Serlo: EN: Monomorphisms

{{#invoke:Mathe für Nicht-Freaks/Seite|oben}}

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

Motivation

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 $V$ over a field $K$, a finite set of vectors $\{v_1,\dots ,v_n\}\subset V$ is linearly independent if and only if the only linear combination by ${\lambda }_1,...,{\lambda }_n\in K$, which leads to zero (${\lambda }_1{v}_1+\dots +{\lambda }_n{v}_n=0\in V$) is the trivial one, i.e., ${\lambda }_1=\dots ={\lambda }_n=0$.

An alternative characterization is that if Vorlage:Einrücken then the set of coefficients ${\lambda }_i,{\mu }_i\in K$ must be equal as ${\lambda }_i={\mu }_i$.

Is this property preserved? Certainly, there are linear maps, which do not preserve linear independence, e.g. the map to zero: $f(v)=0\mathrm{\forall }v\in V$. 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 $\lambda =1$: $1\cdot f(v)=1\cdot 0=0$. 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 $v_1,\dots ,v_n$. For a linear map $f$ to preserve linear independence, it needs to satisfy:

We transform:

Therefore $f$ must have the following property to preserve linear independence:

By setting $x:={\sum }_{i=1}^n{\lambda }_i{v}_i$ and $y:={\sum }_{i=1}^n{\mu }_i{v}_i$, it becomes clearer what this property is. We get that

for all $x,y\in V$ which can be written as linear combination of $v_1,\dots ,v_n$.

This statement should be valid for all linear independent sets and therefore also for bases. In the case of a basis, however, all $x,y$ can be written as such a linear combination, which means that $f$ 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 $f$ an injective linear map and $v_1,\dots ,v_n\in V$ linearly independent vectors. We are to find out whether $f(v_1),\dots ,f(v_n)$ are also linearly independent. According to our considerations above, it is enough to show the following for scalars ${\lambda }_i$ and ${\mu }_i\in K$

Vorlage:Einrücken Let Vorlage:Einrücken Then, we have from the injectivity of $f$ that Vorlage:Einrücken Because $v_1,\dots ,v_n$ are linearly independent, we have that ${\lambda }_i={\mu }_i$ for all $i$. Thus we have shown the above statement and $f$ preserves linear independence.

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

Equivalent characterization of monomorphisms

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:

We can derive a different criterion for a linear map being a monomorphism: Suppose we have linearly independent vectors $v_1,\dots ,v_n\in V$. 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 $f:V\to W$, another vector space $U$ and maps $a,b:U\to V$ such that $f\circ a=f\circ b$ holds. Since no information was lost by the application of $f$, the maps $a$ and $b$ must have been the same before the application. So we have that for a monomorphism $f$, from $f\circ a=f\circ b$ one cam imply $a=b$. 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.

Mathe für Nicht-Freaks: Vorlage:Satz

Relation to the kernel

Alternative derivation of a monomorphism

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: $v_1,\dots ,v_n$ are linearly independent if it follows from ${\sum }_{i=1}^n{\lambda }_i{v}_i=0$ that all coefficients are ${\lambda }_i=0$.

What if, with this definition, we tried to derive the definition of monomorphism? Again we are looking for a property for a linear map $f$ with which we can infer from the linear independence of $v_i$ the linear independence of $f(v_i)$. Let for this $v_1,\dots ,v_n$ be linearly independent. Let us now show that:

This is equivalent to

Our desired property must guarantee that ${\sum }_{i=1}^n{\lambda }_i{v}_i=0$. Then we can show with the linear independence of $v_i$ that all ${\lambda }_i=0$, which also proves the linear independence of $f(v_i)$.

So $f$ needs to fulfil the property: $f(v)=0⟹v=0$ for all vectors $v$. By the principle of contraposition, this property is equivalent to $v\ne 0⟹f(v)\ne 0$. 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 $0$ 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

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.

Reading off injectivity from the kernel

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)

{{#lst:Serlo: EN: Kernel|InjektivitätSatz}}

Alternative definition of a monomorphism

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:

Exercises

{{#invoke:Mathe für Nicht-Freaks/Seite|unten}}