# Properties of linear maps – Serlo

We consider some properties of linear maps between vector spaces.

## Overview[Bearbeiten]

A linear map, also called vector space homomorphism, preserves the structure of the vector space. This is shown in the following properties of a linear mapping :

- The zero vector is mapped to the zero vector: .
- Inverses are mapped to inverses: .
- Linear combinations are mapped to linear combinations.
- Compositions of linear maps are again linear
- Images of subspaces are subspaces
- The image of a span is the span of the individual image vectors: ( is supposed to be an arbitrary set)

## Zero vector is mapped to the zero vector[Bearbeiten]

The zero vector / origin has a central meaning in our view of vector spaces. And indeed, the origin is sent to the origin by any linear map. Mathematically, "origin" means the neutral element of addition.

And the -property can be shown as a mathematical theorem:

**Theorem** (Zero vector is mapped to the zero vector)

Any linear map between two -vector spaces maps the neutral element of to the neutral element in . Formally, this means .

**How to get to the proof?** (Zero vector is mapped to the zero vector)

We first start with the vector , which is the neutral element of addition of the vector space . Thus, it does not change a vector to which it is added. Therefore we have that in particular .

We need the additivity of a linear map. Thus, follows.

Using these two properties, we get . This equation is satisfied only of . We can subtract on both sides and get .

**Proof** (Zero vector is mapped to the zero vector)

We have that

So we have that

Now, let us add on both sides:

Hence, we have that .

## Inverses are mapped to inverses[Bearbeiten]

Another important structure of vector space is that there is an additive inverse to every element . We now want to show that inverses are preserved by linear maps.

**Theorem** (Inverses are mapped to inverses)

Each linear map sends the inverse of an element to the inverse of the image of the element. Or within one formula, for all in we have that .

**How to get to the proof?** (Inverses are mapped to inverses)

Our goal is to show that . We already know that in a vector space the inverse of any vector is given by . Likewise we have that for any vector that . This simplifies the statement to be shown to .

We can transform the expression into using the homogeneity of the linear map .

**Proof** (Inverses are mapped to inverses)

Let be any element of the vector space .

Thus we have that indeed .

above, we have used that for and for all . This holds in every vector space. You can find the proof here (link missing).

**Alternative proof** (Inverses are mapped to inverses)

Let again be any element of the vector space . Our goal is to show that holds. Let's start with a statement that we know is true: .

The addition of an element with its inverse always gives . So we have that . Now, we show that also holds. Since we are working with maps, we should use their properties. For instance, additivity:

thus, . It follows that is the additive inverse of with respect to . Or, in other words, .

The statement of this theorem also holds in any abelian group. However, scalar multiplication does not exist there. Hence, the alternative (2nd) version of the proof must be used in this case.

## Linear combinations are mapped to linear combinations[Bearbeiten]

Linear mappings preserve the structure of a linear combination and thus map linear combinations in the domain of definition to their corresponding linear combinations in the range of values:

**Theorem** (Linear combinations are mapped to linear combinations)

A map between two -vector spaces and is a linear map if and only if it preserves linear combinations. That is, every linear map sends the linear combination of elements to the linear combination of the images of the elements. Put in a formula, this means that for finitely many and we have that:

**How to get to the proof?** (Linear combinations are mapped to linear combinations)

We want to show that for all and we have that: is a linear map.

We know from the definition of the linear map that additivity and homogeneity hold, and we make use of then.

For the direction "left to right" within the proof we choose two linear combinations in such a way that we get the two properties by inserting them into the above formula.

For the "right to left" direction we know that is a linear map. We can show by induction that above formula holds true for all elements. This way, we reduce the linear combination to single addition and scalar multiplication, to which we can apply additivity and homogeneity.

**Proof** (Linear combinations are mapped to linear combinations)

**Proof step:** is a linear map.

Let and . The terms and are two linear combinations in . If we plug them into the formula , we obtain

So is by definition a linear map.

