KerV* has codimension 1.

Any Ideas?

Thanks.
 
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Let W be a linear subspace of V which strictly contains ker(L). Our goal is to show that W=V.

Take w in W such that L(w)=1. This exists, since W contains the kernel strictly.
Now, take v in V, and let a=L(v). Then v-aw is an element of the kernel, and is thus contained in W. Thus v=aw+(v-aw) is an element of W. This shows that V=W.
 
Sorry, I don't get it; you showed that the kernel is the whole space. Then the map
should be the zero map, right?. Also: how did you use the fact that V is infinite-dimensional?
 
Never mind, sorry, I jumped the gun; I got it. Thanks.
 

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