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.
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.
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 .
Proof (The image is the span of the images of a generating system)
We show the two inclusions.
Then there are , and coefficients , such that
Since the are in ,
there exist some with for .
Then, because of the linearity of , we have
Then there is a with .
Since is a generator of ,
there are an , and coefficients , such that
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 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 .
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.
: The subspace spanned by in
: The line spanned by
: A plane covering all of the two-dimensional space
: A line spanned by
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)
Consider the matrix and the mapping induced by it. What is the image ?
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
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