If T is injective, then $v_1,\ldots,v_n$ spans $V$ I need help in a Linear Algebra demonstration. I am doing the Exercise 6 from Section 3.F of the book Linear Algebra Done Right by Axler (3rd edition). I am doing the first problem that says:

Suppose V is finite-dimensional and $v_1,..., v_m \in V$. Define a linear map T: V' to $F^m$ by: $T(\phi) = (\phi(v_1),...,\phi(v_m))$. Show that $gen(v_1,...,v_m)$ = V $\leftrightarrow$ T is injective.

I believe a have done the first implication, but I am not sure about the second one. I was able to show that if T is injective then, ker(T) = {$\phi$ : T($\phi$) = 0}, so if $\phi \in$ ker(T), T($\phi$) = ($\phi(v_1),...,\phi(v_2)) = 0$. If we evaluate by component, then $\phi(v_1) = 0,...,\phi(vn) = 0$. Now I think that is correct, but I am not sure if it is safe to say this next thing:
From the previous results we can assume that ker(T) = $(gen(v_1,...,v_n))^o$
Can someone tell me if the last thing is correct, please?
 A: If $\operatorname{span}\bigl(\{v_1,\ldots,v_n\}\bigr)\ne V$, take $v\in V\setminus\operatorname{span}\bigl(\{v_1,\ldots,v_n\}\bigr)$ and take $\phi\in V'$ such that $\phi(v)\ne0$ and $\phi\left(\operatorname{span}\bigl(\{v_1,\ldots,v_n\}\bigr)\right)=\{0\}$. Then $T$ is not injective, since $\phi\ne0$, but $T(\phi)=0$.
Now, assume that $\operatorname{span}\bigl(\{v_1,\ldots,v_n\}\bigr)=V$ and let $\phi\in V'$ such that $T(\phi)=0$. Asserting that $T(\phi)=0$ is the same thing as asserting that $\phi(v_k)=0$ for each $k\in\{1,2,\ldots,n\}$. But, since $\operatorname{span}\bigl(\{v_1,\ldots,v_n\}\bigr)=V$ , it follows from this that $(\forall v\in V):\phi(v)=0$. So, $\phi=0$.
A: Here is a more “abstract” proof:
Consider the maps
\begin{align*}
(F^m)’ & \stackrel \Omega \longrightarrow F^m \\
\phi & \longmapsto (\phi(e_1),\dots,\phi(e_m)) \\[3mm]
F^m & \stackrel \Psi \longrightarrow V \\
(a_1,\dots,a_m) & \longmapsto a_1v_1+\cdots+a_mv_m \\[3mm]
V & \stackrel \Lambda \longrightarrow (V’)’ \\
v & \longmapsto (\varphi \mapsto \varphi(v))
\end{align*}
where $e_1,\dots,e_m$ is the standard basis of $F^m$. It is easy to prove that $\Omega$ is an isomorphism. And, because $V$ is finite-dimensional, $\Lambda$ is too.
Now, the dual map $T’ \colon (F^m)’ \to (V’)’$ of $T$ can be written as $\Lambda \circ \Psi \circ \Omega$, so $\Psi = \Lambda^{-1} \circ T’ \circ \Omega^{-1}$ and then we have that
\begin{align*}
v_1,\dots,v_n \textrm{ spans } V
& \iff \Psi \textrm{ is surjective} \\
& \iff T’ \textrm{ is surjective} \\
& \iff T \textrm{ is injective.} \\
\end{align*}
