Proofs for linear maps – Serlo

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We will give here a proof structure that shows how to prove linearity of a map.

General procedure[Bearbeiten]

Recap: Definition of a linear map[Bearbeiten]

We recall that a linear map (or homomorphism) is a structure-preserving map of a -vector space into a -vector space . That is, for the map , the following two conditions must hold:

  1. must be additive, i.e., for we have that:
  2. must be homogeneous, i.e., for we have that: .

So for a linear map it doesn't matter if we first do the addition or scalar multiplication in the vector space and then map the sum into the vector space , or first map the vectors into the vector space and perform the addition or scalar multiplication there, using the images of the map.

Proving that a map is linear[Bearbeiten]

The proof that a map is linear can be done according to the following structure. First, we assume that a map is given between vector spaces. That is, and are -vector spaces and is well-defined. Then for the linearity of we have to show:

  1. additivity:
  2. homogeneity:

Exercise (Introductory example)

We consider the following map

and show that it is linear.

Proof (Introductory example)

First, and are vector spaces over the field . Moreover, the map is well-defined.

Proof step: proving additivity

Let .

Thus we have proved the additivity of .

Proof step: proving homogeneity

Let and . Then, we have

Thus we have proved the homogeneity of .

The map to zero[Bearbeiten]

The map to zero is the map which sends every vector to zero. For instance, the map to zero of to looks as follows:

Exercise (The map to zero linear)

Show that the map is linear.

Proof (The map to zero linear)

We already know that and are both -vector spaces and that the map to zero is well-defined.

Proof step: additivity

For all we have that

Proof step: homogeneity

For all we have that

Thus, the map to zero is linear.

An example in [Bearbeiten]

We consider an example for a linear map of to :

with

Exercise (Linearity of )

Show that the map is linear.

Proof (Linearity of )

is an -vector space. In addition, the map is well-defined.

Proof step: additivity

Let and be any vectors from the plane . Then, we have:


Proof step: homogeneity

Let and . Then:

Thus the map is linear.

A linear map in the vector space of sequences[Bearbeiten]

Next, we consider the space of all sequences of real numbers. This space is infinite-dimensional, because there are not finitely many sequences generating this sequence space. But it is a vector space, as we have shown in the chapter about sequence spaces.

Exercise (Sequence space)

Let be the -vector space of all real-valued sequences. Show that the map

is linear.

How to get to the proof? (Sequence space)

To show linearity, two properties need to be checked:

  1. is additive: for all
  2. is homogeneous: for all and

The vectors and are sequences of real numbers, i.e. they are of the form and with for all .

Proof (Sequence space)

Proof step: additivity

Let and . Then, we have

It follows that is additive.

Proof step: homogeneity

Let and . Then, we have

So is homogeneous.

Thus it was proved that is a -linear map.

Abstract example[Bearbeiten]

In this chapter, we deal with somewhat more abstract vectors. Let be arbitrary sets; a field and a -vector space. We now consider the set of all maps/ functions of the set into the vector space and denote this set with . Furthermore, we also consider the set of all maps of the set into the vector space and denote this set with . The addition of two maps is defined for by

Die scalar multiplication is defined for via

Analogously, we define addition scalar multiplication for .

Exercise (The set is a -vector space)

Show that is a -vector space.

How to get to the proof? (The set is a -vector space)

Simply check the vector space axioms.

We now consider a function/ map

where , so is arbitrary.

Exercise (The map is linear.)

Show that:

is a linear map.

It is important that you exactly follow the definitions. Note that is a map that assigns to every map of to a map of to . These maps, which are elements of and respectively, need not themselves be linear, since there is no vector space structure on the sets and .

Summary of proof (The map is linear.)

To prove the linearity of , we need to check the two properties again:

  1. is additive: for all
  2. is homogeneous: for all and

So at both points an equivalence of maps is to be shown. For this we evaluate the maps at every m element .

Proof (The map is linear.)

Let .

Proof step: additivity

For all we have that

Thus we have shown , i.e., is additive.

Let and .

Proof step: homogeneity

For all we have that

Thus we have shown , i.e., is homogeneous.

Now, additivity and homogeneity of implies that is a linear map.