Serlo: EN: Cosets of a subspace

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Derivation: coset or affine subspace

Lines in ${ℝ}^2$ Vorlage:Anker

You probably already know the concept of a straight line. But how do we describe a line in ${ℝ}^2$ mathematically? You know from school that you can parameterise straight lines by $v+tu$, where $v,u\in {ℝ}^2,u\ne 0$ are two fixed vectors and $t$ takes all values in $ℝ$. That is, all points on the straight line form the set $G:=\{v+tu\mid t\in ℝ\}$. Geometrically described, this is the (infinitely long) line running through $v$ in the direction of $u$.

An affine line is described by the support vector v and the direction vector u.

In general, a line does not pass through the origin $(0,0)$. Thus $G$ is not a subspace of ${ℝ}^2$, since by definition every subspace contains the origin. However, the line $G$ is a displaced version of the line $U:=\{tu\mid t\in ℝ\}$ by the vector $v$. Here $U$ is a line passing through the origin. This is a subspace because it contains the origin and is closed under addition and scalar multiplication. That is, every straight line is given by the choice of a (one-dimensional) subspace $U\subset {ℝ}^2$ and a vector $v\in {ℝ}^2$. This justifies the notation $G=v+U$. This notation can also be formalised:

For a subspace $W$ consider a vector $v$. Let $v+W$ be by $v+W:=\{v+w\mid w\in W\}$. Then the following applies for the sets $G$ and $U$ defined above, that $G=\{v+tu\mid t\in ℝ\}=\{v+w\mid w\in U\}=v+U$.

Planes in ${ℝ}^3$

Let's increase the dimension and consider ${ℝ}^3$. We can describe a line in analogy to the set $G:=\{v+tu\mid t\in ℝ\}$ with vectors $v$ and $u\in {ℝ}^3,u\ne 0$. This is a displaced version of a line through the origin by a vector $v$. So formally again, any line is of the form $v+U$ for a vector $v\in {ℝ}^3$ and a one-dimensional subspace $U\subset {ℝ}^3$.

What about the planes in ${ℝ}^3$? We parameterise them by $v+t_1{u}_1+t_2{u}_2$, where $v,u_1,u_2\in {ℝ}^3$ are fixed vectors and $t_1,t_2$ pass through all values in $ℝ$. The vectors $u_1$ and $u_2$ must not be scalar multiples of each other - otherwise we would get a line. All points on the plane form the set $E:=\{v+t_1{u}_1+t_2{u}_2\mid t_1,t_2\in ℝ\}$. As in the case of lines, the plane $E$ is generally not a subspace, since the origin need not lie in $E$. However, the plane is a displaced version of the subspace $U:=\{t_1{u}_1+t_2{u}_2\mid t_1,t_2\in ℝ\}$ by the vector $v$. It is therefore analogously true that every plane is given by a two-dimensional subspace and a vector, i.e. that $E=v+U$.

An affine plane is described by the support vector v and the direction vectors u_1 and u_2.

Lines in $(\mathbb{Z}/5\mathbb{Z}{)}^2$

We can also look at certain straight lines in a more complicated space: We consider the $\mathbb{Z}/5\mathbb{Z}$-vector space $(\mathbb{Z}/5\mathbb{Z}{)}^2$. In the article vector space we have already seen that we can think of this vector space as regular points on a torus. Now what is a "straight line" on this torus? We have seen in the previous two sections how we can describe straight lines in the vector spaces ${ℝ}^2$ and ${ℝ}^3$: There a straight line is the same as a set $G=\{v+tu\mid t\in ℝ\}$ with a support vector $v$ and a direction vector $u$. In other words, it is the set $G=v+U$, where $U=\{tu\mid u\in ℝ\}$ is a one-dimensional subspace. We can transfer this construction to $(\mathbb{Z}/5\mathbb{Z}{)}^2$, that is, we can consider a straight line as $G=v+U$, where $U$ is a one-dimensional subspace of $(\mathbb{Z}/5\mathbb{Z}{)}^2$. That is, $G$ is of the form $G=\{v+tu\mid t\in \mathbb{Z}/5\mathbb{Z}\}$. We can visualise this set on a torus:

points of an affine line in (Z/5Z)^2 on a torus

The points appear to lie on a line. If we connect the points each in the shortest way, we get a closed line that feels like a straight line on the torus.

