Dimension of Vector Space on Cartesian Product

Theorem

Let $\struct {K, +, \circ}$ be a division ring.

Let $n \in \N_{>0}$.

Let $\mathbf V := \struct {K^n, +, \times}_K$ be the $K$-vector space $K^n$.

Then the dimension of $\mathbf V$ is $n$.

Let the unity of $K$ be $1$, and the zero of $K$ be $0$.

Consider the vectors:

$\ds \mathbf e_1$

$:=$

$\ds \underbrace {\tuple {1, 0, \ldots, 0} }_{n \text { coordinates} }$

$\ds \mathbf e_2$

$:=$

$\ds \underbrace {\tuple {0, 1, \ldots, 0} }_{n \text { coordinates} }$

$\ds $

$\vdots$

$\ds $

$\ds \mathbf e_n$

$:=$

$\ds \underbrace {\tuple {0, 0, \ldots, 1} }_{n \text { coordinates} }$

Thus $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $\mathbf V$.

From Standard Ordered Basis is Basis, $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is a basis of $\mathbf V$.

$\blacksquare$

1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Vector Spaces: $\S 34$. Dimension: Example $66$