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Thread: Dual space problem

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    Member kezman's Avatar
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    Dual space problem

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

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