Show that $\Gamma: V' \mapsto F^m$ is injective/surjective if and only if... 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?
 A: 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 $a_1,\ldots,a_m$).
A: I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
This problem is the same problem as Exercise 3.F.6 in this book.

Exercise 3.F.6
Suppose $V$ is finite-dimensional and $v_1,\dots,v_m\in V.$ Define a linear map $\Gamma:V'\to\mathbb{F}^m$ 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.
(b) Prove that $v_1,\dots,v_m$ is linearly independent if and only if $\Gamma$ is surjective.

My solution is here:

Let $u_1,\dots,u_n$ be a basis of $V$.
Let $\varphi_1,\dots,\varphi_n$ be the dual basis of $u_1,\dots,u_n$.
Let
$v_1=b_{11}u_1+\dots+b_{1n}u_n,$
$\dots$
$v_m=b_{m1}u_1+\dots+b_{mn}u_n.$
Then, $x_1v_1+\dots+x_mv_m=y_1u_1+\dots+y_nu_n,$
where $\begin{pmatrix}y_1\\\vdots\\y_n\end{pmatrix}=\begin{pmatrix}b_{11}&\cdots&b_{m1}\\\vdots&\cdots&\vdots\\b_{1n}&\cdots&b_{mn}\end{pmatrix}\begin{pmatrix}x_1\\\vdots\\x_m\end{pmatrix}=B^t\begin{pmatrix}x_1\\\vdots\\x_m\end{pmatrix}.$
The matrix of $\Gamma$ with respect to the basis $\varphi_1,\dots,\varphi_n$ of $V'$ and the standard basis of $\mathbb{F}^m$ is $B=\begin{pmatrix}b_{11}&\cdots&b_{1n}\\\vdots&\cdots&\vdots\\b_{m1}&\cdots&b_{mn}\end{pmatrix}$.
Then,
$\Gamma$ is injective
$\Leftrightarrow$
$x\mapsto Bx$ is injective
$\Leftrightarrow$
$x\mapsto B^tx$ is surjective
$\Leftrightarrow$
$v_1,\dots,v_m$ spans $V$.
Then,
$\Gamma$ is surjective
$\Leftrightarrow$
$x\mapsto Bx$ is surjective
$\Leftrightarrow$
$x\mapsto B^tx$ is injective
$\Leftrightarrow$
$v_1,\dots,v_m$ is linearly independent.

