Are the coefficients of a vector according to a basis unique? If I have a vector space $V$ ( of dimension $n$ ) over real numbers such that $\{v_1,v_2...v_n\}$ is the basis for the space ( not orthogonal ). Then I can write any vector $l$ in this space as $l=\sum_i\alpha_iv_i$. Here $\alpha_1,\alpha_2...\alpha_n$ are the coefficients that define the vector  $l$ according to this basis. Can another set of coefficients $\beta_1,\beta_2...\beta_n$ give the same vector $l$ ? If the basis was orthogonal the answer would be no, but I can't prove for a non orthogonal basis.
 A: They cannot.  Suppose so.  Then $\sum \alpha_i v_i = \sum \beta_i v_i$.  This implies:
$$0 = \sum (\alpha_i - \beta_i) v_i$$
which violates the basis being a set of linearly independent vectors.
A: Here is an alternative approach based on the existence and uniqueness of the matrix inverse for square matrices with full column rank. It is slightly circular in the sense that proofs of the  existence and uniqueness of this matrix inverse usually use arguments similar to the other answers here. However, hopefully this approach will provide some intuition about the meaning of changing from one basis to another.
The operation of taking a vector of coefficients and building a linear combination of vectors from it is a linear operation. In particular, it is equivalent to multiplying the coefficient vector with a matrix where the basis vectors are stacked in each column,
$$\underbrace{\sum_{i=1}^n \alpha_i v_i}_{l} = \underbrace{\begin{bmatrix}v_1 & v_2 & \dots & v_n\end{bmatrix}}_{P}\underbrace{\begin{bmatrix}\alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n \end{bmatrix}}_{\alpha}.$$
More generically, 
$$l = P \alpha,$$
where $l$ is the linear combination, $P$ is the matrix of basis vectors, and $\alpha$ is the vector of coefficients. If the basis vectors are linearly independent, then $P$ is invertible, so the only possible map going the other direction is given by $P^{-1}$;
$$\alpha = P^{-1} l.$$
There can't be other inverse maps by the uniqueness of matrix inverses, which is the desired result.
More generally, if you want to go from a vector written in the standard basis $\{e_i\}_{i=1}^n$ to an alternative basis $\{v_i\}_{i=1}^n$, one multiplies by $P^{-1}$, whereas if you want to go from the $v_i$ basis back to the standard basis you multiply by $P$. 

A: They are unique. Think about what happens if they aren't and you subtract two different representations from each other.
