In this article we consider the quotient space of a -vector space with respect to a subspace . The quotient space is a vector space in which we can do computations as in , up to an addition of arbitrary terms from .
Computations with solutions of a linear system[Bearbeiten]
We consider the matrix
We now want to solve the linear system of equations for different vectors . For example, taking , we get a solution and for , we get a solution . That is, and hold. What is then the solution for ? To find this out, we can use linearity of : We just have to add our previous solutions together, since . Thus, a solution to is given by .
The solution to the above system of equations is not unique. For instance, the system is also solved by and the system is also solved by . The solutions and , as well as and differ from each other. Their differences are and . Both and are solutions to the (homogeneous) linear system . That is, they lie in the kernel of .
Link kernel of a matrix
This "kernel property" is true in general: if and are two different solutions of , they differ exactly by an element in the kernel of , because . Since the kernel of is important, we give it the separate name in the following.
Conversely, whenever we have two solutions and of , then their difference is in the kernel . 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 as equivalent, since if one vector solves , then the other also does.
For scalar multiplication by , we can use linearity of again: We have a solution of and we want to solve without recalculating. Again, we can obtain a solution by using our already determined solution : We have , so is a solution to . For the second solution this also works: is a solution of . Again, the difference of both (equivalent) solutions and is in . So we can scale solutions of linear systems to find solutions to scaled systems. While scaling, the differences stay in , so both solutions stay equivalent.
A different way to say that two vectors are equivalent is to say that they are the same modulo whenever they differ only by some vector in . For example, the solutions and of the system of equations are equal modulo , since . When calculating with solutions of systems of linear equations, we therefore calculate modulo .
In the example, we made calculations in a vector space , but only looked at the results up to differences in a subspace . That is, we considered two vectors and in as equivalent, whenever . To formalise these "calculations up to some element in ", we identify vectors which using an equivalence relation that is defined by
This is exactly the relation we used to define cosets of a subspace. In this article, we have also checked that is an equivalence relation. Mathematically, the set of all equivalence classes is denoted by .
We will now show that on , we can define a natural vector space structure. To do so, we introduce an addition and a scalar multiplication on : For and we define
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 is a vector space with this addition and scalar multiplication, which we will do below.
In the previous section, we considered what a vector space might look like, in which we can calculate modulo .
The elements of are the cosets . 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 from that on , we refer to the operations on as "" and "" in this article. Other articles and sources mostly use "" and "" for the vector space operations.
A short explanation concerning the brackets appearing in and : To define the addition in , we need two vectors from . Vectors in are cosets, so and denote cosets given by . The expression is also a coset, namely the one associated with :
The scalar multiplication works similarly: For a scalar and a coset with we want to define . For this we first calculate the scalar product in and then turn to the associated coset :
So we first execute the addition or scalar multiplication of the representatives in and then turn to the coset to get the addition or scalar multiplication on . Mathematically, we also say that the vector space structure on "induces" the structure on .
Well-defined operations in the quotient space [Bearbeiten]
We want to check whether the operations of and are independent of the choice of representatives - that is, they are well-defined.
Theorem (Well-defined operations in the quotient space)
Let be a -vector space and a subspace of vectors.
Then addition and scalar multiplication on are well-defined.
Proof (Well-defined operations in the quotient space)
For well-definedness, we need to show the following:
If in the definition, we plug in different representatives of the coset(s) on the left-hand side, we end up with the same coset on the right-hand side.
Mathematically, we have to show :
For : If and , then .
For : If and , then .
Proof step: Well-defined addition
By definition of a coset we have to show that holds.
Since , this is equivalent to .
Now and or and each represent the same coset modulo .
Since is a subspace of , it follows that .
So the addition is indeed independent of the choice of representatives.
Proof step: Well-defined scalar multiplication
Well-definedness of the scalar multiplication can be seen in same way:
In the above notation, we have to show that .
Since and represent the same coset modulo , we have .
And since is a subspace, we also have .
So the scalar multiplication is also independent of the choice of representative.
We show that the quotient space is again a -vector space by taking the axioms valid for and inferring those axioms of . Hence, taking quotient spaces is a way to generate new vector spaces from an existing -vector space, just like taking subspaces.
Exercise (Proof of the vector space axioms in quotient space)
Let be a -vector space and a subspace of it, then with the operations and defined above is also a -vector space
Solution (Proof of the vector space axioms in quotient space)
Proof step: Establishing properties of a commutative additive group (also called an Abelian group).
We first consider the properties of addition. For this let .
We trace back the associativity to associativity in
We also trace commutativity back to commutativity in
3. Existence of a neutral element
Since we are considering displacements of , the coset should be the neutral element with respect to addition. We can verify this by using that is the neutral element in :
4. Existence of an inverse
We consider the coset . For the inverse of , we need that
Thus, the addition of an element with its inverse indeed yields the neutral element .
We also trace the inverse of back to inverse in . Let be a representative of and its inverse in . Then,
Thus, the element inverse to is .
