The problem is this: Let $V$ be a vector space over $\mathbb{K}$ and $v \in V$. Show that if $f(v) = 0, \forall$ $f \in L(V, \mathbb{K})$, then $v=0$.

It's a problem of a book I'm using to study Dual Spaces. It seems easy to solve, but I'm out of ideas... I tried write $v$ as a linear combination of the basis of $V$ and $f$ as a combination of the dual basis, but then I realized dimension of $V$ can be infinity.


Assume $v\ne0$ so let $V'$ such that $$V=V'\oplus\operatorname{span}(v)$$ and let $g$ the linear form such that $g(v)=1$ and $g(x)=0,\;\forall x\in V'$ and this contradicts that $f(v)=0,\;\forall f\in L(V,\Bbb K)$.

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    $\begingroup$ It is worth pointing out that the existence of such a $V'$ uses the fact that $v$ is a member of some basis of $V$. $\endgroup$ Feb 18 '14 at 18:11
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    $\begingroup$ I didn't understand why you can assume that there is a linear form $g$ such that $g(v)=1$ if the hypothesis is precisely that $g(v)=0$ $\forall$ $g$... I'm confused here, sorry.. $\endgroup$
    – Anna
    Feb 18 '14 at 18:47
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    $\begingroup$ @Carol I didn't assume the existence of $g$ but I constructed it. Notice that we can construct a linear transformation $f$ on $E=E_1\oplus E_2$ if we give the expression of its restriction on each subspace $E_1$ and $E_2$. $\endgroup$
    – user63181
    Feb 18 '14 at 18:51
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    $\begingroup$ You have a talent for giving excellent concise answers! $\endgroup$
    – amWhy
    Feb 19 '14 at 13:18

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