Theorem – Uniqueness of coefficients in linear combination.

Theorem

Consider a vector space $V$ over $F$ . Let $\mathbf{v}_1\in{V}$ be a particular vector in $V$ , and let $\mathbf{x}_i$ be in a basis for $V$ for all $\mathbb{N}\cap{[1, n]}$ . Then, the equation

$\sum\limits_{i=1}^{n} \alpha_i \mathbf{x}_i = \mathbf{v}_1$

where the left hand side is a linear combination of the vectors in the basis, has a unique list of coefficients $\alpha_i$ for each element $\mathbf{v}_1\in{V}$ .

Proof

Let us assume that there are two distinct sets of coefficients corresponding to the same vector $\mathbf{v}_1\in{V}$ . I.e, there exist $\{\alpha_1, ..., \alpha_n\}, \{\beta_1, ..., \beta_n\}$ such that

$\sum\limits_{i=1}^{n} \alpha_i \mathbf{x}_i = \sum\limits_{i=1}^{n} \beta_i \mathbf{x}_i = \mathbf{v}_1$

Since the two sums are equal, by axiom 1.4 of a vector space, their difference is the zero vector,

$\begin{aligned} &\sum\limits_{i=1}^{n} \alpha_i \mathbf{x}_i - \sum\limits_{i=1}^{n} \beta_i \mathbf{x}_i = \mathbf{0} \\ \implies & \sum\limits_{i=1}^{n} (\alpha_i - \beta_i)\mathbf{x}_i = \mathbf{0} \end{aligned}$

But we know that the $x_i$ ‘s are a basis for $V$ , so they are linearly independent. It follows then that

$\sum\limits_{i=1}^{n} (\alpha_i - \beta_i)\mathbf{x}_i = \mathbf{0} \implies (\alpha_i - \beta_i) = 0, ~~~\forall{i} = 1, ..., n$

Which contradicts our assumption that the sets $\{\alpha_1, ..., \alpha_n\}, \{\beta_1, ..., \beta_n\}$ are distinct. $\square$