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Let $V$ be a finite-dimensional vector space and $v_1, \dots, v_m \in V$. Let $F$ indiscriminately denote either the field $\mathbb{R}$ or $\mathbb{C}$. Define the linear map $\Gamma: V' \mapsto F^m$ by $\Gamma(\varphi) = \big(\varphi(v_1),\dots,\varphi(v_m)\big)$.

Show that $v_1,\dots,v_m$ spans $V$ if and only if $\Gamma$ is injective and that $v_1,\dots,v_m$ is linearly independent if and only if $\Gamma$ is surjective.

In both cases, I am only able to prove the only if implication.

In the first case we can reduce the vectors $v_i$ to a basis $v_1,\dots,v_n$ of $V$ (with $n \leq m$). Now $\varphi$ can be written as $\sum_{i=1}^{n} a_i \varphi_i$ (with $\varphi_i$ representing the dual basis of $v_1,\dots,v_n$) and it is not difficult to show that $\Gamma$ injective implies $a_i=0$ for each $i$ (which means $\ker{\Gamma}=0$).

In the second case we procede similarly by extending the vectors $v_i$ to a basis $v_1,\dots,v_n$ (here we have $n \geq m$ instead) and by considering the dual basis $\varphi_1,\dots,\varphi_n$. For each $(a_1,\dots,a_m) \in F^m$ then we have $\Gamma(\sum_{i-1}^{m} a_i\varphi_i) = (a_1,\dots,a_m)$ as desired.

Any hints about how to tackle the other direction?

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  • $\begingroup$ I don't understand your first proof. $\endgroup$ – amsmath Dec 11 '17 at 19:13
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Hint 1: If $v_1,\ldots,v_m$ does not span $V$, then there exists a nonzero functional $\varphi$ that maps $v_1,\ldots,v_m$ to zero. (Think about vectors outside the span of these $m$ vectors.) Then $\Gamma(\varphi) = 0$.

Hint 2: If $v_1,\ldots,v_m$ are not linearly independent, then some linear combination $a_1 v_1 + \cdots + a_m v_m$ of them equals zero. That means $a_1 \varphi(v_1) + \cdots + a_m \varphi(v_m) = 0$ for any functional $\varphi$, so the image of $\Gamma$ in $F^m$ lies in a strict subspace of $F^m$ (namely it lies in the hyperplane specified by this linear constraint involving $a_1,\ldots,a_m$).

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