Image of a linear map – Serlo

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The image of a linear map is the set of all vectors in that are "hit by ". This set of vectors forms a subspace of and can be used to make the linear map surjective.


Image of the linear map
Visualization of the linear map

We consider a linear map between two -vector spaces and . A vector is transformed by into a vector . The mapping does not necessarily hit all elements from , because is not necessarily surjective. The mapped vectors form a subset . This set is called image of .

Since is linear, preserves the structure of the vector spaces and . Therefore, we conjecture that maps the vector space into a vector space. Consequently, the image of , i.e., the set should be a subspace of . We will indeed prove this in a theorem below.


Definition (Image of a linear map)

Let and be two -vector spaces and a linear map. Then we call the image of .


In the literature, the notation is also often used instead of for the image of .

In the derivation we already considered that should be a subspace of . We now prove this as a theorem.

Theorem (The image is a subspace)

Let a linear map between the -vector spaces and . Then is a subspace of .

Proof (The image is a subspace)

To show that is a subspace, we need to check the subspace criteria:

  1. For all we have .
  2. For all and for all we have .

Proof step:

For every we have . So .

Proof step:

Since is a linear map, it holds that . Thus .

Proof step: For all we have .

Consider as given. That means, we can choose vectors and from with and . We now show that . To do this, we need to find a vector in that is mapped by to . Now

As and we have that is inside the image of .

Proof step: For all and for all we have .

Let and . Then there is a vector with . We need to show that there is a vector in that is mapped to . It holds:

Now, since we have that .

Image and surjectivity[Bearbeiten]

We already know that a mapping is surjective if and only if the mapping "hits" all elements of . Formally, this means that is surjective if and only if . Now if is a linear map, then is a subspace of . In particular, if is finite-dimensional, then is surjective exactly if .


The identity is a linear map. It is surjective, because every element has the preimage . Hence, we have and in particular .

The map is also linear. Further, each element has a preimage, for example . Thus we have shown and thus, is surjective. In particular .

The embedding is also linear, but not surjective. The vector is not contained in . Thus must hold. And indeed .

Sometimes it is useful to show the surjectivity of by proving .


We consider the linear map and ask if is surjective. We want to answer the question by determining the dimension of and comparing it with . To do this, we first look for linearly independent vectors in the image of . The vectors and are linearly independent. Therefore, . Now from which we get . Thus, we obtain and is surjective.

The relationship between image and generating system[Bearbeiten]

We have seen in the article on epimorphisms, that a linear map preserves generators of if and only if it is surjective. In this case, the image of each generator of generates the entire vector space . In particular, the image of each generator of generates the image of . The last statement holds also for non-surjective linear maps:

Theorem (The image is the span of the images of a generating system)

Let be a linear map between two -vector spaces and . Let be a generator of . Then:

Proof (The image is the span of the images of a generating system)

We show the two inclusions.

Proof step:

Let . Then there are , and coefficients , such that

Since the are in , there exist some with for . Then, because of the linearity of , we have

Proof step:

Let . Then there is a with . Since is a generator of , there are an , and coefficients , such that

Now linearity of finally implies:

Image and linear system [Bearbeiten]

Let be an matrix and . The associated system of linear equations is . We can also interpret the matrix as a linear map . In particular, the image of is a subset of .

If , there is some such that . By definition of we have . Thus, the linear system of equations is solvable. Conversely, if is solvable, then there exists an with . For this , we now have . Thus .

So the image gives us a criterion for the solvability of systems of linear equations: A linear system of equations is solvable if and only if lies in the image of . However, the criterion makes no statement about the uniqueness of solutions. For this, one can use the kernel.


We will now look at how to determine the image of a linear map.


Let us consider the linear map

This is a projection to the axis. Intuitively, then, the image of should be the -axis, i.e.

We now want to prove this:

If , then there exists some with . So .

Conversely, because every vector of the form has a preimage under . So every such vector lies in .

This proves the desired statement.


Let be a field. We consider the linear map

We want to determine the image of . To do this, we exploit the fact that is a basis of , so in particular it is a generator. We have seen in the last section that then .

We can specify this space explicitly by calculating the span:

After considering two examples in finite-dimensional vector spaces, we can venture to an example with an infinite-dimensional vector space. We consider the same function in the examples for determining the kernel of a linear map.


Our goal is to determine the image of the linear map of the derivative of polynomials over . The set is a basis of . The derivative function is defined by for all .

We now want to know whether is surjective. To do this, we note that holds for every . Thus every basis element of is hit. So , and is indeed surjective.

When solving systems of linear equations, we will see many more examples. We will also learn a methodical way of solving for the determination of images.

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link as soon as it is written.

Making linear maps "epic"[Bearbeiten]

We now want to construct a surjective linear map from a given linear map . If we consider to be a mapping of sets, we already know how to accomplish this: We restrict the target set of to and get some restricted mapping . Now, we just need to check that is linear. But this is clear because is a subspace of . So all we need to do to make surjective (i.e., an epi-morphism) is to restrict the objective of to .

