Expression of Vector as Linear Combination from Basis is Unique

Theorem
Let $V$ be a vector space of dimension $n$.

Let $\mathcal B = \left\{{\mathbf x_1, \mathbf x_2, \ldots, \mathbf x_n}\right\}$ be a basis for $V$.

Let $\mathbf x \in V$ be any vector of $V$.

Then $\mathbf x$ can be expressed as a unique linear combination of elements of $\mathcal B$.

Proof of Existence
By the definition of basis, $\mathcal B$ is a spanning set.

Hence the result, by the definition of a spanning set.

Proof of Uniqueness
Seeking a contradiction, suppose otherwise, that:
 * $\displaystyle \sum_{k \mathop = 1}^n \alpha_k \mathbf x_k = \mathbf x = \sum_{k \mathop = 1}^n \beta_k \mathbf x_k$

where $\alpha_i \ne \beta_i$ for some $1 \le i \le n$.

Then:

However, we have that $\mathcal B = \left\{{\mathbf x_1, \mathbf x_2, \ldots, \mathbf x_n}\right\}$ is a basis for $V$.

So, by definition, $\mathcal B$ is a linearly independent set.

This means that, for $1 \le i \le n$:


 * $\alpha_i - \beta_i = 0$

and hence $\alpha_i = \beta_i$ for all $1 \le i \le n$.

This contradicts our assumption that $\alpha_i \ne \beta_i$ for some $i$.

Hence the result, from proof by contradiction.