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Inner direct sum – Serlo

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Derivation and definition

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We have already learned about sums of two subspaces. If and are two subspaces, then the sum of and is again a subspace . So for each vector we may find two vectors and , such that . Now the question arises: Are there several ways to write as such a combination?

The answer is yes, there can be several possibilities. As an example, let's look at the vector space . This space can be viewed as the sum of the -plane and the -plane. This means that if , then there are actually several ways to represent as the sum of vectors from the -plane and the -plane. For the vector , for example, we have .

Such representations are therefore generally not unique. We now want to find a criterion for uniqueness.

Suppose we have two different representations of , i.e. and with and (if one of them is the same, then so is the other). In particular, we know that and . If we now rearrange the equation , we get . Because the left-hand side is in and the right-hand side is in , this is an element in which is not a zero vector at the same time. So is not just . (The zero vector is in the intersection because and are both subspaces). This means that if the representation is not unique, then the intersection does not only contain the zero vector.

Conversely, if the intersection is not , we do not have a unique representation: Let with . Then there are two representations of , namely (on the one hand with and and on the other hand with and ). Because of , these representations are different from each other.

We can therefore conclude an equivalence: The intersection is exactly if the representation of all vectors in is unique.

In this case, we give the sum a special name: We call the sum of and , in the case , the direct sum of and and write .

Definition (Direct sum)

Let and be two subspaces of a vector space . We call the sum direct if holds. The subspace is called the direct sum of and and we write .

Examples

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Sum of two lines in ℝ²

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The lines and

We consider the following two lines in :

So is the -axis and is the line that runs through the origin and the point . Their sum is

Question: Why do wa have ?

By the definition we can describe the set as

We can write each vector in as with matching . Specifically, for each vector we can find scalars and such that , namely and . We conclude .

Intuitively, you can immediately see that . This is because is a subspace of , which contains the lines and . The only subspaces of are the null space, lines that run through the origin and . As the lines and do not coincide but are different, cannot be a line. Therefore, we must have .

Let us now investigate whether this sum is direct. To do so, we need to determine . If , then we know the following: Because , we have . And because , we have . Therefore and we get . Because also contains , we get . This means that the sum of and is direct and we can write .

Sum of two lines in ℝ³

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The lines and

We have the following lines in :

Then is a line in that runs through the origin and the point , and is a line that runs through the origin and . The sum is a plane spanned by the vectors and , i.e.

Question: Why is this the sum?

So is a plane that is spanned by the vectors and .

Also here, we want to determine whether the sum is direct. To do so, we consider a vector . Then, because , we have . And because , we obtain . Therefore, and the sum is direct. This means that we can write .

Sum of a line and a plane in ℝ³

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The line and the plane

We consider the subvector spaces and of .

The subspace is the line through the origin and the point , while represents the y-z-plane. Together, and span the entire , i.e. .

Question: Why is the sum of and the entire space ?

Since and are subvspaces of , the sum is also a subspace of . We still have to show that is contained in . To do so, we prove that any vector lies in . To accomplish this, we show that there is a and a with .

We choose and . Then . In addition, and apply.

Therefore, the entire is contained in . Thus .

One may now ask whether the sum is direct. To check this, we need to analyze the intersection . If only contains the zero vector , then the sum is direct.

Let be a vector in . Since , we have . Consequently, we can write as . Furthermore, , which implies . We have therefore shown that .

It follows that . Since the intersction only contains the zero vector, the sum is direct. Therefore, we can conclude .

Sum of even and odd polynomials

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We will now look at an example of a direct sum in the vector space of real polynomials . Let us consider the following subspaces and of : consists of all odd polynomials over , while is the space of the even polynomials over . In formulas, this is

The odd polynomials only contain monomials with odd exponents, while the even polynomials only contain monomials with even exponents. For example, is an even polynomial, while is neither even nor odd. We now show that the even and odd polynomials together generate the entire polynomial space . Expressed in formulas: .

To show this, we need to prove that every polynomial in can be written as the sum of an odd and an even polynomial. To do so, we consider any polynomial from . We must write as the sum of an even and an odd polynomial.

Therefore, is contained in the sum .

Now we want to check whether the sum is direct. That is, we need to check whether the intersection of the two subspaces only contains the zero vector, i.e. the zero polynomial. Let be a polynomial in the intersection . Then lies both in and in . We can write as . Since lies in , only consists of odd monomials. Therefore, the prefactors of the even monomials must be equal to . So for all even . Since lies in , only consists of even monomials. So for all odd . This means that all coefficients are equal to zero and is therefore the zero polynomial. Thus , and the sum of and is direct.

