# Dual space problem

• Sep 8th 2007, 07:15 PM
kezman
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
• Sep 8th 2007, 07:18 PM
ThePerfectHacker
Quote:

Originally Posted by kezman
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 ?
• Sep 8th 2007, 07:29 PM
kezman
Transpose of f
• Sep 8th 2007, 08:29 PM
ThePerfectHacker
Quote:

Originally Posted by kezman
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.
• Sep 9th 2007, 09:57 AM
kezman
Let $V, W$ K-vector spaces and let $f : V \to W$ linear transf.
The function $f^t : W^* \to V^*$is defined:
$f^t( \psi) = \psi o f \forall \psi \in W^*$
• Sep 12th 2007, 10:20 PM
Rebesques
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 $f^t$.