- #1
Bacle
- 662
- 1
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.
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.