Fix Annihilator
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Fix Annihilator
Dual vector spaces, dual maps, annihilators?
If you saw this question before, I'm afraid I made a slight error (should've said annihilator of kernel, not of image). Fixed now!
Anyway, the following claim seems to need finite dimensionality:
If T is a linear map from V to W (where V, W are vector spaces) then the annihilator of the kernel of T is the image of the dual map T'
It's easy to show that the image of T' is a subspace of the annihilator of the kernel of T, but the converse I think needs finite dimensionality. However, I've been unable to think of an infinite dimensional counter-example.
Can you think of one (or prove it holds even in infinite dimensional spaces)?
If the category is bounded linear maps on normed spaces, the result is true by the Hahn-Banach Theorem. If the category is just algebraic linear maps on general vector spaces (which it is, if I take the strict meaning of your question), then the result is again true, but not for a reason anywhere near as deep as the H-B theorem.
The proof boils down to the fact that any functional defined on any subspace M of a vector space W extends to a functional defined on the whole space W. See the link for a proof -- its pretty elementary; it begins by constructing an algebraic complement N of the given subspace M (using Hamel bases), and then extending the given functional on M to all of W=M+N by composing it with the (algebraic) projection along N to M.
To see how to reduce your question to this extension result, argue as follows. If a:V→C (C is the scalar field) annihilates ker(T), then we can define b:V/ker(T)→C by b(v+ker(T)) = a(v) (trivial to check that b is well-defined). Also T:V→W defines S:V/ker(T)→W by S(v+ker(T))=T(v) (well-defined, too). Now S is 1-to-1, so we can define c:Image(S)→C by c(w)=b(v+ker(T)), where S(v+ker(T))=w. Use the extension result above to extend c (defined on Image(S)=image(T)) to a functional d:W→C. The claim is that a=doT. Well, if vεV, then a(v) = b(v+ker(T)) = c(w), where S(v+ker(T))=w. Note S(v+ker(T)) = T(v). So a(v) = c(T(v)) = d(T(v)). So a is in the image of T'.
Fix a windshield like a pro with the Annihilator