We have seen that . In other words, the polynomial space can be written as the direct sum of the subspaces and , where is the subspace of odd polynomials and is the subspace of even polynomials.

Counterexamples

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Two planes in ℝ³

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The planes and

We consider the following two planes:

The two planes together span all of . However, the sum is not direct, as the intersection is a line and therefore does not only contain the zero vector. That is, .

We want to check this mathematically. This requires looking for a vector in the intersection of and which is not zero. We consider a vector that lies in the intersection . Because this vector lies in , we have so . In addition, there must be so , since .

We now look for suitable values for to fulfill both conditions. From and , we get . Because , we also have . Furthermore, results from . Finally, we conclude .

One possible solution is , and . The vector therefore lies in the intersection of and . Hence, .

Various polynomials in polynomial space

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Let be a field. We consider two subspaces in the polynomial space : Let be the space of polynomials of degree less or equal to two, and let

be the space of polynomials whose sum of coefficients is . We want to investigate whether the sum is direct. To find this out, we need to decide whether .

An element is a polynomial , which has a maximum degree of and for which applies. Because the polynomial has degree two, we have . Therefore, we get . This means consists of all polynomials for which . Thus, we can find a non-zero element of if we use the equation

with non-trivial . One possibility for this is , i.e. . So the intersection of and is not zero, and the sum is therefore not direct.

Unique decomposition of vectors

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We have already considered in the derivation that the decomposition of vectors is unique for the direct sum. We will prove this result here rigorously.

Theorem (Equivalent characterizations of the direct sum)

Let be subspaces of . Then the following statements are equivalent:

  1. The sum of and is direct (that is, ).
  2. and have trivial intersection (that is, is the trivial subspace).
  3. The representation of all elements of is unique (that is, if with and , then we already have and ).
  4. The representation of the zero is unique (that is, if with and , then we already have ).

Proof (Equivalent characterizations of the direct sum)

The definition of the inner direct sum is just . We now show the implications . The statement then readily follows.

Proof step:

Let . We must prove that can be written uniquely as the sum of two elements of and .

Let and with the property that . In order to prove uniqueness, we must show that these two representations of are equal. "Equal" means that and .

Because , we have . This element lies in (because of the representation on the left of "") and in (because of the representation on the right of ""). So lies in the intersection . According to the prerequisite, . This means . So and . This is exactly what we wanted to show.

Proof step:

Let and with . This is a representation of .

On the other hand, is also a representation of .

Since representations are unique according to the requirements, we conclude and .

Proof step:

Let . Then, of course, and . Since is a subspace, for each element also its additive inverse element msut be in this space, i.e., . Therefore, .

This gives us . From the uniqueness of the representation of the zero, we conclude . The intersection is therefore trivial, i.e. .

Inner direct sum and disjoint union of sets

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We can imagine the sum of two subspaces as a structure-preserving union: Forming the sum is "structure-preserving" because the result is again a subspace. This means that the vector space structure is preserved when forming the sum. We can also think of this construction as a union because the sum contains both subspaces. The subspaces and are subsets of the sum . The sum is the smallest subspace that contains the two subspaces and . Just as you can form unions with sets, the sums of subspaces also work in the same way.

The direct sum is a special case of the sum of subspaces. This means that every direct sum is also a structure-preserving union. "Being direct" is a property of a sum of subspaces. We now want to see whether there is a property of the union of sets that corresponds to the directness of a sum.

Direct sums are characterized by the fact that the decomposition of the vectors in the sum is unique. If we have a vector with , where and , then the vectors and are unique. For a union of sets and , each element lies in or in . The element can also lie in both, which means that we generally do not clearly know where they lie. We cannot assign unambiguously if , i.e. in the intersection. This means that the assignment of elements is unique if is empty. In fact, this criterion corresponds exactly to the criterion for a sum to be direct: We want , which is the smallest possible vector space, so the intersection contains nothing more from and (except the zero, which it must contain anyway as a vector space). This is exactly the definition of a disjoint union. In other words, the direct sum of subspaces intuitively corresponds to the disjoint union of sets.

Basis and dimension

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We have seen that the direct sum is a special case of a sum of subspaces. So we can transfer everything we know about the vector space sum to the direct sum. We have already seen that the union of bases of and is a generating system of . This means if is a basis of and if is a basis of , then is a generating system of . If and are finite dimensional, we can use the dimension formula.

If the sum is direct, i.e. if , then we even have . Since , the following sum formula applies in the finite-dimensional case:

So the dimension of the sum space is exactly the sum of the dimensions and . If is a basis of and if is a basis of , then we can conclude

Since , the union of the bases of and is disjoint, i.e. . Therefore, we get . Because is a generating system of and because , we conclude that is a basis of .

