# Vector space/Finitely generated/Dimensiontheory/No proofs/Introduction/Section

A finitely generated vector space has many quite different bases. However, the number of elements in a basis is constant and depends only on the vector space. We will formulate this important property now and take it as the departure for the definition of dimension of a vector space.

## Theorem

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

### Proof

$\Box$
This theorem enables the following definition.

## Definition

Let $K$ be a field, and let $V$ be a $K$-vector space with a finite generating system. Then the number of vectors in any basis of $V$, is called the dimension of $V$, written
    $\dim _{\left(V\right)}.$
Due to the preceding theorem, the dimension is well-defined. If a vector space is not finitely generated, then one puts $\dim _{\left(V\right)}=\infty$. The null space $0$ has dimension $0$. A one-dimensional vector space is called a line, a two-dimensional vector space a plane, a three-dimensional vector space a space (in the strict sense) but every vector space is called a space.

## Corollary

Let $K$ be a field and $n\in \mathbb {N}$. Then the standard space $K^{n}$ has the
$n$

### Proof

The standard basis $e_{i}$, $i=1,\ldots ,n$, consists of $n$ vectors, hence the dimension is $n$.
$\Box$

## Example

The complex numbers form a two-dimensional real vector space, a basis is $1$ and ${\mathrm {i} }$.

## Example

The polynomial ring $R=K[X]$ over a field $K$ is not a finite-dimensional vector space. To see this, we have to show that there is no finite generating system for the polynomial ring. Consider $n$ polynomials $P_{1},\ldots ,P_{n}$. Let $d$ be the maximum of the degrees of these polynomials. Then every $K$-linear combination $\sum _{i=1}^{n}a_{i}P_{i}$ has at most degree $d$. In particular, polynomials of larger degree can not be presented by $P_{1},\ldots ,P_{n}$, so these do not form a generating system for all polynomials.

## Corollary

Let $V$ denote a finite-dimensional vector space over a field $K$. Let $U\subseteq V$ denote a linear subspace. Then $U$ is also finite-dimensional, and the estimate
    $\dim _{\left(U\right)}\leq \dim _{\left(V\right)}\,$

### Proof

$\Box$

## Corollary

Let $K$ be a field, and let $V$ be a $K$-vector space with finite dimension $n=\dim _{\left(V\right)}$. Let $n$ vectors $v_{1},\ldots ,v_{n}$ in $V$ be given. Then the following properties are equivalent.
* $v_{1},\ldots ,v_{n}$ form a basis of $V$.
* $v_{1},\ldots ,v_{n}$ form a generating system of $V$.
* $v_{1},\ldots ,v_{n}$ are linearly independent.

### Proof

$\Box$

## Example

Let $K$ be a field. It is easy to get an overview over the linear subspaces of $K^{n}$, as the dimension of a linear subspace equals $k$ with $0\leq k\leq n$, due to fact. For $n=0$, there is only the null space itself, for $n=1$ there is the null space and $K$ itself. For $n=2$, there is the null space, the whole plane $K^{2}$, and the one-dimensional lines through the origin. Every line $G$ has the form
    $G=Kv={\left\{sv\mid s\in K\right\}}\,,$
with a vector $v\neq 0$. Two vectors different from $0$ define the same line if and only if they are linearly dependent. For $n=3$, there is the null space, the whole space $K^{3}$, the one-dimensional lines through the origin, and the two-dimensional planes through the origin.