This method also gives us an approach for making functions between other structures surjective: We need to check that the restriction on the image preserves the structure. For example, for a group homomorphism we can show that is again a group and is again a group homomorphism.

Outlook: How surjective is a linear map? - The cokernel[Bearbeiten]

In the article about the kernel we see that the kernel "stores" exactly that information which a linear map "eliminates". Further, is injective if and only if and the kernel intuitively represents a "measure of the non-injectivity" of .

We now want to construct a similar measure of the surjectivity of . The image of is not sufficient for this purpose: For example, the images of and are isomorphic, but is surjective and is not. From the image alone, no conclusions can be drawn as of whether is surjective, because surjectivity also depends on the target space . To measure "non-surjectivity," on the other hand, we need a vector space that measures, which part of is not hit by .

The space contains the information, which vectors are hit by . The goal is to "remove this information" from . We have already realized this "removal of information" in the article on the factor space by taking the quotient space . We call this space the cokernel of . It is indeed suitable for characterizing the non-surjectivity of , because is equal to the null space if and only if is surjective: A vector in that is not hit by yields a nontrivial element in and, conversely, a nontrivial element in yields an element in that is not hit by .

The kokernel even measures how non-surjective is exactly: if is larger, more vectors are not hit by . If is finite dimensional, we can measure the size of using the dimension. Thus, is a number we can use to quantify how non-surjective is. However, unlike , this number does not allow us to reconstruct the exact vectors that are not hit by .


Exercise (Associating image spaces to figures)

We consider the following four subspaces from the vector space , given as images of the linear maps

Match these four subspaces to the subspaces shown in the figures below.

Solution (Associating image spaces to figures)

First we look for the image of : To find , we can apply a theorem from above: If is a generator of , then holds. We take the standard basis as the generator of . Then

Now we apply to the standard basis

The vectors generate the image of . Moreover, they are linearly independent and thus a basis of . Therefore . So .

Next, we want to find the image of . However, it is also possible to compute the image directly by definition, which we will demonstrate here.

So the image of is spanned by the vector . Thus .

Now we determine the image of using, for example, the same method as for . That means we apply to the standard basis:

Both vectors are linearly dependent. So it follows that and thus .

Finally, we determine the image of . For this we proceed for example as with .

So the image of is spanned by the vector . Thus is the -axis, so .

Exercise (Surjectivity and dimension of and )

Let and be two finite-dimensional vector spaces. Show that there exists a surjective linear map if and only if .

How to get to the proof? (Surjectivity and dimension of and )

We want to estimate the dimensions of and against each other. The dimension is defined as the cardinality of a basis. That is, if is a basis of and is a basis of , we must show that holds if and only if there exists a surjective linear map. "if and only if" means that we need to establish two directions ().

Given a surjective linear map , we must show that the dimension of is at least . Now bases are maximal linearly independent subsets. That is, to estimate the dimension from below, we need to construct a linearly independent subset with elements. In the figure, we have already a linearly independent subset with elements, which is the basis . Because is surjective, we can lift these to vectors with . Now we need to verify that are linearly independent in . We see this, by converting a linear combination via into a linear combination and exploiting the linear independence of .

Conversely, if holds, we must construct a surjective linear map . Following the principle of linear continuation, we can construct the linear map by specifying how acts on a basis of . For this we need elements of on which we can send . We have already chosen a basis of above. Therefore, it is convenient to define as follows:

Then the image of is spanned by the vectors . However, these vectors also span all of and thus is surjective.

Solution (Surjectivity and dimension of and )

Proof step: ""

Suppose there is a suitable surjective mapping . We show that the dimension of cannot be larger than the dimension of (this is true for any linear map). Because of the surjectivity of , it follows that .

So let be linearly independent. There exists with for . We show that are also linearly independent: Let with . Then we also have that

By linear independence of , it follows that . So are also linearly independent. Overall, we have shown that

In particular, it holds that a basis of (a maximal linearly independent subset of ) must contain at least as many elements as a basis of , that is, .

Proof step: ""

Assume that . We use that a linear map is already uniquely determined by the images of the basis vectors. Let be a basis of and be a basis of . Define the surjective linear map by

This works, since by assumption, holds. The mapping constructed in this way is surjective, since by construction, . As the image of is a subspace of , the subspace generated by these vectors, i.e., , also lies in the image of . Accordingly, holds and is surjective.

Exercise (Image of a matrix)

  1. Consider the matrix and the mapping induced by it. What is the image ?
  2. Now let be any matrix over a field , where denote the columns of . Consider the mapping induced by . Show that holds. So the image of a matrix is the span of its columns.

Solution (Image of a matrix)

Solution sub-exercise 1:

We know that the image of the linear map is a subspace of . Since the -vector space has dimension , a subspace can only have dimension or . In the first case the subspace is the null vector space, in the second case it is already all of . So has only the two subspaces and . Since holds, we have that . Thus, .

Solution sub-exercise 2:

Proof step: ""

Let . Then, there is some with . We can write as . Plugging this into the equation , we get.

Since , we obtain .

Proof step: ""

Let with for . We want to find with . So let us define . The same calculation as in the first step of the proof then shows