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 $V$.

Let $\mathbf x$ be expressed as a linear combination of elements of $\mathcal B$.

Then such an expression is unique.

Proof
Suppose otherwise, that:
 * $\displaystyle \sum_{k=0}^n \alpha_k \mathbf x_k = \mathbf x = \sum_{k=0}^n \beta_k \mathbf x_k$

Then:
 * $\displaystyle \sum_{k=0}^n \left({\alpha_k - \beta_k}\right) \mathbf x_k = \mathbf 0$

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:
 * $\forall k: \alpha_k - \beta_k = 0$

and so the expression for $\mathbf x$ is unique.