Expression of Vector as Linear Combination from Basis is Unique
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 $\BB$.
The definition we have of linear combination doesn't really work for this. We probably need to expand that definition.
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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
- 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.