Monomorphisms – Serlo

Linear maps preserve linear combinations. We now learn about special linear maps that preserve linear independence. These are called monomorphisms.

Motivation Bearbeiten

We have introduced linear maps as functions between vector spaces that preserve linear combinations. Thus, they satisfy the property that a linear combination is preserved nuder the mapping:

 

Using linear combinations, we have defined the property of linear independence. Recall: For a vector space   over a field  , a finite set of vectors   is linearly independent if and only if the only linear combination by  , which leads to zero ( ) is the trivial one, i.e.,  .

An alternative characterization is that if

 

then the set of coefficients   must be equal as  .

Is this property preserved? Certainly, there are linear maps, which do not preserve linear independence, e.g. the map to zero:  . Any set of vectors containing the zero vector is linearly dependent, as there is a non-trivial linear combination leading to the zero vector, e.g. with  :  . Now are there even linear maps which preserve linear independence?

The answer is: yes and they are called monomorphisms.

What additional property does a linear map need to have in order to preserve linear independence? We take some linearly independent vectors  . For a linear map   to preserve linear independence, it needs to satisfy:

 

We transform:

 

Therefore   must have the following property to preserve linear independence:

 

By setting   and  , it becomes clearer what this property is. We get that

 

for all   which can be written as linear combination of  .

This statement should be valid for all linear independent sets and therefore also for bases. In the case of a basis, however, all   can be written as such a linear combination, which means that   must be injective. Thus injectivity is a necessary condition for a linear map to preserve linear independence.

Is injectivity also a sufficient condition for this property? Let for this   an injective linear map and   linearly independent vectors. We are to find out whether   are also linearly independent. According to our considerations above, it is enough to show the following for scalars   and  

 

Let

 

Then, we have from the injectivity of   that

 .

Because   are linearly independent, we have that   for all  . Thus we have shown the above statement and   preserves linear independence.

Thus, a linear map preserve linear independence if and only if it is injective. We call injective linear maps monomorphisms.

Definition Bearbeiten

Definition (monomorphism)

A monomorphism is an injective linear map   between two  -vector spaces   and  .

That is,   is a linear map such that for all   the statement   implies that also  .

Equivalent characterization of monomorphisms Bearbeiten

We have considered in the motivation that monomorphisms should be exactly those linear maps, which preserve linear independence of vectors. We now prove this mathematically:

Theorem (monomorphisms preserve linear independence)

Let   be a linear map. Then, we have that   is injective if and only if the image of every linearly independent subset   is again linearly independent.

Thus, the linear map   preserves linear independence exactly if   is a monomorphism.

How to get to the proof? (monomorphisms preserve linear independence)

We follow the preliminary considerations from the motivation. What we would like to show are two implications: "  is injective   the image every linearly independent subset   is linearly independent." and "The image of every linearly independent subset   is linearly independent     is injective."

However, it is easier to prove linear dependence than linear independence, because with linear dependence of a set we only need to find one example for a non-trivial combination to 0. With linear independence, we need to prove that every finite subset of the set is linearly independent. Therefore, we do not directly show the above implications, but use a proof by contradiction.

Proof (monomorphisms preserve linear independence)

We show "There exists a linearly independent subset   such that   is linearly dependent"   "  is not injective"

Proof step: "

So let   be linearly independent, but   be linearly dependent.

Then   contains a finite linearly dependent subset  . Let   be the preimages of the vectors  , so   with  . Since   are linearly dependent, there exist scalars   which are not all zero but

 

Then, we have  , since at least one   and because of   these vectors are linearly independent. Now on the one hand  , but we also know that  . Because of  ,   is not injective.

Proof step: 

Since   is not injective, there are some   with  , but  . For   we have that then  .

Now define the set   as the span  . Because of  ,   is linearly independent, but   is linearly dependent.

We can derive a different criterion for a linear map being a monomorphism: Suppose we have linearly independent vectors  . The linear independence means that the vectors describe "independent information". We have seen above that monomorphisms preserve linear independence. This means that monomorphisms map independent information to independent information. So monomorphisms preserve all information. Suppose we have a monomorphism  , another vector space   and maps   such that   holds. Since no information was lost by the application of  , the maps   and   must have been the same before the application. So we have that for a monomorphism  , from   one cam imply  . One also says that the monomorphism can be left shortened. The next theorem verifies that the ability to being left shortened is equivalent to a linear map being a monomorphism.


Theorem (monomorphisms can be "left shortened")

Let   be linear map. Then, we have:   is a monomorphism if and only if for all vector spaces   and for all   with   we have that  .

One also says that   can be left shortened.

