Let be a field and two -vector spaces. We want to investigate the set of all -linear mappings from to . We give this set (like "homeomorphism", which is another word for "linear map"). So:
We will now prove that this set is again a -vector space, so it has some very convenient properties. More precisely, is a subspace of the set of functions from to , called .
Theorem
Let be a field and and two -vector spaces. Then the set of linear maps from to is again a -vector space.
How to get to the proof?
It is sufficient to show that is a -subspace of . For this we have to show:
If then also .
If and then also .
Proof
We show that is a subspace of .
Proof step:
The map is -linear, so .
Proof step: If then also .
We have to show that the map
is -linear.
Proof step: Additivity of
Let . Then
Proof step: Homogeneity of
Let and . Then
Proof step: If and then .
Let be a linear map and . We must show that the map
is -linear.
Proof step: Additivity of
Let . Then
Proof step: Homogeneity of
Let and . Then
Therefore, is a subspace of .
The dimension of the vector space of linear mappings
In the following, we calculate the dimension of the vector space of linear maps between two finite-dimensional vector spaces.
Theorem (Dimension of the vector space of linear maps)
Let be a field and two vector spaces with .
Then .
How to get to the proof? (Dimension of the vector space of linear maps)
To find the dimension of , we have to construct a basis of this space.
If we want to define a map from to , we have to define its image in for different basis vectors of . This can again be represented as a linear combination of the different basis vectors of .
Proof (Dimension of the vector space of linear maps)
At first, we chose two bases:
Let be a basis of ,
Let be a basis of .
For each we want to define a linear map . By the principle of linear continuation, we can define these uniquely by their values on the basis :
Let .
This set has elements.
To prove the statement of the theorem, we must therefore justify that is a basis of .
To do sop, we have to show that is linearly independent and a generating system.
Proof step: is linearly independent
Let , such that .
We must now show that for all .
Let .
Then:
As the form a basis of , they are linearly independent. Therefore, already follows for all and our fixed .
Since this was chosen arbitrarily, now follows for all .
The are therefore linearly independent.
Proof step: is a generating system
Let .
For every we have .
Since is a basis of , there is a unique decomposition such that .
We will now show that
Due to the linearity, this can be verified on the basis vectors . Let be arbitrary. Then,
Therefore, is a generatin system.
This proves the statement of the theorem.
Hint
A similar statement also applies to infinite-dimensional vector spaces:
If or , and ,
then we also have .
However, if and is arbitrary (or vice versa), then we always get .