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Let $V$ be a finite-dimensional vector space. Let $f=(f_1,f_2,\dots)\colon V\to \mathbb{R}^{\mathbb{N}}$ be an injective linear map. Do $f_1,f_2,\dots$ span $V^*$?

I know that the answer is yes when instead I have $f\colon V\to \mathbb{R}^d$, with $d<\infty$. This follows directly from the fact that the dual map $\mathbb{R}^d\ni x\mapsto x_i f_i$ is surjective. However, here we use an identification of $\mathbb{R}^d$ with its dual which we do not have in the infinite-dimensional case.

Furthermore, I know that the answer is false when instead $V$ is assumed to be infinite-dimensional: By assumption it is then necessarily countably infinite dimensional. Let $\{v_i\}_{i\in\mathbb{N}}$ be a basis of $V$ and take as $f_i$ the corresponding dual set defined by $f_i(v_j)=\delta_{ij}$. The map $f$ is injective, it is even bijective. However, $g\in V^*$ defined by $$g(v_i):=1\;\forall i$$ is not in the span of the $f_i$ because every element in the span of $f_i$ has finite support.

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You are given that $ker f={0}$, Hence $\cap ker f_i={0}.$ Now you have to use a result that if $g,f_1,..f_N$ are linear functional on $V$ with $\cap ker f_i\subset ker g,$ then $g$ is a linear combination of $f_i$'s. Let us do it for $N=1$ i.e. $ker f_1\subset ker g.$ If $f=0$ then $ker f=V$, hence $ker g=v$. So $f_1=g=0$. Otherwise there exists $x_0$ such that $f_1(x_0)=1.$ Let $y\in V$ be arbitrary. Then $f_1(y-x_0f_1(y))=f(y-f_1(x_0)f_1(y)=0$, This implies $g(y-x_0(f_1(y))=0 \implies g(y)=g(x_0)f_1(y)$

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  • $\begingroup$ Does this hold for infinitely many $f_i$? $\endgroup$
    – Bananach
    Nov 3 '15 at 10:06
  • $\begingroup$ Unless $V^*$ has any topology , infinite linear combinaton won't make sense. $\endgroup$ Nov 3 '15 at 10:17
  • $\begingroup$ But a linear generating system still makes sense, meaning that: Every element can be written as a finite linear combination of the possibly infinitely many elements of the generating system. $\endgroup$
    – Bananach
    Nov 3 '15 at 10:20
  • $\begingroup$ You will see from my latest edit that your answer cannot be correct, since it does not use that $V$ is finite-dimensional. $\endgroup$
    – Bananach
    Nov 3 '15 at 17:38
  • $\begingroup$ Your example is correct...But if you take $dim V\leq \infty$ but number of $f_i$'s finite , then my answer is correct. $\endgroup$ Nov 4 '15 at 5:05
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The answer is yes.

Let $d$ be the dimension of $V$. If the $f_i$ do not span the dual, their span would be at most $d-1$ dimensional. The intersection of their kernels would thus be at least 1 dimensional. Since $\ker f=\bigcap \ker f_i$ we obtain that $f$ is not injective.

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