Exercise: Suppose $V$ is finite-dimensional and $v_1,\dots,v_m \in V$. Define a linear map $\Gamma : V'\to F^m$ by $$\Gamma(\varphi)=(\varphi(v_1),\dots,\varphi(v_m))$$
$(a)$ Prove that $v_1,\dots,v_m$ spans $V$ if and only if $\Gamma$ is injective.
Here $V'$ denotes the dual space of $V$ and $F$ denotes the field of real or complex numbers. I managed to prove the the implication in one direction. The following is that part of the proof.
Proof: Suppose that $v_1,\dots,v_m$ spans $V$. Then for all $\varphi\in V'$ such that $\varphi \not\equiv0$ we have that $\varphi(v_j)\ne 0$ for at least one $j$ as $\varphi(v_1),\dots,\varphi(v_m)$ spans range $\varphi$. Note that range $\varphi\ne \{0\}$ as every linear functional is either the zero map or surjective. Thus, $\Gamma(\varphi)=(0, \dots,0)$ if and only if $\varphi \equiv 0$. Hence, null $\Gamma=\{0\}$ and $\Gamma$ is injective.
Could someone give me some pointers as to how to proceed proving the implication in the other direction? I am assuming it has something to do with the fact that if $\Gamma$ is injective then for the dual basis $\varphi_1,\dots,\varphi_n$ the list $\Gamma(\varphi_1),\dots,\Gamma(\varphi_n)$ is linearly independent. But I can't seem to find a connection between $\Gamma$ and $v_1,\dots,v_m$.
Let $w_1,\ldots,w_k$ be a basis of the span of $v_1,\ldots,v_m$. If the span is not equal to $V$, extend it to a basis $w_1,\ldots,w_k,w_{k+1},\ldots,w_n$ of $V$. Let $\varphi$ be the coordinate functional of $w_{k+1}$ (so $\varphi(w_i)=\delta_{i,k+1}$ for $1\le i\le n$, where $\delta$ is the Kronecker delta). Then $\varphi\ne 0$ but $\varphi(v_i)=0$ for all $1\le i\le m$ since the $v_i$ are in the span of $w_1,\ldots,w_k$, so $\Gamma(\varphi)=0$ and $\Gamma$ is not injective.