Expression of Vector as Linear Combination from Basis is Unique

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Theorem

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

Let $\BB = \set {\mathbf x_1, \mathbf x_2, \ldots, \mathbf x_n}$ 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 $\BB$.


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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:

$\ds x = \sum_{\tuple {r, v} \mathop \in C} r \cdot v$
$\forall \tuple {r, v} \in C: r \ne 0_R$


Proof

Proof of Existence

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

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

$\Box$


Proof of Uniqueness

Aiming for a contradiction, suppose otherwise, that:

$\ds \sum_{k \mathop = 1}^n \alpha_k \mathbf x_k = \mathbf x = \sum_{k \mathop = 1}^n \beta_k \mathbf x_k$

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

Then:

\(\ds \sum_{k \mathop = 1}^n \alpha_k \mathbf x_k - \sum_{k \mathop = 1}^n \beta_k \mathbf x_k\) \(=\) \(\ds \mathbf x - \mathbf x\)
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 1}^n \paren {\alpha_k - \beta_k} \mathbf x_k\) \(=\) \(\ds \mathbf 0\)

However, we have that $\BB = \set {\mathbf x_1, \mathbf x_2, \ldots, \mathbf x_n}$ is a basis for $V$.

So, by definition, $\BB$ 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.

$\blacksquare$


Also see


Sources

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