Serlo: EN: Dimension

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In this article we define the dimension of a vector space and show some elementary properties like the dimension formula.

Motivation

In this article we define the notion of a dimension of a vector space. It would be nice to do it in a way that common vector spaces like $ℝ,{ℝ}^2,{ℝ}^3,...$ have the "obvious dimensions" $1,2,3,...$ . The vector space ${ℝ}^3$ has a basis with $3$ elements, for example the standard basis $B_3=\{e_1,e_2,e_3\}$.

Similarly, for any field $K$, we want the vector space $K^n$ to have dimension ${\mathrm{d}\mathrm{i}\mathrm{m}}_K(K^n)=n$. Again, with the standard basis $B_n=\{e_1,...,e_n\}$ we find a basis with exactly $n$ vectors.

This suggests to define the dimension of a vector space $V$ as the number of vectors of a basis of $V$. At this point, it is not yet clear that every basis has the same cardinality. We will prove this in the following.

Definition of the dimension

From this definition it is not clear that the dimension is independent of the choice of the basis of our vector space. For example, it could happen that a vector space has different bases with different numbers of elements. This does actually never happen, as the next theorem shows:

Properties of the dimension

We now prove some properties of the "dimension" notion: Mathe für Nicht-Freaks: Vorlage:Satz

In order to show that it is important to assume $V$ to be finite-dimensional, consider an example of an infinite-dimensional vector space that has a proper infinite-dimensional subspace:

Dimension formula

Proof of the dimension formula

The following dimensional formula gives how to calculate the dimension of the sum of two finite dimensional subspaces $U,W\subseteq V$ of a $K$-vector space $V$.

Next, we consider a conclusion of the dimension formula that makes a statement about the sum of subspaces (missing). Visually, this means that the complement of a subspace in terms of dimension is the missing "remainder".

Exercises: Dimension formula

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