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.
 A: Choose $w \in W$, $w \neq 0$. Define $T_k$ by $T_k(v_i) = \delta_{ik} w$. By assumption $T_k$  exists.
We want to show that $v_1,...,v_n$ are linearly independent. So suppose $\sum_i \alpha_i v_i = 0$. Then $T_k(\sum_i \alpha_i v_i) = \alpha_k w = 0$. Hence $\alpha_k = 0$.
Now we need to show that $v_1,...,v_n$ spans $V$. Define the transformation $Z(v_i) = 0$. By assumption, $Z$ is the unique linear transformation mapping all $v_i$ to zero. It follows that $V = \operatorname{sp}\{v_1,...,v_n\}$.
To see this, suppose $x \notin \operatorname{sp}\{v_1,...,v_n\}$. Then we can extend $Z$ to $\operatorname{sp}\{x,v_1,...,v_n\}$: Define $Z_+(v_i) = 0$, $Z_+(x) = w$, $Z_-(v_i) = 0$, $Z_-(x) = -w$. However, by uniqueness, we must have $Z = Z_+ = Z_-$, which is a contradiction. Hence no such $x$ exists.
A: 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:
$$0=T(a_1v_1+\ldots+a_nv_n)=\sum_{k=1}^na_kTv_k=a_iv_i=a_iw_i\ldots$$
