Let be a field. We now consider as a vector space over itself.
From school we already know the vector space over the field . The vectors in have the form with . We can consider the vectors in a 3-dimensional coordinate system. Since is a vector space, we can add and scale vectors.
We also know the vector space . The vectors in have the form with . We can get from by deleting one of the coordinates (e.g., the last one). Illustratively, we then go from the 3-dimensional coordinate system to the plane. So when omitting a coordinate from , the vector space structure is conserved. What happens if we delete another coordinate?
For example, if we omit the second coordinate of , only remains and we get an element in . Illustratively, we thus go from the plane to the axis. Again, when deleting a coordinate, the vector space structure should not be broken.
We can add and scale the elements in (just like vectors), because for all we have and for all and it holds that .
Addition of the vectors and on the real line
Scalar multiplication of the vector with the scalar on the real line
Now our field should be an -vector space.
Visually, this vector space is the number line.
We can apply this idea to an arbitrary field , since also in an arbitrary we can add elements and multiply them by scalars in . Therefore, we conjecture that is a -vector space.
Definition of the vector space structure[Bearbeiten]
Let be a field.
Then we can define an addition and a scalar multiplication.
Definition (Vector space structure on )
We define an addition on by
Similarly, we define a scalar multiplication via
The field is a vector space over itself[Bearbeiten]
Theorem ( is a vector space)
is a -vector space .
How to get to the proof? ( is a vector space)
We proceed as in the article Proofs for vector spaces.
Proof ( is a vector space)
So now we have to establish the eight Vector space axioms.
Proof step: Associativity of addition
This shows the associativity of the addition.
Proof step: Commutativity of addition
This shows the commutativity of the addition.
Proof step: Neutral element of addition
We now have to show that there is a neutral element with respect to , that is, for all . It is natural to use the zero element of the field as the neutral element.
Thus we have shown that is the neutral element of the addition.
In the following we will therefore simply write for the neutral element.
Proof step: Inverse with respect to addition
We have to show that there is a such that .
It is natural to choose as the inverse in with respect to , i.e., we choose
Thus we have shown that for any there is a with .
Proof step: Scalar distributive law
Let and .
Thus the scalar distributive law is shown.
Proof step: Vectorial distributive law
Let and .
Thus the vectorial distributive law is shown.
Proof step: Assoziativität bezüglich Multiplikation
Let and .
This shows the associative law for multiplication.
Proof step: Unitary law
Thus we have shown the unitary law.
With this we have established all eight vector space axioms and thus is a vector space.