Proof (monomorphisms can be "left shortened")

Proof step:  , by a direct proof

Let   be a monomorphism, i.e. an injective, linear map. Let   be another vector space and   with  . Let  . Then  . Since   is injective, it follows that  . Since we have chosen   arbitrary, we obtain  .

Proof step:   , proof by contradiction

Suppose that   is not a monomorphism, i.e.   is not injective. Then there exist   and   with   and  .

Without loss of generality, let   (otherwise swap   and  ).

We extend   to a basis   of  .

Then we consider the two linear maps   and   for all  . (the second linear map is given by linear continuation starting from the basis vectors).

We now show that   holds. It suffices to check this identity on our basis  : For all   we have that   and  . In addition, we have that  , since this relation holds for all basis elements of  .

But we also have that  , since  .

This is a contradiction to the assumption and it follows that   is a monomorphism.

Hint

This theorem is useful because sometimes it is easier to show that   holds, instead of directly proving  . This theorem gives us a kind of "rule of calculation" for linear maps. Moreover, we do not use concrete elements for left shortening. This allows us to generalize the concept of monomorphism to categories that you may encounter in further study.

Examples Bearbeiten

Example

The map   with of the following mapping rule is a vector space monomorphism:

 

Indeed, from  , it follows:

 

But then   and   must hold and so the equality of the arguments follows  . This shows that   is injective.

Relation to the kernel Bearbeiten

Alternative derivation of a monomorphism Bearbeiten

Linear maps preserve linear independence if and only if they are injective. We call these maps monomorphisms. To derive this, we have first clarified how linear independence is defined, namely via the uniqueness of the representation of vectors as a linear combination. As mentioned before that, instead of considering all these vectors, however, linear independence can also be defined only by the representation of the zero vector:   are linearly independent if it follows from   that all coefficients are  .

What if, with this definition, we tried to derive the definition of monomorphism? Again we are looking for a property for a linear map   with which we can infer from the linear independence of   the linear independence of  . Let for this   be linearly independent. Let us now show that:

 

This is equivalent to

 

Our desired property must guarantee that  . Then we can show with the linear independence of   that all  , which also proves the linear independence of  .

So   needs to fulfil the property:   for all vectors  . By the principle of contraposition, this property is equivalent to  . So the property we are looking for is: "The set of elements that are mapped to zero consists only of the zero vector." This property, by the way, is the special case of injectivity at the point   and states that only the zero element of the domain vector space is mapped to the zero element of the image vector space.

Definition of the kernel Bearbeiten

So the set of elements that are mapped to zero has a special meaning in this context. That is why it has its own name, one speaks of the kernel of the map.

Definition (Kernel of a linear map)

Let   be a linear map between two  -vector spaces   and  . The kernel of the map   is the set of all vectors from   that are mapped to   by   and is denoted  . In mathematical terms:

 

Reading off injectivity from the kernel Bearbeiten

We now know two properties of linear maps which guarantee that they preserve linear independence: On the one hand the injectivity and on the other hand that the kernel of the linear map being trivial (i.e., only including the zero vector). Both properties have the same effect. So it can be assumed that both properties are equivalent. As the following proof will show, this assumption is correct: (this part is still missing)

Serlo: EN: Kernel

Alternative definition of a monomorphism Bearbeiten

So we have learned a second property with which one can characterize monomorphisms. A linear map is a monomorphism if its kernel consists only of the zero vector. We also say that the kernel is "trivial". We can thus formulate an alternative definition for monomorphisms:

Definition (monomorphism)

A monomorphism is a linear map   between two  -vector spaces   and   for which one (or all) of the following equivalent statements hold:

  •   is injective.
  • For all   we have that  .
  • For all   we have that  .
  • The kernel of   is trivial, i.e.,  .

Exercises Bearbeiten

Exercise (Verification of a monomorphism)

Show that for  , the map   is a monomorphism. This shows that one can map every "smaller" vector space   injectively into a "bigger" vector space  , as long as  .

Solution (Verification of a monomorphism)

Let   and  , as well as  . By definition of the map  , we have that:

 

So   is linear. It remains to be shown that   is injective. To show the injectivity of  , there are (at least) two ways:

1st way

From the definition of the linear map   it is clear that only the zero vector of   is mapped by   to the null element of  . Thus

 

Thus the kernel of   contains only the zero vector. By the theorem on the relation between kernel and injectivity of a linear map, it follows that   is injective. Together with the linearity of   it is thus shown that   is a monomorphism.

2nd way

A second way to prove the injectivity of the map   is to directly recalculate the definition of injectivity:

Let  . This is equivalent to the statement  . In other words

 

From this representation one recognizes immediately that   must hold. Thus   and hence,   is injective. Together with the linearity of   we have therefore shown that   is a monomorphism.