We have thus seen that in finite dimensions, the union of the bases of and is a basis of . This also applies in general:

Theorem (Basis of the direct sum)

Let and be two subspaces of a -vector space . Assume that the sum of and is direct, that is, we can write . Let be a basis of and a basis of . Then the union of and is disjoint and is a basis of .

Proof (Basis of the direct sum)

We have already seen that is a generating system of . Therefore, we only have to show that the union is disjoint and linearly independent.

Proof step:

Suppose we have . Then , so . However, this is a contradiction to and , as a basis cannot contain the zero vector. Therefore, there can be no , i.e. .

Proof step: is linearly independent

Let

for any , and pairwise different, as well as . We must show that all and are equal to . This corresponds exactly to the definition of the linear independence of .

From

we conclude

This term is in (as a linear combination of elements in ) as well as in (as a linear combination of elements in ). Since is a direct sum, we obtain

From the linear independence of we conclude for all and from the linear independence of we conclude for all .

From this theorem, we may immediately conclude that

Exercises

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Exercise

Let and let . Consider the two subspaces and . Show that and determine and so that holds.

Solution

To show , we need to prove two things: First, that the sum of and is direct, i.e. . Secondly, we must show that the sum of and equals , i.e. .

Proof step:

Because and contain the zero vector as subspaces, is obvious. For the proof of the reverse inclusion, let be arbitrary. Then

for certain . From the first line of the vectors we get . So and therefore .

Proof step:

By definition, . The two vectors that span are obviously linearly independent, so . Furthermore, and . The dimension formula for subspaces then renders

The dimensions of the subspaces are therefore equal and follows from .

Alternatively, you could prove the equality by showing that every can be written as the sum of a and a .

We want to write as the sum of a vector in and a vector in . That means, we are looking for with

We may write this as a linear system:

From the first line we conclude . Plugging this into the second line gives . Again plugging this into the third line finally yields . Therefore, holds with

For the following two exercises, you should know what a linear map is.

Exercise (Self-inverse linear maps and subspaces)

Let be a -vector space and a linear map.

  1. Show that the subsets and are subspaces of .
  2. Let additionally , where denotes the identity map on . (A linear mapping with this property is called self-inverse.) Show that then holds for the two subspaces from the first exercise part.

Solution (Self-inverse linear maps and subspaces)

Solution sub-exercise 1:

We use the subspace criterion and show that and are non-empty subsets of that are closed under linear combinations. We only provide the proof for . The proof for works in the same way, you just have to replace all equations of the form "" with "".

Proof step:

This holds by definition of .

Proof step: is nonempty.

Since we have . So is nonempty.

Proof step: is closed under linear combinations.

Let and be arbitrary. Then

So the linear combination also lies in .

Solution sub-exercise 2:

In order to show , we need to prove two things: First, that the sum of and is direct, i.e. . Secondly, we must show that the sum of and equals , i.e. that every vector can be written as the sum of a vector and a vector .

Proof step:

Because and contain the zero vector as subspaces, is obvious. For the proof of the reverse inclusion, let be arbitrary. Then

i.e., , so . Because was arbitrary, we have thus shown that .

Proof step:

Because and are subsets of , is obvious. For the reverse inclusion, let be arbitrary. Then the following then applies:

Because , it follows from the linearity of that

So . Analogously, one shows :

So is a sum of a vector from and a vector from . Since was arbitrary, we conclude .

For this exercise, you need to know what the kernel and the image of a linear map are.

Exercise (Idempotent mappings)

Let be a linear map with . (A linear map with this property is called idempotent or a projection). Show: .

Solution (Idempotent mappings)

We show that und . By definition of the direct sum , the sum of is therefore indeed direct.

Proof step:

Since both the kernel and the image of are subspaces of , we immediately get . Let us now show the reverse inclusion .

Let be arbitrary. From the condition , we get , or in other words . Due to the linearity of , we conclude . Therefore, the element lies in the kernel of . Furthermore, lies in the image of by definition. Thus

is the sum of an element from and an element from . So is in . Because was arbitrary, we have shown .

Proof step:

Because and contain the zero vector as subspaces, is obvious. For the proof of the reverse inclusion, let be arbitrary. Then is an element of the kernel of and applies. Because is also in the image of , there is a so that . Because , we have

Since was arbitrary, we have thus shown .

In we can illustrate the statement from the previous exercise:

Example (Projektions in )

Let be . Then is linear. In addition, applies: For each vector , we have

The map is therefore a projection. Clearly, projects vectors in along the -axis onto the first angle bisector . In particular, . Further, maps the -axis to the zero vector, i.e. . Therefore, we indeed have as proven in the exercise.