Linear Algebra/Combining Subspaces/Solutions

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Solutions[edit]

This exercise is recommended for all readers.
Problem 1

Decide if is the direct sum of each and .

  1. ,
  2. ,
  3. ,
  4. ,
Answer

With each of these we can apply Lemma 4.15.

  1. Yes. The plane is the sum of this and because for any scalars and
    shows that the general vector is a sum of vectors from the two parts. And, these two subspaces are (different) lines through the origin, and so have a trivial intersection.
  2. Yes. To see that any vector in the plane is a combination of vectors from these parts, consider this relationship.
    We could now simply note that the set
    is a basis for the space (because it is clearly linearly independent, and has size two in ), and thus ther is one and only one solution to the above equation, implying that all decompositions are unique. Alternatively, we can solve
    to get that and , and so we have
    as required. As with the prior answer, each of the two subspaces is a line through the origin, and their intersection is trivial.
  3. Yes. Each vector in the plane is a sum in this way
    and the intersection of the two subspaces is trivial.
  4. No. The intersection is not trivial.
  5. No. These are not subspaces.
This exercise is recommended for all readers.
Problem 2

Show that is the direct sum of the -plane with each of these.

  1. the -axis
  2. the line
Answer

With each of these we can use Lemma 4.15.

  1. Any vector in can be decomposed as this sum.
    And, the intersection of the -plane and the -axis is the trivial subspace.
  2. Any vector in can be decomposed as
    and the intersection of the two spaces is trivial.
Problem 3

Is the direct sum of and ?

Answer

It is. Showing that these two are subspaces is routine. To see that the space is the direct sum of these two, just note that each member of has the unique decomposition .

This exercise is recommended for all readers.
Problem 4

In , the even polynomials are the members of this set

and the odd polynomials are the members of this set.

Show that these are complementary subspaces.

Answer

To show that they are subspaces is routine. We will argue they are complements with Lemma 4.15. The intersection is trivial because the only polynomial satisfying both conditions and is the zero polynomial. To see that the entire space is the sum of the subspaces , note that the polynomials , , , etc., are in and also note that the polynomials , , etc., are in . Hence any member of is a combination of members of and .

Problem 5

Which of these subspaces of

: the -axis,      :the -axis,      :the -axis,     
:the plane ,      :the -plane

can be combined to

  1. sum to ?
  2. direct sum to ?
Answer

Each of these is .

  1. These are broken into lines for legibility.

    , , , ,
        , , ,
    , , ,
    , ,
    ,
    , ,
    , ,
    , ,

  2. , , , ,
This exercise is recommended for all readers.
Problem 6

Show that .

Answer

Clearly each is a subspace. The bases for the subspaces, when concatenated, form a basis for the whole space.

Problem 7

What is if ?

Answer

It is .

Problem 8

Does Example 4.5 generalize? That is, is this true or false:if a vector space has a basis then it is the direct sum of the spans of the one-dimensional subspaces ?

Answer

True by Lemma 4.8.

Problem 9

Can be decomposed as a direct sum in two different ways? Can ?

Answer

Two distinct direct sum decompositions of are easy to find. Two such are and , and also and . (Many more are possible, for example and its trivial subspace.)

In contrast, any partition of 's single-vector basis will give one basis with no elements and another with a single element. Thus any decomposition involves and its trivial subspace.

Problem 10

This exercise makes the notation of writing "" between sets more natural. Prove that, where are subspaces of a vector space,

and so the sum of subspaces is the subspace of all sums.

Answer

Set inclusion one way is easy: is a subset of because each is a sum of vectors from the union.

For the other inclusion, to any linear combination of vectors from the union apply commutativity of vector addition to put vectors from first, followed by vectors from , etc. Add the vectors from to get a , add the vectors from to get a , etc. The result has the desired form.

Problem 11

(Refer to Example 4.19. This exercise shows that the requirement that pariwise intersections be trivial is genuinely stronger than the requirement only that the intersection of all of the subspaces be trivial.) Give a vector space and three subspaces , , and such that the space is the sum of the subspaces, the intersection of all three subspaces is trivial, but the pairwise intersections , , and are nontrivial.

Answer

One example is to take the space to be , and to take the subspaces to be the -plane, the -plane, and the -plane.

This exercise is recommended for all readers.
Problem 12

Prove that if then is trivial whenever . This shows that the first half of the proof of Lemma 4.15 extends to the case of more than two subspaces. (Example 4.19 shows that this implication does not reverse; the other half does not extend.)

