Showing a linear transformation $T\colon V^{*}\to\mathbb R^{m}$ is surjective 
Let $V$ be a finite dimension vector space and let $V^{*}=\operatorname{Hom}(V,\mathbb R)$ be
its dual space.
Let $v_{1},\ldots,v_{m}\in V$, and let
$T:V^{*}\to\mathbb R^{m}, \ \ T(\phi)=\pmatrix{\phi(v_{1})\\.\\.\\.\\\phi(v_{m})}$ be a linear
tranformation.
Prove that $T$ is surjective if $\left\{v_{1},...,v_{m}\right\}$ is
linearly independant and not surjective if
$\left\{v_{1},...,v_{m}\right\}$ is linearly dependant.

I'm having trouble with this one.. my attempt for the first direction:
Let $\phi\in V^{*}$. Then $T(\phi)=\pmatrix{\phi(v_{1})\\.\\.\\.\\\phi(v_{m})}=\phi(v_{1})\pmatrix{1\\0\\.\\.\\.\\0}+...+\phi(v_{m})\pmatrix{0\\.\\.\\.\\0\\1}\in\operatorname{span}\left\{e_{1},...,e_{m}\right\}=\mathbb{R}^{n}$
Note that as $e_{1},...,e_{m}$ are linear independant the equation $c_{1}e_{1}+...+c_{m}e_{m}=0$ has only the trivial solution, thus $T(\phi)=0$ only if $\phi$ is the zero transformation, therefore $\operatorname{ker}(T)=\left\{0\right\}$ therefore $T$ is injective and $\dim V^{*}\leq\dim\mathbb{R}^{m}=m$, but as $v_{1},...,v_{m}\in V$ are linearly independant $m\leq\dim V$, which means $m\leq \dim V=\dim V^{*}\leq \dim\mathbb R^{m}=m \Rightarrow  \dim V^{*}=\dim\mathbb R^{m}$ and thus $T$ is surjective.
I'm not sure about my explanations and if they valid, and I couldn't solve the second direction and I'm looking for ideas on how to approach it..
 A: For the first: Let $x\in R^m$. Define $\phi$ on $\text{span}(v_1,\dots,v_n)$ via $\phi(v_i) = x_i$ and extending linearly. Then $\phi$ is well defined as $v_1,\dots,v_n$ are linear independent. Extend $\phi$ to a map on $V$ by setting $\phi(v) = 0$ if $v$ is not in the span of $v_1,\dots,v_n$. Then clearly $\phi \in V^*$ and $T(\phi) = x$.
A: You proved that if $T$ is injective and $v_1,\dots,v_m$ are linearly independent, then $\dim(V^*) = \dim(\mathbb R^m)$, and hence $T$ is also surjective. That's fine. But there is a problem in your proof: $T(\phi) = 0$ doesn't implies $\phi = 0$, instead, $T(\phi) = 0$ just implies that $\phi(v_1)=\cdots=\phi(v_m)=0$ (as you said, $c_1e_1+\cdots+c_me_m = 0$ implies $c_1=\cdots=c_m=0$).
Note, your problem reduces to show that $T$ is surjective if and only if $v_1,\dots,v_m$ are linearly independent. Also, is easy to show the following:

*

*A linear map $f : X \to Y$ is surjective if and only if its dual/transpose $f^* : Y^* \to X^*$ ($f^*(\phi) = \phi \circ f$ for all $\phi \in Y^*$) is injective.

*The vectors $v_1,\dots,v_m$ are linearly independent if and only if the linear map $\varphi : \mathbb R^m \to V$ defined by $\varphi(a_1,\dots,a_m) = a_1v_1+\cdots+a_mv_m$ is injective.

Now, prove that the maps $c : (\mathbb R^m)^* \to \mathbb R^m$ and $\operatorname{ev} : V \to (V^*)^*$ defined by $$c(g) = (g(e_1),\dots,g(e_m)) \quad \& \quad \operatorname{ev}(v)(\phi) = \phi(v)$$ for all $g \in (\mathbb R^m)^*$, $v \in V$ and $\phi \in V^*$, are linear isomorphisms such that $T^* = \operatorname{ev} \circ\, \varphi \circ c$ (see the diagram below). Hence, $T^*$ is injective if and only if $\varphi$ is injective, so we conclude that $T$ is surjective if and only if $v_1,\dots,v_m$ are linearly independent.
$$
\require{AMScd}
\begin{CD}
(\mathbb R^m)^* @>{T^*}>> (V^*)^* \\
@VcVV @AA{\operatorname{ev}}A \\
\mathbb R^m @>>\varphi> V
\end{CD}
$$