**Proof step:** is a linear map .

Let be a linear map. We prove this equation by induction in :

**Theorem whose validity shall be proven for the :**

**1. Base case:**

We start with the induction with and find that the property of homogeneity is sufficient:

**1. inductive step:**

**2a. inductive hypothesis:**

**2b. induction theorem:**

**2b. proof of induction step:**

Let and . Then

## Compositions of linear maps are again linear[Bearbeiten]

Let us take two linear maps and . Both are compatible with the vector space structure and preserve linear combinations. This preservation should also hold for the consecutive execution of both maps with . This is mathematically established by the following theorem:

**Theorem** (Composition of linear maps)

Let and be two linear maps between the -vector spaces , and . Then the composition of these two maps with for is also a linear map.

**How to get to the proof?** (Composition of linear maps)

We know that the composition (missing) of two maps is again a well-defined map. So we just need to show that is linear. To do this, we need to prove that satisfies additivity and homogeneity.

- For all we have that and
- For all and we have that .

To prove this, we exploit the additivity and homogeneity of the individual maps and .

**Proof** (Composition of linear maps)

Let first be any two vectors. We have that

For the proof of homogeneity we choose any and any :

## Subspaces are mapped to subspaces[Bearbeiten]

That linear maps preserve the vector space structure can also be seen in the following property: The images of subspaces of a linear map are again subspaces.

**Theorem** (Subspaces are mapped to subspaces)

Let be a linear map between two -vector spaces and . Then the image of every subspace is again a subspace of .

**Proof** (Subspaces are mapped to subspaces)

Let be a subspace of . The image is the set of all function values of arguments from and thus a subset of the range of values . To show that is a subspace, the following criteria must be shown:

- For all we have that .
- For all and for all we have that .

**Proof step:**

Since is a subspace of we have . With , we have at least one element in and so .

**Proof step:** For all we have that

Let us take any two vectors . Because these vectors lie in the image, there are at least two vectors with and . Now

Thus is the image of (the vector is mapped to ). Because is a subspace of , we have that and thus lies in .

**Proof step:** For all and for all we have that

Let and . Since lies within the image of , we can find a with . Now

Thus is the image of (the vector is mapped to ). Because is a subspace of , we have that and thus lies in .

**Hint**

The above theorem also proves that **the image of a linear map is always a vector space**. This follows from the fact that the vector space is also a subspace of itself. According to the theorem above is a subspace of .

## Spans are mapped to spans[Bearbeiten]

Now suppose we have a subset . For this subset, it does not matter whether we first calculate the span and then apply the map or vice versa. This is content of the following theorem:

**Theorem** (Spans are mapped to spans)

Let be any subset (not necessarily a subspace!) of the vector space . Then we have that for the span of :

**How to get to the proof?** (Spans are mapped to spans)

Since we want to show the equality of two sets, we must show that the sets are contained in the other. Once we have shown this, it follows that the two sets are equal.

To show that , we first choose an arbitrary vector . Because this is in the span, it can be written as a linear combination of elements from the set :

where .

We then apply the linear map to both and the linear combination of elements from . This yields the following expression:

Then, using the properties of linear maps, we reshape the expression:

Since the right-hand side is contained in , we have that .

Similarly, we show that . Now we need to prove that for any there exists a vector with . We know

with . Using the linearity of , we can transform the expression to:

But the right hand side is in and thus we have that the same holds for .

**Proof** (Spans are mapped to spans)

**Proof step:**

We first take any vector , for which we have that: .

Since we know that the vector is in the span of , there are coefficients and vectors , such that:

If we now apply our linear map to this expression, we get:

We now further transform the right-hand side of this expression using the properties of linear maps:

On the right-hand side we now have .

So we have shown that and thus .

**Proof step:**

We again choose an arbitrary vector . Since it is in the span of the map, we can write it as:

We can now start transforming the expression in the same way as in the previous proof step:

On the right-hand side there is now a vector in . We have thus shown that holds.

Moreover, we have shown the equality of the sets and , as both are contained within each other.