Show that $\left<\mathcal{M}\right>=V$ iff $v_i$ is a linear combination of the elements of $\mathcal{M}$. I'm struggling with the following problem:

Let $V$ be a finitely generated vector space over the field $F$.  Let
  $\left( v_1, \ldots, v_n \right)$ be a basis of $V$. Let 
  $\mathcal{M}$ be a family of vectors in $V$. 
Show that $\left<\mathcal{M}\right>=V$ if and only if, for
  $i=1,\ldots,n$; $v_i$ is a linear combination of the elements of
  $\mathcal{M}$.

I have the following:
$\Rightarrow$) Suppose $\left<\mathcal{M}\right>=V$ such that $m_i \in \mathcal{M}$ for $i=1,\ldots,n$. Which means that $\left<m_1,\ldots,m_n\right>=V$. Given that $\left( v_1, \ldots, v_n \right)$ is a basis of $V$, $\left< v_1, \ldots, v_n \right>=V=\left<m_1,\ldots,m_n\right>$. So  does this mean that $v_i$ is a linear combination of the elements of $\mathcal{M}$?
I think I might have some right idea for the above but it might not be sufficient.
$\Leftarrow$) If I assume that $v_i$ is a linear combination of the elements of $\mathcal{M}$ for $i=1,\ldots,n$, this would mean $\lambda_1 m_1+\ldots+\lambda_n m_n=\left( v_1, \ldots, v_n \right)$ for $\lambda_1,...,\lambda_n \in F$. How should I continue to show that $\left<\mathcal{M}\right>=V$?

Do note that it is not necessary for $M$ to have $n$ elements. It is only a generating set, so all you know is that it has at least $n$ elements.
Suppose $\langle M\rangle=V$. So, every element of $V$ can be written as a linear combination of elements of $M$. Now, $(v_i)_1^n$ is a basis of $V$, so each $v_i$ must be an element of $V$, so each $v_i$ can be written as a linear combination of elements of $M$.
Now, conversely, suppose that each $v_i$ can be written as a linear combination of elements of $M$. So,
$$v_i=\sum_{m\in M}\lambda_mm~~~\textrm{for some scalars }\lambda_m$$
Now, since $(v_i)$ is a basis of $V$, every $v\in V$ can be written as a linear combination of the $v_i$'s, so,
$$v=\sum_i\alpha_i v_i=\sum_i\alpha_i\sum_{m\in M}\lambda_mm=\sum_i\sum_{m\in M}\alpha_i\lambda_m m$$
with the last equality showing that each $v\in V$ is a linear combination of the elements of $M$, so $\langle M\rangle=V$
