Prove that $B$ is a basis of the space $V$. Let $V,W$ be nonzero spaces over a field $F$ and suppose that a set $B =\lbrace v_1, . . . , v_n \rbrace \subset V$ has the following property:
For any vectors $w_1, . . . ,w_n \in W$, there exists a unique linear transformation $T : V \rightarrow W$ such that $T(v_i) = w_i$  for all $i = 1, . . . , n$.
Prove that $B$ is a basis of the space $V$. 
Can anyone help me with this? I have no idea how to prove this.

Hints:
Suppose we have scalars $\,a_1,\ldots a_n\,$ s.t.
$$a_1v_1+\ldots +a_nv_n=0$$
Now, for $\,1\le i\le n\,$ , choose 
$$w_i\neq 0\;,\;\;w_1=w_2=\;\ldots =w_{i-1}=w_{i+1}=\ldots=w_n=0\;,\;\;$$
It's given there's a unique linear transformation $\,T:V\to W\,$ s.t. $\,Tv_j=w_j\;,\;\;\forall j=1,...,n\,$ , so:
$$0=T(a_1v_1+\ldots+a_nv_n)=\sum_{k=1}^na_kTv_k=a_iv_i=a_iw_i\ldots$$
