# Vector space/Basis/Characterization/Existence/Section

The following theorem gives an important characterization for a family of vectors to be basis.

## Theorem

Let $K$ be a field, and let $V$ be a $K$-vector space. Let $v_{1},\ldots ,v_{n}\in V$
* The family is a basis of $V$.
* The family is a minimal generating system, that is, as soon as we remove one vector $v_{i}$, the remaining family is not a generating system any more.
* For every vector $u\in V$, there is exactly one representation
    $u=s_{1}v_{1}+\cdots +s_{n}v_{n}\,.$
* The family is maximal linearly independent, that is, as soon as some vector is added, the family is not linearly independent any more.

### Proof

Proof by ring closure. $(1)\Rightarrow (2)$. The family is a generating system. Let us remove a vector, say $v_{1}$, from the family. We have to show that the remaining family, that is $v_{2},\ldots ,v_{n}$, is not a generating system anymore. So suppose that it is still a generating system. Then, in particular, $v_{1}$ can be written as a linear combination of the remaining vectors, and we have
    $v_{1}=\sum _{i=2}^{n}s_{i}v_{i}\,.$
But then
    $v_{1}-\sum _{i=2}^{n}s_{i}v_{i}=0\,$
is a nontrivial representation of $0$, contradicting the linear independence of the family. $(2)\Rightarrow (3)$. Due to the condition, the family is a generating system, hence every vector can be presented as a linear combination. Suppose that for some $u\in V$, there is more than one representation, say
    $u=\sum _{i=1}^{n}s_{i}v_{i}=\sum _{i=1}^{n}t_{i}v_{i}\,,$
where at least one coefficient is different. Without loss of generality we may assume $s_{1}\neq t_{1}$. Then we get the relation
    ${\left(s_{1}-t_{1}\right)}v_{1}=\sum _{i=2}^{n}{\left(t_{i}-s_{i}\right)}v_{i}\,$
Because of $s_{1}-t_{1}\neq 0$, we can divide by this number and obtain a representation of $v_{1}$ using the other vectors. In this situation, due to exercise, also the family without $v_{1}$ is a generating system of $V$, contradicting the minimality. $(3)\Rightarrow (4)$. Because of the unique representability, the zero vector has only the trivial representation. This means that the vectors are linearly independent. If we add a vector $u$, then it has a representation
    $u=\sum _{i=1}^{n}s_{i}v_{i}\,$
and therefore
    $0=u-\sum _{i=1}^{n}s_{i}v_{i}\,$
is a non-trivial representation of $0$, so that the extended family $u,v_{1},\ldots ,v_{n}$ is not linearly independent. $(4)\Rightarrow (1)$. The family is linearly independent, we have to show that it is also a generating system. Let $u\in V$. Due to the condition, the family $u,v_{1},\ldots ,v_{n}$ is not linearly independent. This means that there exists a non-trivial representation
    $0=su+\sum _{i=1}^{n}s_{i}v_{i}\,.$
Here $s\neq 0$, because else this would be a non-trivial representation of $0$ with the original family $v_{1},\ldots ,v_{n}$. Hence, we can write
    $u=-\sum _{i=1}^{n}{\frac {s_{i}}{s}}v_{i}\,,$
yielding a representation for $u$.
$\Box$

## Remark

Let a basis ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ of a $K$-vector space $V$ be given. Due to fact, this means that for every vector $u\in V$ there exists a unique representation (a linear combination)
    $u=s_{1}v_{1}+s_{2}v_{2}+\cdots +s_{n}v_{n}\,.$
Here, the uniquely determined elements $s_{i}\in K$ (scalars) are called the coordinates of $u$ with respect to the given basis. This means that for a given basis, there is a correspondence between vectors and coordinate tuples $(s_{1},s_{2},\ldots ,s_{n})\in K^{n}$. We say that a basis determines a linear coordinate system of $V$. To paraphrase, a basis gives in particular a bijective mapping
    $\Psi _{\mathfrak {v}}\colon K^{n}\longrightarrow V,{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}\longmapsto s_{1}v_{1}+s_{2}v_{2}+\cdots +s_{n}v_{n}.$
The inverse mapping
    ${\left(\Psi _{\mathfrak {v}}\right)}^{-1}\colon V\longrightarrow K^{n}$
is also called the coordinate mapping.

## Theorem

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

### Proof

Let $v_{i}$, $i\in I$, be a finite generating system of $V$, with a finite index set $I$. We argue with the characterization from fact. If the family is minimal, then we have a basis. If not, then there exists some $k\in I$, such that the remaining family where $v_{k}$ is removed, that is $v_{i}$, $i\in I\setminus \{k\}$, 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 $J\subseteq I$ such that $v_{i}$, $i\in J$, is a minimal generating set, hence a basis.
$\Box$

## Remark

In general, the Theorem of Hamel says that every vector space has a basis. The proof of this theorem uses strong set-theoretical methods, in particular the axiom of choice and the Lemma of Zorn. This is the reason why many statements about finite-dimensional vector spaces passes over to vector spaces of infinite dimension. In this course we will concentrate on the finite-dimensional case.
* ↑ 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.