linear combination of given basis vectors

Show that in an $n$-dimensional vector space $X$, the representation of any $x$ as a linear combination of a given basis $e_1,\ldots,e_n$ is unique.

I only know that this can be proved by contradiction, I think, where we assume the basis $e_1,\ldots,e_n$ is not unique.

A basis for a vector space $X$ is by definition a linearly independent set of vectors which spans $X$.
Let $x$ be any vector in $X$. Since the basis $\{e_1,\ldots,e_n\}$ spans $X$, there exists a linear combination $a_1 e_1 + \ldots + a_n e_n$ equal to $x$.
To show that this linear combination is unique, suppose that we have another linear combination $b_1 e_1 + \ldots + b_n e_n$ also equal to $x$. Then $$a_1 e_1 + \ldots + a_n e_n = b_1 e_1 + \ldots + b_n e_n$$ Therefore, $$(a_1 - b_1) e_1 + \ldots + (a_n - b_n) e_n = 0$$ Since $\{e_1,\ldots,e_n\}$ is linearly independent, this means that all of the $a_i - b_i$ must be zero. Equivalently, $a_i = b_i$ for every $i$.
• @user175418: It is part of the definition of a basis. By definition, a basis for a vector space $X$ is (1) a linearly independent set of vectors which (2) spans $X$. – Bungo Sep 11 '14 at 23:43
Suppose $$\alpha_1\cdot e_1+\dotsb+\alpha_n\cdot e_n=\beta_1\cdot e_1+\dotsb+\beta_n\cdot e_n$$ Then $$(\alpha_1-\beta_1)\cdot e_1+\dotsb+(\alpha_n-\beta_n)\cdot e_n=\mathbf 0$$ What does this equation tell us about the $\alpha_i$'s and $\beta_i$'s?