Question regarding basis in vector spaces How can one prove the following proposition ? 

$
B = (e_{1,...,} e_n )\, $ forms a basis for a space $V$ if and only if each vector of $V$ can only be written as an unique linear combination of elements from $B$ .

I'm really confused here, any ideas?
 A: If $(e_1,\dots,e_n)$ is a base then $x\in V\implies x=a_1e_1+\cdots+a_ne_n$ for some reals $a_i$. If there was other $a'_i$ with $x=a'_1e_1+\cdots+a'_ne_n$ then 
$(a_1-a'_1)e_1+\cdots(a_n-a'_n)e_n=0$, so $a_i=a'_i$.
The other way is also simple.
A: For the first part, suppose $B$ is a basis. By definition a basis is a set of linearly independent vectors that spans $V$. Take then $v\in V$: $v$ can be written as a linear combination of elements of $B$ in this way $v=\sum_{i=1}^n\alpha_ie_i$. The problem here is to prove that this way is unique. Suppose then that it can be written with other scalars like $v=\sum_{i=1}^n\beta_ie_i$. Then, $$0=v-v=\sum_{i=1}^n\alpha_ie_i-\sum_{i=1}^n\beta_ie_i=\sum_{i=1}^n(\alpha_i-\beta_i)e_i$$ but, given that the $e_i's$ are linealy independent, $(\alpha_i-\beta_i)=0$ for each $i=1,...,n$. Then $\alpha_i=\beta_i$ for each $i$ and the linear combination is unique.

For the converse, it is clear that the elements of $B$ span the whole space. To prove that these elements are in fact l.i, consider the linear combination $\sum_{i=1}^n0\cdot e_i=0$. That is a way to write $0$ as linear combination of element of $B$, and the hypothesis says that this way is unique, then in another linear combination of the form $\sum_{i=1}^n\alpha_ie_i=0$, each $\alpha_i=0$, and this is the definition of linear independence.
