dual vector spaces, dual maps, annihilators?

If you've seen this question before, I'm afraid I made a small error (should have said to W (where V, W are vector spaces) and then the annihilator of the kernel of T is the dual image of the map T "is easy to show that the image of T is a subspace of the annihilator of the kernel of T, but otherwise I think the needs of finite dimensionality. However, I have been unable to think of an infinite-dimensional counter-example. Can you think of one (or show that is maintained even in infinite dimensional spaces)?

If the category is limited in normed linear spaces, the result is true by the Hahn-Banach theorem. If the category is only linear algebra in vector spaces in general (that is, if I take the sense of your question), then the result is true, but not for a reason not nearly as deep as the theorem of HB. The proof is reduced to the fact that any functional defined in any subspace M W of a vector space extends to a functional defined on the whole space W. See link for a test – its very basic, beginning with the construction of an algebraic complement To see how to reduce your question to the result of extension, 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)) V = a () (trivial to check that b is well defined). Also T: V → W defined by S: V / ker (T) → where S (v + ker (T)) = w. Use the result of extension above to enlarge C (defined in the image (S) = image (T)) to a function d: W → C. The argument is that a = point. 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 one is in the image of T '.

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