Serlo: EN: Quotient space

{{#invoke:Mathe für Nicht-Freaks/Seite|oben}} In this article we consider the quotient space $V/U$ of a $K$-vector space $V$ with respect to a subspace $U$. The quotient space $V/U$ is a vector space in which we can do computations as in $V$, up to an addition of arbitrary terms from $U$.

Introduction

Computations with solutions of a linear system

We consider the matrix Vorlage:Einrücken We now want to solve the linear system of equations $Ax=b$ for different vectors $b\in {ℝ}^2$. For example, taking $b_1=(3,-7{)}^T$, we get a solution $x_1=(1,2,-4{)}^T$ and for $b_2=(1,2{)}^T$, we get a solution $x_2=(2,0,3{)}^T$. That is, $A{x}_1=b_1$ and $A{x}_2=b_2$ hold. What is then the solution for $Ax=b_1+b_2$? To find this out, we can use linearity of $A$: We just have to add our previous solutions together, since $A(x_1+x_2)=A{x}_1+A{x}_2=b_1+b_2$. Thus, a solution to $Ax=b_1+b_2$ is given by $x_1+x_2=(3,2,-1{)}^T$.

The solution to the above system of equations is not unique. For instance, the system $Ax=b_1$ is also solved by ${x_1}^{\prime }=(2,1,-4{)}^T$ and the system $Ax=b_2$ is also solved by ${x_2}^{\prime }=(-1,3,3{)}^T$. The solutions $x_1$ and ${x_1}^{\prime }$, as well as $x_2$ and ${x_2}^{\prime }$ differ from each other. Their differences are $x_1={x_1}^{\prime }+(-1,1,0{)}^T$ and $x_2={x_2}^{\prime }+(3,-3,0{)}^T$. Both $(-1,1,0{)}^T$ and $(3,-3,0{)}^T$ are solutions to the (homogeneous) linear system $Ax=0$. That is, they lie in the kernel of $A$.Vorlage:Todo This "kernel property" is true in general: if $x$ and $x^{\prime }$ are two different solutions of $Ax=b$, they differ exactly by an element in the kernel of $A$, because $A(x-x^{\prime })=Ax-A{x}^{\prime }=b-b=0$. Since the kernel of $A$ is important, we give it the separate name $U$ in the following. Conversely, whenever we have two solutions $x_1+x_2$ and ${x_1}^{\prime }+{x_2}^{\prime }$ of $Ax=b_1+b_2$, then their difference $(x_1+x_2)-({x_1}^{\prime }+{x_2}^{\prime })$ is in the kernel $U$. So once a single solution is found, then the kernel can be used to find all solutions to the system. Put differently, we can consider two vectors whose difference is in $U$ as equivalent, since if one vector solves $Ax=b$, then the other also does.

For scalar multiplication by $\lambda \in ℝ$, we can use linearity of $A$ again: We have a solution $x_1$ of $Ax=b_1$ and we want to solve $Ax=\lambda b_1$ without recalculating. Again, we can obtain a solution by using our already determined solution $x_1$: We have $A(\lambda x_1)=\lambda (A{x}_1)=\lambda b_1$, so $\lambda x_1$ is a solution to $Ax=\lambda b_1$. For the second solution ${x_1}^{\prime }$ this also works: $x=\lambda {x_1}^{\prime }$ is a solution of $Ax=\lambda b$. Again, the difference of both (equivalent) solutions $\lambda x_1$ and $\lambda {x_1}^{\prime }$ is in $U$. So we can scale solutions of linear systems to find solutions to scaled systems. While scaling, the differences stay in $\ker (A)=U$, so both solutions stay equivalent. A different way to say that two vectors are equivalent is to say that they are the same modulo $U$ whenever they differ only by some vector in $U$. For example, the solutions $x_1+x_2$ and ${x_1}^{\prime }+{x_2}^{\prime }$ of the system of equations $Ax=b_1+b_2$ are equal modulo $U$, since $(x_1+x_2)-({x_1}^{\prime }+{x_2}^{\prime })\in U$. When calculating with solutions of systems of linear equations, we therefore calculate modulo $U$.

Construction of the quotient space

In the example, we made calculations in a vector space $V$, but only looked at the results up to differences in a subspace $U$. That is, we considered two vectors $v$ and $v^{\prime }$ in $V$ as equivalent, whenever $v-v^{\prime }\in U$. To formalise these "calculations up to some element in $U$", we identify vectors which using an equivalence relation $\sim$ that is defined by Vorlage:Einrücken This is exactly the relation we used to define cosets of a subspace. In this article, we have also checked that $\sim$ is an equivalence relation. Mathematically, the set of all equivalence classes is denoted by $V/U$.

