Expression of Vector as Linear Combination from Basis is Unique
Contents
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 $\mathcal B$.
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:
- $\displaystyle x = \sum_{\left({r, v}\right) \mathop \in C} r \cdot v$
- $\forall \left({r, v}\right) \in C: r \ne 0_R$
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.
$\blacksquare$
Also see
Sources
- 1964: Iain T. Adamson: Introduction to Field Theory ... (previous) ... (next): $\S 1.4$
- For a video presentation of the contents of this page, visit the Khan Academy.
