# Vector space/Finitely generated/Dimension/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 ${\displaystyle {}K}$ be a field, and let ${\displaystyle {}V}$ be a ${\displaystyle {}K}$-vector space with a finite generating system. Then any two bases of ${\displaystyle {}V}$ have the same number of vectors.

### Proof

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

This theorem enables the following definition.

## Definition

Let ${\displaystyle {}K}$ be a field, and let ${\displaystyle {}V}$ be a ${\displaystyle {}K}$-vector space with a finite generating system. Then the number of vectors in any basis of ${\displaystyle {}V}$, is called the dimension of ${\displaystyle {}V}$, written

${\displaystyle \dim _{}{\left(V\right)}.}$

Due to the preceding theorem, the dimension is well-defined. If a vector space is not finitely generated, then one puts ${\displaystyle {}\dim _{}{\left(V\right)}=\infty }$. The null space ${\displaystyle {}0}$ has dimension ${\displaystyle {}0}$. 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 ${\displaystyle {}K}$ be a field and ${\displaystyle {}n\in \mathbb {N} }$. Then the standard space ${\displaystyle {}K^{n}}$ has the dimension ${\displaystyle {}n}$.

### Proof

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

${\displaystyle \Box }$

## Example

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

## Example

The polynomial ring ${\displaystyle {}R=K[X]}$ over a field ${\displaystyle {}K}$ 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 ${\displaystyle {}n}$ polynomials ${\displaystyle {}P_{1},\ldots ,P_{n}}$. Let ${\displaystyle {}d}$ be the maximum of the degrees of these polynomials. Then every ${\displaystyle {}K}$-linear combination ${\displaystyle {}\sum _{i=1}^{n}a_{i}P_{i}}$ has at most degree ${\displaystyle {}d}$. In particular, polynomials of larger degree can not be presented by ${\displaystyle {}P_{1},\ldots ,P_{n}}$, so these do not form a generating system for all polynomials.

## Corollary

Let ${\displaystyle {}V}$ denote a finite-dimensional vector space over a field ${\displaystyle {}K}$. Let ${\displaystyle {}U\subseteq V}$ denote a linear subspace. Then ${\displaystyle {}U}$ is also finite-dimensional, and the estimate

${\displaystyle {}\dim _{}{\left(U\right)}\leq \dim _{}{\left(V\right)}\,}$

holds.

### Proof

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

## Corollary

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

1. ${\displaystyle {}v_{1},\ldots ,v_{n}}$ form a basis of ${\displaystyle {}V}$.
2. ${\displaystyle {}v_{1},\ldots ,v_{n}}$ form a generating system of ${\displaystyle {}V}$.
3. ${\displaystyle {}v_{1},\ldots ,v_{n}}$ are linearly independent.

### Proof

${\displaystyle \Box }$

## Example

Let ${\displaystyle {}K}$ be a field. It is easy to get an overview over the linear subspaces of ${\displaystyle {}K^{n}}$, as the dimension of a linear subspace equals ${\displaystyle {}k}$ with ${\displaystyle {}0\leq k\leq n}$, due to fact. For ${\displaystyle {}n=0}$, there is only the null space itself, for ${\displaystyle {}n=1}$ there is the null space and ${\displaystyle {}K}$ itself. For ${\displaystyle {}n=2}$, there is the null space, the whole plane ${\displaystyle {}K^{2}}$, and the one-dimensional lines through the origin. Every line ${\displaystyle {}G}$ has the form

${\displaystyle {}G=Kv={\left\{sv\mid s\in K\right\}}\,,}$

with a vector ${\displaystyle {}v\neq 0}$. Two vectors different from ${\displaystyle {}0}$ define the same line if and only if they are linearly dependent. For ${\displaystyle {}n=3}$, there is the null space, the whole space ${\displaystyle {}K^{3}}$, the one-dimensional lines through the origin, and the two-dimensional planes through the origin.