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Here, we consider the vector space of linear mappings between two given vector spaces.
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
.
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: 
Proof step: If
then also
.
We have to show that the map
is
-linear.
Proof step: Additivity of 
Let
. Then
Proof step: Homogeneity of 
Therefore,
is a subspace of
.
In the following, we calculate the dimension of the vector space
of linear maps between two finite-dimensional vector spaces.
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.
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