# 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 $\mathcal B$.

### General Result

Let $V$ be a vector space over a division ring $R$.

Let $B$ be a basis for $V$.

Let $x \in V$.

Then there is a unique finite subset $C$ of $R \times B$ such that:

$\displaystyle x = \sum_{\left({r, v}\right) \mathop \in C} r \cdot v$
$\forall \left({r, v}\right) \in C: r \ne 0_R$

## Proof

### Proof of Existence

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

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

$\Box$

### 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:

 $\displaystyle \sum_{k \mathop = 1}^n \alpha_k \mathbf x_k - \sum_{k \mathop = 1}^n \beta_k \mathbf x_k$ $=$ $\displaystyle \mathbf x - \mathbf x$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle \sum_{k \mathop = 1}^n \left({\alpha_k - \beta_k}\right) \mathbf x_k$ $=$ $\displaystyle \mathbf 0$ $\quad$ $\quad$

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.

$\blacksquare$