Points of an affine straight line in (Z/5Z)^2 on a torus, connected by a line

Thus, displaced one-dimensional subspaces also correspond to straight lines here.

We consider another example of a straight line in $(\mathbb{Z}/5\mathbb{Z}{)}^2$. Consider the one-dimensional subspace $U:=\{n(1,1)\mid n\in \mathbb{Z}/5\mathbb{Z}\}$. We shift this by the vector $(2,0)\in (\mathbb{Z}/5\mathbb{Z}{)}^2$. Thus we obtain the line $G:=(2,0)+U$. Here a line consists of only five vectors. In our case $G=\{(2,0),(3,1),(4,2),(0,3),(1,4)\}$.

We have characterised geometric objects (e.g. lines and planes) as displaced subspaces in various vector spaces. Let's give them a name.

Derivation: set of cosets of a subspace

We have defined cosets as displaced subspaces. Consider the following example of a displaced subspace $U$ of ${ℝ}^2$ by two different vectors $v$ and $v^{\prime }$:

Different displacements of a subspace leading to the same affine subspace

In the example above, we see that different displacements of a subspace can lead to the same affine subspace. So we ask ourselves the following question:

When are two shifted subspaces $v+U$ and $v^{\prime }+U^{\prime }$ the same?

Let us first imagine the whole thing in ${ℝ}^2$, where both shifted subspaces are lines. If they are equal, they have the same slope. This characterises the lines passing through the origin $U$ and $U^{\prime }$. It follows that $U$ and $U^{\prime }$ must be equal.

Let us now consider the question for general vector spaces. So let $V$ be a vector space, $U,U^{\prime }\subseteq V$ be subspaces of vectors, $v,v^{\prime }\in V$ be vectors, and let $v+U=v^{\prime }+U^{\prime }$ be sets. We would like to first conclude (as in ${ℝ}^2$) that $U=U^{\prime }$. To do this, it would be nice to get $U$ from $v+U$. This is done by taking all vectors of $v+U$ and subtracting $v$ , which indeed gives us $U$ . Hence, we can write $U$ as: Vorlage:Einrücken

Since $U$ is a subspace, we have $0\in U$. The above equation thus implies $0\in (v^{\prime }-v)+U^{\prime }$, i.e., there is a $u^{\prime }\in U^{\prime }$, such that $(v^{\prime }-v)+u^{\prime }=0$, i.e., $v^{\prime }-v=-u^{\prime }$. In particular, $v^{\prime }-v\in U^{\prime }$.

More generally, for each subspace $W$ and vector $w\in W$, we have $w+W=W$. The reason is that each $w^{\prime }\in W$ can be written as $w^{\prime }=w+(w^{\prime }-w)$. Since $w^{\prime }-w\in W$ , we have $w^{\prime }\in w+W$. Geometrically, you can also imagine the whole thing like this: If you move the subspace $W$ in a direction in which it already lies, it is mapped onto itself.

Back to our original question: Since $v^{\prime }-v\in U^{\prime }$ , we know that $(v^{\prime }-v)+U^{\prime }=U^{\prime }$. So all in all we get the desired $U=U^{\prime }$. On the way we have also seen that $v^{\prime }-v\in U^{\prime }=U$ is also a necessary criterion for $v+U=v^{\prime }+U^{\prime }$.

Are these criteria also sufficient? Yes: Suppose we have $v,v^{\prime }\in V$ and $U,U^{\prime }\subset V$ with $U=U^{\prime }$ and $v^{\prime }-v\in U^{\prime }=U$ . Then $U=U^{\prime }=(v^{\prime }-v)+U^{\prime }$ and hence, by adding $v$ on both sides, we have $v+U=v^{\prime }+U^{\prime }$.

