Proof that if a set of vectors is linearly independent over R the it is linearly independent over C I need to prove this for an exercise but I don't seem to understand why the first implication is true.

$$
Let \hspace{2mm} v_1,...,v_n \in {\Bbb R}^{n}. Prove \ that \hspace{2mm} \{v_1,...,v_n \} \ is \ linearly \ independent \ over \ {\Bbb R} \iff \{v_1,...,v_n \} \ is \ linearly \ independent \ over \ {\Bbb C}.
$$

I was able to prove the second implication. What I did was :

Let's suppose that $\{v_1,...,v_n\}$ is linearly independent over ${\Bbb C}$ then $$0=a_1*v_1+...+a_n*v_n$$
with $a_k=0 \hspace{2mm} \forall \hspace{1mm} 1\leq k\leq n$ and $a_k \in {\Bbb C}$. This is the only possible solution for the $a_k$ over ${\Bbb C}$ and also over ${\Bbb R}$, because ${\Bbb R} \subset {\Bbb C}$.

I'd like you to help me prove the first implication. A hint or the complete proof are ok. Thank you :D
As the set of vectors $v_i, 1\le i\le n$ is LI over $\mathbb R$, so

$\sum_{i=1}^n c_iv_i=0$

$\implies c_i=0\in \mathbb R \forall i$

or  $ c_i=0+i.0\in \mathbb C \forall i$

hence $v_i, 1\le i\le n$ is LI over $\mathbb C$.

For converse part you can proceed backwards.
As the set of vectors $v_i, 1\le i\le n$ is LI over $\mathbb R$, so

$\sum_{i=1}^n c_iv_i=0$

$\implies c_i=0\in \mathbb R \forall i$

or  $ c_i=0+i.0\in \mathbb C \forall i$

hence $v_i, 1\le i\le n$ is LI over $\mathbb C$.

For converse part you can proceed backwards.