Dimension – Serlo

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


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 have the "obvious dimensions" . The vector space has a basis with elements, for example the standard basis .

Similarly, for any field , we want the vector space to have dimension . Again, with the standard basis we find a basis with exactly vectors.

This suggests to define the dimension of a vector space as the number of vectors of a basis of . 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[Bearbeiten]

Definition (Dimension of a vector space)

Let be a -vector space and let be a basis of . If is finite, we define the dimension of by . (We usually omit the specification of the field if this is obvious from the context). In this case, we say that is finite dimensional. If is infinite instead, we say that is infinite-dimensional and write .

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:

Theorem (Well-definedness of the dimension)

Let be a -vector space and two bases of . If is finite, then is also finite and we have that .

Proof (Uniqueness of the dimension)

Let be finite. Suppose that the basis was infinite. Then, we could choose any -elementary subset of . Note that is linearly independent, as it is a subset of the linearly independent set . This contradicts Steinitz's exchange theorem because of . Thus is finite. It remains to show the equality of cardinalities. Since is linearly independent and is a basis of , it follows from Steinitz's exchange theorem that . Analogously, we can prove . This establishes the theorem.

Examples of dimensions[Bearbeiten]

Dimension of [Bearbeiten]


In this example we want to verify that the dimension of is indeed . A possible basis would be the canonical basis:

This basis is finite and we have that . We get .

Dimension of a polynomial space[Bearbeiten]


The polynomial space over a field is defined as

with coefficient-wise addition and scalar multiplication. From this we see that is a generator of . Since the vector space identifications operate coefficient-wise, this set is also linearly independent. From we finally get .

Dimension of as an -vector space[Bearbeiten]


We would like to determine the dimension of the complex numbers, understood as an -vector space. Each complex number can be written uniquely as , with . From this we see that is a basis of over . So , i.e., as a real vector space, is 2-dimensions. (Note that as a complex vector space, is 1-dimensional).

Dimension of the null space[Bearbeiten]


For every field , the ("trivial") null space is a vector space. To determine its dimension, we need to find a basis. As we have already seen in the article on the null space, the null space is generated by the empty set. Furthermore, is linearly independent by definition and therefore a basis of the null space. Thus we have .

Properties of the dimension[Bearbeiten]

We now prove some properties of the "dimension" notion:


Let be a finite-dimensional -vector space and a subspace. Then we have that:

  1. .
  2. If , then follows.


Let be a basis of . Then is a linearly independent subset of . Therefore, according to the basis completion theorem, there exists a basis of , with . It follows immediately that , so . Now, if in addition is assumed, then we have that . Since is finite, we get . So we can conclude .

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


Let be the polynomial space over a field and the subspace of polynomials without a constant term. One can easily show (as above) that is a basis of . Thus we see . But , since the constant polynomial is .

Dimension formula[Bearbeiten]

Proof of the dimension formula[Bearbeiten]

The following dimensional formula gives how to calculate the dimension of the sum of two finite dimensional subspaces of a -vector space .

Theorem (Dimension formula)

Let be a -vector space and let be finite-dimensional subspaces. Then, we have:

Proof (Dimension formula)

Since are finite dimensional, the spaces are also finite dimensional. set . Then . So there are some , such that and . Furthermore, let be a basis of . Since is a subspace of and of , according to the basis completion theorem there exist vectors and vectors , such that is a basis of and is a basis of .

We now show that is a basis of .

Proof step: First, we show that is a generator.

To do this, we show that any vector can be represented as a linear combination of elements from .

So let . Then there are some with . Since is a linear combination of the basis of , we have

And is a linear combination of the basis of , so


So is linear combination of and a generator of .

Proof step: Now we show the linear independence of .

Let , with

We have to show that for all . Let for this . Then, we have and because of the above condition.

So also , i.e., .

Thus can be represented as a linear combination of the basis of and there exist such that

Now, we further have

Since is a Basis of , it is linearly independent and


Since is a Basis of and hence the vectors are linearly independent, we have that

Thus all coefficients are zero and the vectors are linearly independent. is therefore actually a basis.

Now, we have that


This is the claim to be proven.

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".


Let be a finite-dimensional -vector space and let be subspaces, with . Then, we have:


First, because of , both subspaces are finite-dimensional. Using the dimensional formula, we conclude

As shown in the example above, we have that and get

Exercises: Dimension formula[Bearbeiten]

Exercise (Dimension formula)

Let be a -vector space and let be subspaces of . Further, let , , . What dimension can and have?

How to get to the proof? (Dimension formula)

Bound from above and apply the dimension formula.

Solution (Dimension formula)

We have the dimensional formula

Further, we have that , since is a subspace of . So

Since holds, we also have that . Both results together yield

From the dimensional formula we now conclude

In total we obtain:

Exercise (Dimension formula)

Consider the -vector space and the subspaces , . Show that holds.

Solution (Dimension formula)

We first show . Let for this be . Then there exist with

It follows that and hence . So we have that .

From the dimension formula we now obtain

Using the theorem above about properties of the dimension, it follows that . Together we get .