Basis for Finite Submodule of Function Space

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $A$ be a set.


For each $a \in A$, let $f_a: A \to R$ be defined as:

$\forall x \in A: \map {f_a} x = \begin{cases} 1 & : x = a \\ 0 & : x \ne a \end{cases}$


Then $B = \set {f_a: a \in A}$ is a basis of the Finite Submodule of Function Space $R^{\paren A}$.


Proof

Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of distinct terms of $A$.

Let $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ a sequence of scalars.

Then: $\displaystyle \sum_{k \mathop = 1}^n \lambda_k f_{a_k}$ is the mapping whose value at $a_k$ is $\lambda_k$ and whose value at any $x$ not in $\set {a_1, a_2, \ldots, a_n}$ is zero.

Hence $B$ is a generator of $R^{\left({A}\right)}$ which is linearly independent.

Thus, by definition, $B$ is a basis of $R^{\paren A}$.


If $A = \closedint 1 n$, then $B$ is the standard basis of $R^n$.

$\blacksquare$


Sources