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

Proof

This proof was not presented in the lecture.

$\Box$

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 dimension ${}n$.

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

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

holds.

Proof

This proof was not presented in the lecture.

$\Box$

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.

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

Proof

See exercise.

$\Box$

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

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.