Linear Algebra/Combining Subspaces/Solutions

Solutions

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This exercise is recommended for all readers.
Problem 1

Decide if   is the direct sum of each   and  .

  1.  ,  
  2.  ,  
  3.  ,  
  4.  
  5.  ,  
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.