# Vector space/Basis/Introduction/Section

## Definition

Let ${\displaystyle {}K}$ be a field, and let ${\displaystyle {}V}$ be a ${\displaystyle {}K}$-vector space. Then a linearly independent generating system ${\displaystyle {}v_{i}\in V}$, ${\displaystyle {}i\in I}$,

of ${\displaystyle {}V}$ is called a basis of ${\displaystyle {}V}$.

## Example

The standard vectors in ${\displaystyle {}K^{n}}$ form a basis. The linear independence was shown in example. To show that they also form a generating system, let

${\displaystyle {}v={\begin{pmatrix}b_{1}\\b_{2}\\\vdots \\b_{n}\end{pmatrix}}\in K^{n}\,}$

be an arbitrary vector. Then we have immediately

${\displaystyle {}v=\sum _{i=1}^{n}b_{i}e_{i}\,.}$

Hence, we have a basis, which is called the standard basis of ${\displaystyle {}K^{n}}$.

## Theorem

Let ${\displaystyle {}K}$ be a field, and let ${\displaystyle {}V}$ be a ${\displaystyle {}K}$-vector space. Let ${\displaystyle {}v_{1},\ldots ,v_{n}\in V}$

be a family of vectors. Then the following statements are equivalent.
1. The family is a basis of ${\displaystyle {}V}$.
2. The family is a minimal generating system, that is, as soon as we remove one vector ${\displaystyle {}v_{i}}$, the remaining family is not a generating system any more.
3. For every vector ${\displaystyle {}u\in V}$, there is exactly one representation
${\displaystyle {}u=s_{1}v_{1}+\cdots +s_{n}v_{n}\,.}$
4. The family is maximal linearly independent, that is, as soon as some vector is added, the family is not linearly independent any more.

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

## Remark

Let a basis ${\displaystyle {}v_{1},\ldots ,v_{n}}$ of a ${\displaystyle {}K}$-vector space ${\displaystyle {}V}$ be given. Due to fact, this means that for every vector ${\displaystyle {}u\in V}$, there exists a uniquely determined representation

${\displaystyle {}u=s_{1}v_{1}+s_{2}v_{2}+\cdots +s_{n}v_{n}\,.}$

The elements ${\displaystyle {}s_{i}\in K}$ (scalars) are called the coordinates of ${\displaystyle {}u}$ with respect to the given basis. Thus, for a fixed basis, we have a (bijective) correspondence between the vectors from ${\displaystyle {}V}$, and the coordinate tuples ${\displaystyle {}(s_{1},s_{2},\ldots ,s_{n})\in K^{n}}$. We express this by saying that a basis determines a linear coordinate system.[1]

## Theorem

Let ${\displaystyle {}K}$ be a field, and let ${\displaystyle {}V}$ be a ${\displaystyle {}K}$-vector space with a finite generating system. Then ${\displaystyle {}V}$ has a finite basis.

### Proof

Let ${\displaystyle {}v_{i}}$, ${\displaystyle {}i\in I}$, be a finite generating system of ${\displaystyle {}V}$, with a finite index set ${\displaystyle {}I}$. We argue with the characterization from fact. If the family is minimal, then we have a basis. If not, then there exists some ${\displaystyle {}k\in I}$, such that the remaining family where ${\displaystyle {}v_{k}}$ is removed, that is ${\displaystyle {}v_{i}}$, ${\displaystyle {}i\in I\setminus \{k\}}$, is also a generating system. In this case, we can go on with this smaller index set. With this method, we arrive at a subset ${\displaystyle {}J\subseteq I}$ such that ${\displaystyle {}v_{i}}$, ${\displaystyle {}i\in J}$, is a minimal generating set, hence a basis.

${\displaystyle \Box }$
1. Linear coordinates give a bijective relation between points and number tuples. Due to linearity, such a bijection respects the addition and the scalar multiplication. In many different contexts, also nonlinear (curvilinear) coordinates are important. These put points of a space and number tuples into a bijective relation. Examples are polar coordinates, cylindrical coordinates and spherical coordinates. By choosing suitable coordinates, mathematical problems, like the computation of volumes, can be simplified.