Answer

Of course, the zero vector is in all of the subspaces, so the intersection contains at least that one vector. By the definition of direct sum the set is independent and so no nonzero vector of is a multiple of a member of , when . In particular, no nonzero vector from equals a member of .

Problem 13

Recall that no linearly independent set contains the zero vector. Can an independent set of subspaces contain the trivial subspace?

Answer

It can contain a trivial subspace; this set of subspaces of is independent: . No nonzero vector from the trivial space is a multiple of a from the -axis, simply because the trivial space has no nonzero vectors to be candidates for such a multiple (and also no nonzero vector from the -axis is a multiple of the zero vector from the trivial subspace).

This exercise is recommended for all readers.
Problem 14

Does every subspace have a complement?

Answer

Yes. For any subspace of a vector space we can take any basis for that subspace and extend it to a basis for the whole space. Then the complement of the original subspace has this for a basis: .

This exercise is recommended for all readers.
Problem 15

Let be subspaces of a vector space.

  1. Assume that the set spans , and that the set spans . Can span ? Must it?
  2. Assume that is a linearly independent subset of and that is a linearly independent subset of . Can be a linearly independent subset of ? Must it?
Answer
  1. It must. Any member of can be written where and . As spans , the vector is a combination of members of . Similarly is a combination of members of .
  2. An easy way to see that it can be linearly independent is to take each to be the empty set. On the other hand, in the space , if and and and , then their union is not independent.
Problem 16

When a vector space is decomposed as a direct sum, the dimensions of the subspaces add to the dimension of the space. The situation with a space that is given as the sum of its subspaces is not as simple. This exercise considers the two-subspace special case.

  1. For these subspaces of find , , , and .
  2. Suppose that and are subspaces of a vector space. Suppose that the sequence is a basis for . Finally, suppose that the prior sequence has been expanded to give a sequence that is a basis for , and a sequence that is a basis for . Prove that this sequence
    is a basis for for the sum .
  3. Conclude that .
  4. Let and be eight-dimensional subspaces of a ten-dimensional space. List all values possible for .
Answer
  1. The intersection and sum are
    which have dimensions one and three.
  2. We write for the basis for , we write for the basis for , we write for the basis for , and we write for the basis under consideration. To see that that spans , observe that any vector from can be written as a linear combination of the vectors in , simply by expressing in terms of and expressing in terms of . We finish by showing that is linearly independent. Consider
    can be rewritten in this way.
    Note that the left side sums to a vector in while right side sums to a vector in , and thus both sides sum to a member of . Since the left side is a member of , it is expressible in terms of the members of , which gives the combination of 's from the left side above as equal to a combination of 's. But, the fact that the basis is linearly independent shows that any such combination is trivial, and in particular, the coefficients , ..., from the left side above are all zero. Similarly, the coefficients of the 's are all zero. This leaves the above equation as a linear relationship among the 's, but is linearly independent, and therefore all of the coefficients of the 's are also zero.
  3. Just count the basis vectors in the prior item:, and , and , and .
  4. We know that . Because , we know that must have dimension greater than that of , that is, must have dimension eight, nine, or ten. Substituting gives us three possibilities or or . Thus must be either eight, seven, or six. (Giving examples to show that each of these three cases is possible is easy, for instance in .)
Problem 17

Let and for each index suppose that is a linearly independent subset of . Prove that the union of the 's is linearly independent.

Answer

Expand each to a basis for . The concatenation of those bases is a basis for and thus its members form a linearly independent set. But the union is a subset of that linearly independent set, and thus is itself linearly independent.

Problem 18

A matrix is symmetric if for each pair of indices and , the entry equals the entry. A matrix is antisymmetric if each entry is the negative of the entry.

  1. Give a symmetric matrix and an antisymmetric matrix. (Remark. For the second one, be careful about the entries on the diagional.)
  2. What is the relationship between a square symmetric matrix and its transpose? Between a square antisymmetric matrix and its transpose?
  3. Show that is the direct sum of the space of symmetric matrices and the space of antisymmetric matrices.
Answer
  1. Two such are these.
    For the antisymmetric one, entries on the diagonal must be zero.
  2. A square symmetric matrix equals its transpose. A square antisymmetric matrix equals the negative of its transpose.
  3. Showing that the two sets are subspaces is easy. Suppose that .To express as a sum of a symmetric and an antisymmetric matrix, we observe that
    and note the first summand is symmetric while the second is antisymmetric. Thus is the sum of the two subspaces. To show that the sum is direct, assume a matrix is both symmetric and antisymmetric . Then and so all of 's entries are zeroes.
Problem 19

Let be subspaces of a vector space. Prove that . Does the inclusion reverse?