We will now show that on $V/U$, we can define a natural vector space structure. To do so, we introduce an addition $⊞$ and a scalar multiplication $⊡$ on $V/U$: For $v+U,w+U\in V/U$ and $\lambda \in K$ we define Vorlage:Einrücken These definitions make use of representatives. That is, we took one element from each involved coset to define $⊞$ and $⊡$. However, we still have to show that the definitions are independent of the chosen representative.

That is, we must show that this definition is independent of the choice of representative and thus makes sense. We give this proof further below. The property that a mathematical definition makes sense is also called well-definedness.

We also need to show that $V/U$ is a vector space with this addition and scalar multiplication, which we will do below.

Definition

In the previous section, we considered what a vector space $V/U$ might look like, in which we can calculate modulo $U$. The elements of $V/U$ are the cosets $v+U$. We want to define the vector space structure using the representatives. Further below , we then show that the definition makes mathematical sense, that is, the vector space structure is proven to be well--defined.

To distinguish addition and scalar multiplication on $V/U$ from that on $V$, we refer to the operations on $V/U$ as "$⊞$" and "$⊡$" in this article. Other articles and sources mostly use "$+$" and "$\cdot$" for the vector space operations.

Explanation of the definition

A short explanation concerning the brackets appearing in $(v+U)⊞(w+U):=(v+w)+U$ and $\lambda ⊡(v+U):=(\lambda \cdot v)+U$: To define the addition $⊞$ in $V/U$, we need two vectors from $V/U$. Vectors in $V/U$ are cosets, so $(v+U)$ and $(w+U)$ denote cosets given by $v,w\in V$. The expression $(v+w)+U$ is also a coset, namely the one associated with $v+w$: Vorlage:Einrücken The scalar multiplication works similarly: For a scalar $\lambda \in K$ and a coset $v+U$ with $v\in V$ we want to define $\lambda ⊡(v+U)$. For this we first calculate the scalar product $\lambda \cdot v$ in $V$ and then turn to the associated coset $(\lambda \cdot v)+U$: Vorlage:Einrücken So we first execute the addition or scalar multiplication of the representatives in $V$ and then turn to the coset to get the addition or scalar multiplication on $V/U$. Mathematically, we also say that the vector space structure on $V$ "induces" the structure on $V/U$.

Well-defined operations in the quotient space Vorlage:Anker

We want to check whether the operations of $⊞$ and $⊡$ are independent of the choice of representatives - that is, they are well-defined. Mathe für Nicht-Freaks: Vorlage:Satz

Establishing the vector space axioms

We show that the quotient space is again a $K$-vector space by taking the axioms valid for $V$ and inferring those axioms of $V/U$. Hence, taking quotient spaces is a way to generate new vector spaces from an existing $K$-vector space, just like taking subspaces.

Examples

Example in finite vector space

Now, we turn to a more abstract mathematical example, that will involve some donuts.

Relationship between quotient space and complement

In the quotient space $V/U$ we calculate with vectors in $V$ up to arbitrary modifications from $U$. We know another construction that can be interpreted similarly: The complement. A complement of a subspace $U\subseteq V$ is a subspace $W\subseteq V$ such that $U\oplus W=V$. Here $U\oplus W$ denotes the inner direct sum of $U$ and $W$ in $V$, that is, $U\oplus W=U+W$ and $U\cap W=\{0\}$. A vector $v\in V$ can then be decomposed uniquely as $v=u+w$, where $u\in U$ and $w\in W$. But the complement itself is then not unique! There can be different subspaces $W,W^{\prime }\subseteq V$, with $U\oplus W=V=U\oplus W^{\prime }$.

For the quotient space, we "forget" the part of $v$ that is in $U$ by identifying $v$ with the coset $v+U$: Vorlage:Einrücken If $W$ is a complement of $U$ and $v=u+w$ for distinct $u\in U$ and $w\in W$, then we can analogously forget the $U$-part by mapping $v$ to the $W$-part, called $w$: Vorlage:Einrücken Apparently $V/U$ and the complement $W$ are similar. Can we identify the two vector spaces $V/U$ and $W$, i.e., are they isomorphic? Yes, they are, as we prove in the following theorem.

Mathe für Nicht-Freaks: Vorlage:Satz We have seen that $V/U$ is isomorphic to any complement of $U$. So it should also behave like a complement, i.e. $U\oplus V/U=V$ should hold. But be careful: Because $V/U$ is not a subspace of $V$, we cannot form the inner direct sum with $U$. However, we can still consider the outer direct sum of $U$ and $V/U$: Vorlage:Einrücken This may not be equal to $V$, but it may be isomorphic to $V$. And we will show that it indeed is isomorphic. Mathe für Nicht-Freaks: Vorlage:Satz

Exercises

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