Let us summarise: Two shifted subspaces $v+U,v^{\prime }+U^{\prime }$ are equal exactly if the (non-shifted) subspaces are equal, i.e. $U=U^{\prime }$, and the difference of the shifts lie in $U$ , i.e., $v-v^{\prime }\in U$.

Given a subspace, we can now find out whether two displacements by $v$ or $v^{\prime }$ give the same affine subspace. We can thus construct a kind of "new equality" by considering $v$ and $v^{\prime }$ to be "equal" if they produce the same affine subspace. Such new equalities behave reasonably if they are equivalence relations.

Recall the definition of an equivalence relation. Mathe für Nicht-Freaks: Vorlage:Definition

Two elements that are in relation with respect to an equivalence relation are called equivalent. If two elements $x$ and $y$ are equivalent to each other with respect to an equivalence relation $R$, one often writes $x{\sim }_{R}y$ or simply $x\sim y$.

To formally write down the "new equality" mentioned above, we define a relation $\sim$ given by $v\sim v^{\prime }:⟺v+U=v^{\prime }+U⟺v-v^{\prime }\in U$. Intuitively, our relation should be an equivalence relation, since it says when two shifted subspaces are equal. We now check this formally:

We can now consider the equivalence classes of this relation, that is, to $v\in V$ we consider the set $[v]:=\{w\in V\mid v\sim w\}$. So the set$[v]$ consists of all vectors $w$, that displace $U$ to the same affine subspace $v+U$. How else can we characterise these equivalence classes? We have

$[v]=\{w\in V\mid v\sim w\}=\{w\in V\mid v-w\in U\}=\{v+u\mid u\in U\}=v+U$

That is, the equivalence classes of our relation are precisely the coset classes.

Just as we can look at an equivalence relation and its equivalence classes, we can also construct a space in which the "new equality" of the equivalence relation becomes a real equality. This is the set of equivalence classes to which we now want to give a special name.

Definition: set of cosets of a subspace

We have defined the set of cosets $V/U$ as the set of equivalence classes according to $\sim$. In the last section we saw that the equivalence class generated by $v\in V$ is given exactly by the affine subspace $v+U$. Thus an equivalence class with respect to $\sim$ is the same as a displaced version of $U$. This provides two equivalent views of the set $V/U$: on the one hand, $V/U$ is the set of equivalence classes with respect to $\sim$; on the other hand, it is the set of displaced versions of $U$.

Properties of equivalence classes applied to cosets

We have seen above that cosets of a one-dimensional subspace in ${ℝ}^2$ are parallel straight lines. We can also explain this by characterising cosets as equivalence classes: Two equivalence classes, as sets, are either equal or disjoint. For us, this means that two cosets, i.e. two straight lines, are either equal or that they have no point of intersection. The latter means that they are parallel.

Furthermore, we know about equivalence classes that they cover the whole space, i.e. the union of all equivalence classes results in the whole set. From this we conclude that the union of all cosets (in our case parallel straight lines) gives the whole ${ℝ}^2$. We can therefore decompose the vector space into the cosets - like leaves. This decomposition is also called a partition. So the cosets partition the vector space. In our example, this means that we can decompose the ${ℝ}^2$ into displaced versions of an origin line $U$. This is illustrated in the following picture:

V=\R ^2 is partitioned by a line

Both points mentioned also work in general (not only in ${ℝ}^2$), since we have not used any property of ${ℝ}^2$ in any of our arguments. It is therefore true for a vector space $V$ and a subspace $U$ that: Vorlage:Important

Outlook

Cosets occur when solving systems of linear equations: The solutions of the associated homogeneous system of equations $U$ form a subvector space. If the linear system of equations has a solution, the solutions form an affine subspace with respect to $U$.

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