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Math Help - Dual space problem

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    Member kezman's Avatar
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    Dual space problem

    Let f: V \to V endomorphism of a vector space V of  dim = n.Show that  f has no trivial null space iff f^t has no trivial null space
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    Quote Originally Posted by kezman View Post
    Let f: V \to V endomorphism of a vector space V of  dim = n.Show that  f has no trivial null space iff f^t has no trivial null space
    That does it means f^t ?
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    Transpose of f
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    Quote Originally Posted by kezman View Post
    Let f: V \to V endomorphism of a vector space V of  dim = n.Show that  f has no trivial null space iff f^t has no trivial null space
    I am not sure how "transpose" is defined for non-Euclidean linear transformations. But if V is an Euclidean space then f:V\mapsto V can be tought as f(\bold{x})=A\bold{x} where A is a square matrix. Since this endomorphism has no trivial null space, i.e. A\bold{x} = 0 has non-trivial solutions it means \det (A) = 0. But since \det (A^T) = \det(A)=0 it means A^T \bold{x}=0 has non-trivial solutions. And so f^T has the property mentioned.
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    Let V, W K-vector spaces and let f : V \to W linear transf.
    The function f^t : W^* \to V^* is defined:
    f^t( \psi) = \psi   o   f   \forall  \psi  \in  W^*
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    Suppose Ker(f) is non-trivial: there exists some 0\neq v\in V such that f(v)=0. We are to find a nonzero functional v^*\in V^* such that f^t(v^*)=0; Let v^*\in V^* be the length of the projection on V-Ker(f) (well defined). Then for all v'\in V, we have  v^* (f(v'))= f^t (v^*) (v')=0 or f^t(v^*)=0 identically. So 0\neq v^*\in Ker(f^t) and Ker(f^t) is non-trivial.

    For the converse, interchange between f and f^t.
    Last edited by Rebesques; September 13th 2007 at 12:00 AM. Reason: one Kerf too many
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