# 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.

Let be a field, and let be a -vector space with a finite generating system. Then any two bases of have the same number of vectors.

### Proof

This theorem enables the following definition.

Let be a
field,
and let be a
-vector space
with a finite
generating system.
Then the number of vectors in any
basis
of , is called the *dimension* of , written

Due to the preceding theorem, the dimension is *well-defined*. If a vector space is not finitely generated, then one puts
.
The null space has dimension . 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.

Let be a field and . Then the standard space has the dimension .

The standard basis , , consists of vectors, hence the dimension is .

The complex numbers form a two-dimensional real vector space, a basis is and .

The polynomial ring over a field 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 polynomials . Let be the maximum of the degrees of these polynomials. Then every -linear combination has at most degree . In particular, polynomials of larger degree can not be presented by , so these do not form a generating system for all polynomials.

Let denote a finite-dimensional vector space over a field . Let denote a linear subspace. Then is also finite-dimensional, and the estimate

holds.

### Proof

Let be a field, and let be a -vector space with finite dimension . Let vectors in be given. Then the following properties are equivalent.

- form a basis of .
- form a generating system of .
- are linearly independent.

### Proof

Let be a field. It is easy to get an overview over the linear subspaces of , as the dimension of a linear subspace equals with , due to fact. For , there is only the null space itself, for there is the null space and itself. For , there is the null space, the whole plane , and the one-dimensional lines through the origin. Every line has the form

with a vector . Two vectors different from define the same line if and only if they are linearly dependent. For , there is the null space, the whole space , the one-dimensional lines through the origin, and the two-dimensional planes through the origin.