Expression of Vector as Linear Combination from Basis is Unique

Theorem

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

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

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

Aiming for 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:

\(\displaystyle \sum_{k \mathop = 1}^n \alpha_k \mathbf x_k - \sum_{k \mathop = 1}^n \beta_k \mathbf x_k\)

\(=\)

\(\displaystyle \mathbf x - \mathbf x\)

\(\displaystyle \leadsto \ \ \)

\(\displaystyle \sum_{k \mathop = 1}^n \paren {\alpha_k - \beta_k} \mathbf x_k\)

\(=\)

\(\displaystyle \mathbf 0\)

However, we have that $\mathcal B = \set {\mathbf x_1, \mathbf x_2, \ldots, \mathbf x_n}$ 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.

$\blacksquare$

Also see

• Definition:Coordinate Vector

Sources

• 1964: Iain T. Adamson: Introduction to Field Theory ... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 4$. Vector Spaces
• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Vector Spaces: $\S 35$. Coordinates

• For a video presentation of the contents of this page, visit the Khan Academy.

WARNING: This link is broken. Amend the page to use {{KhanAcademySecure}} and check that it links to the appropriate page.