Standard Ordered Basis is Basis

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Theorem

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

Let $n$ be a positive integer.

For each $j \in \closedint 1 n$, let $e_j$ be the ordered $n$-tuple of elements of $R$ whose $j$th entry is $1_R$ and all of whose other entries is $0_R$.


Then $\sequence {e_n}$ is an ordered basis of the $R$-module $R^n$.


This ordered basis is called the standard ordered basis of $R^n$.

The corresponding set $\set {e_1, e_2, \ldots, e_n}$ is called the standard basis of $R^n$.


Proof

\(\ds \sum_{k \mathop = 1}^n \lambda_k e_k\) \(=\) \(\ds \lambda_1 \tuple {1_R, 0_R, 0_R, \ldots, 0_R}\)
\(\ds \) \(+\) \(\ds \lambda_2 \tuple {0_R, 1_R, 0_R, \ldots, 0_R}\)
\(\ds \) \(+\) \(\ds \ldots\)
\(\ds \) \(+\) \(\ds \lambda_n \tuple {0_R, 0_R, 0_R, \ldots, 1_R}\)
\(\ds \) \(=\) \(\ds \tuple {\lambda_1, \lambda_2, \lambda_3, \ldots, \lambda_n}\)

$\blacksquare$


Also see


Sources

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