Answer

Assume that . Then where and . Note that and, as a subspace is closed under addition, . Thus .

This example proves that the inclusion may be strict: in take to be the -axis, take to be the -axis, and take to be the line . Then and are trivial and so their sum is trivial. But is all of so is the -axis.

Problem 20

The example of the -axis and the -axis in shows that does not imply that . Can and happen?

Answer

It happens when at least one of is trivial. But that is the only way it can happen.

To prove this, assume that both are non-trivial, select nonzero vectors from each, and consider . This sum is not in because would imply that is in , which violates the assumption of the independence of the subspaces. Similarly, is not in . Thus there is an element of that is not in .

This exercise is recommended for all readers.
Problem 21

Our model for complementary subspaces, the -axis and the -axis in , has one property not used here. Where is a subspace of we define the orthogonal complement of to be

(read " perp").

  1. Find the orthocomplement of the -axis in .
  2. Find the orthocomplement of the -axis in .
  3. Find the orthocomplement of the -plane in .
  4. Show that the orthocomplement of a subspace is a subspace.
  5. Show that if is the orthocomplement of then is the orthocomplement of .
  6. Prove that a subspace and its orthocomplement have a trivial intersection.
  7. Conclude that for any and subspace we have that .
  8. Show that equals the dimension of the enclosing space.
Answer
  1. The set
    is easily seen to be the -axis.
  2. The -plane.
  3. The -axis.
  4. Assume that is a subspace of some . Because contains the zero vector, since that vector is perpendicular to everything, we need only show that the orthocomplement is closed under linear combinations of two elements. If then and for all . Thus for all and so is closed under linear combinations.
  5. The only vector orthogonal to itself is the zero vector.
  6. This is immediate.
  7. To prove that the dimensions add, it suffices by Corollary 4.13 and Lemma 4.15 to show that is the trivial subspace . But this is one of the prior items in this problem.
This exercise is recommended for all readers.
Problem 22

Consider Corollary 4.13. Does it work both ways— that is, supposing that , is if and only if ?

Answer

Yes. The left-to-right implication is Corollary 4.13. For the other direction, assume that . Let be bases for . As is the sum of the subspaces, any can be written and expressing each as a combination of vectors from the associated basis shows that the concatenation spans . Now, that concatenation has members, and so it is a spanning set of size . The concatenation is therefore a basis for . Thus is the direct sum.

Problem 23

We know that if then there is a basis for that splits into a basis for and a basis for . Can we make the stronger statement that every basis for splits into a basis for and a basis for ?

Answer

No. The standard basis for does not split into bases for the complementary subspaces the line and the line .

Problem 24

We can ask about the algebra of the "" operation.

  1. Is it commutative; is ?
  2. Is it associative; is ?
  3. Let be a subspace of some vector space. Show that .
  4. Must there be an identity element, a subspace such that for all subspaces ?
  5. Does left-cancelation hold:if then ? Right cancelation?
Answer
  1. Yes, for all subspaces because each side is the span of .
  2. This one is similar to the prior one— each side of that equation is the span of .
  3. Because this is an equality between sets, we can show that it holds by mutual inclusion. Clearly . For just recall that every subset is closed under addition so any sum of the form is in .
  4. In each vector space, the identity element with respect to subspace addition is the trivial subspace.
  5. Neither of left or right cancelation needs to hold. For an example, in take to be the -plane, take to be the -axis, and take to be the -axis.
Problem 25

Consider the algebraic properties of the direct sum operation.

  1. Does direct sum commute: does imply that ?
  2. Prove that direct sum is associative:.
  3. Show that is the direct sum of the three axes (the relevance here is that by the previous item, we needn't specify which two of the three axes are combined first).
  4. Does the direct sum operation left-cancel:does imply ? Does it right-cancel?
  5. There is an identity element with respect to this operation. Find it.
  6. Do some, or all, subspaces have inverses with respect to this operation:is there a subspace of some vector space such that there is a subspace with the property that equals the identity element from the prior item?
Answer
  1. They are equal because for each, is the direct sum if and only if each can be written in a unique way as a sum and .
  2. They are equal because for each, is the direct sum if and only if each can be written in a unique way as a sum of a vector from each and .
  3. Any vector in can be decomposed uniquely into the sum of a vector from each axis.
  4. No. For an example, in take to be the -axis, take to be the -axis, and take to be the line .
  5. In any vector space the trivial subspace acts as the identity element with respect to direct sum.
  6. In any vector space, only the trivial subspace has a direct-sum inverse (namely, itself). One way to see this is that dimensions add, and so increase.