Proof step: Distributive laws
1. Scalar Distributive Law
Multiplication of a vector (in a quotient space, i.e., the vector is a coset) with a sum of scalars yields:
2. Vector Distributive Law
Likewise, we can show that the distributive law also holds for the multiplication of a scalar with the sum of two vectors (i.e., with two cosets in the quotient space):
Proof step: Properties of scalar multiplication
We now show that the scalar multiplication of cosets also satisfies the corresponding vector space axioms. Again, we trace back properties in the quotient space back to the corresponding properties in . To this end, let and . Then the following axioms hold:
1. Associative law for scalars
The scalar multiplication is associative, since
2. Neutral element of scalar multiplication
We want to prove that is also the neutral element for .
That is, must hold. Since 1 is neutral in and since , we get
So is the
neutral element of scalar multiplication and is indeed a -vector space.
We imagine that we are standing on a vantage point in New York City from which we are looking at the skyline. In this situation, we will see the city in three dimensions. So objects (e.g. skyscrapers) can be identified with vectors in . However, there are also situations where we want to look at the city in only two dimensions, for instance, when taking a virtual tour using a map or a satellite image of New York.
If we want to create a map or a satellite image, we need to "project" information from three dimensions into two dimensions. This process can mathematically be described by a reduction to some quotient space .
For example, let us take a look at the edge of a skyscraper. On the oblique image, we see that an edge reaches about 600 feet up into the air. However, on the satellite image, the edge is just displayed as a dot. So all points (= vectors) on the edge are identified with this one dot. The dot is then a coset in . The vector space is given by the (1-dimensional) -axis, since after adding a vector on the -axis (i.e., shifting a point up or down), we end up at the same dot on the sattelite image. The space contains all dots, i.e., it corresponds to the (2-dimensional) map.
Now, we turn to a more abstract mathematical example, that will involve some donuts.
Example (Quotient space in )
In the vector space article, we considered the abstract mathematical set , which can be seen as lattice points on a torus (= surface of a donut). Using the same method, we can think of as 9 lattice points on a torus as well:
We obtain a torus from a square by stretching gluing the edges as follows:
In other words, the surface of a donut is the same as a square, where, if you walk out on one edge, you immediately enter it at the opposite side.
Thus we may visualize as follows: On the torus, we draw nine points in lattice form. We then get the following picture:
The subspace generated by , corresponds to a discrete straight line. We put this line through the above points.
We now have points lying on two different sides directly next to our line. Some points are lying directly to the right of the line; that is, they are displaced from the straight line by . Some other points lie directly to the left of our line; that is, they are displaced by . In the picture it looks like this:
The vector space also allows for "adding the points on the donut": Here, we get the following relations:
If we add a point on the left and a point on the right of the line, we get a point on the line: For example, .
If we add two points on the left of the line, we get a point on the right: For example, we have .
If we add two points on the right of the line, we get a point on the left: For example, .
If we form the quotient space (with 3 cosets), we see that two points have the same position with respect to the line (on/left/right) if they are in the same coset. Each coset then consists of 3 points. Furthermore, our addition relations above just represent the addition on the quotient space .
Relationship between quotient space and complement[Bearbeiten]
In the quotient space we calculate with vectors in up to arbitrary modifications from . We know another construction that can be interpreted similarly: The complement. A complement of a subspace is a subspace such that . Here denotes the inner direct sum of and in , that is, and . A vector can then be decomposed uniquely as , where and . But the complement itself is then not unique! There can be different subspaces , with .
For the quotient space, we "forget" the part of that is in by identifying with the coset :
If is a complement of and for distinct and , then we can analogously forget the -part by mapping to the -part, called :
Apparently and the complement are similar. Can we identify the two vector spaces and , i.e., are they isomorphic? Yes, they are, as we prove in the following theorem.
Theorem (Isomorphism between complement and quotient space)
Let be a complement of in . Then the projection is a linear isomorphism between and the quotient space .
Proof (Isomorphism between complement and quotient space)
We want to show that is linear, i.e., compatible with addition and scalar multiplication, and bijective.
Proof step: Linearity of
Since is a subspace and scalar multiplication and addition is defined on representatives, is compatible with addition and scalar multiplication. That is, for and , we have
Proof step: Surjectivity of
Let . Since is a complement of , we find and with . Then
where we used in that and thus holds. So is surjective.
Proof step: Injectivity of
We show . Let , i.e. with . So holds. Thus . Since is a complement of , we have . Further, and implies , so .
We have seen that is isomorphic to any complement of . So it should also behave like a complement, i.e. should hold. But be careful: Because is not a subspace of , we cannot form the inner direct sum with .
However, we can still consider the outer direct sum of and :
This may not be equal to , but it may be isomorphic to . And we will show that it indeed is isomorphic.
Let be a subspace of a -vector space . Then, holds.
Let be a complement of , i.e. and . From the previous theorem we know that the function
is an isomorphism. We use this to show that
is an isomorphism, where denotes the outer direct sum.
Proof step: is linear
We have for all . It follows directly that is linear, since addition and scalar multiplication on are defined component-wise and as and are linear.
Proof step: is bijective
This also follows from for all , since the identity and are bijective.
Thus we have . By this theorem, the inner direct sum of the subspaces and is isomorphic to their outer direct sum. So , where denotes the inner direct sum of and .
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