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Vector space of a linear map – Serlo

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Here, we consider the vector space of linear mappings between two given vector spaces.

The vector space of linear maps

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Forming the vector space

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:

  1. If then also .
  2. 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

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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 .