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Consider the following statements, given $V$ is a vector space.

(A): Given $v \in V, v \neq 0$, there exists $f \in V'$ such that $f(v) \neq0$.

(B): Given a set of linearly independent vectors $\{v_1, \dotsc, v_m\} \in V$, there exists a reciprocal set $f_1, \dotsc, f_m \in V'$ such that $f_i(v_j) = \delta_{ij}$.

It seems to me I am going in circles with the logic here. To prove (A), I use the fact that there is a reciprocal set, and to prove (B), I start with the fact that a putative reciprocal functional is well defined.

Is this because $V$ is assumed finite dimensional ? Are (A) and (B) correct if $V$ is not finite dimensional ?

me10240
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1 Answers1

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For (a), you may explicitly define $f(v) = 1, f(0) = 0$. Then $f$ is a linear map on the subspace $\operatorname{span}(v)$, so there exists a dual vector $g \in V'$ with $g\restriction_{\operatorname{span}(v)} = f$. (This is a fact that holds for all finite-dimensional $V$ and linear maps on subspaces of $V$) Then (b) follows easily.

Things are not simple on infinite-dimensional vector spaces, but if you assume axiom of choice, you may extend linear maps from subspaces to the whole space for arbitrary vector spaces.

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Henricus V.
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