Linear Algebra/Dimension/Solutions

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Solutions[edit]

Assume that all spaces are finite-dimensional unless otherwise stated.

This exercise is recommended for all readers.
Problem 1

Find a basis for, and the dimension of, .

Answer

One basis is , and so the dimension is three.

Problem 2

Find a basis for, and the dimension of, the solution set of this system.

Answer

The solution set is

so a natural basis is this

(checking linear independence is easy). Thus the dimension is three.

This exercise is recommended for all readers.
Problem 3

Find a basis for, and the dimension of, , the vector space of matrices.

Answer

For this space

this is a natural basis.

The dimension is four.

Problem 4

Find the dimension of the vector space of matrices

subject to each condition.

  1. and
  2. , , and
Answer
  1. As in the prior exercise, the space of matrices without restriction has this basis
    and so the dimension is four.
  2. For this space
    this is a natural basis.
    The dimension is three.
  3. Gauss' method applied to the two-equation linear system gives that and that . Thus, we have this description
    and so this is a natural basis.
    The dimension is two.
This exercise is recommended for all readers.
Problem 5

Find the dimension of each.

  1. The space of cubic polynomials such that
  2. The space of cubic polynomials such that and
  3. The space of cubic polynomials such that , , and
  4. The space of cubic polynomials such that , , , and
Answer

The bases for these spaces are developed in the answer set of the prior subsection.

  1. One basis is . The dimension is three.
  2. One basis is so the dimension is two.
  3. A basis is . The dimension is one.
  4. This is the trivial subspace of and so the basis is empty. The dimension is zero.
Problem 6

What is the dimension of the span of the set ? This span is a subspace of the space of all real-valued functions of one real variable.

Answer

First recall that , and so deletion of from this set leaves the span unchanged. What's left, the set , is linearly independent (consider the relationship where is the zero function, and then take , , and to conclude that each is zero). It is therefore a basis for its span. That shows that the span is a dimension three vector space.

Problem 7

Find the dimension of , the vector space of -tuples of complex numbers.

Answer

Here is a basis

and so the dimension is .

Problem 8

What is the dimension of the vector space of matrices?

Answer

A basis is

and thus the dimension is .

This exercise is recommended for all readers.
Problem 9

Show that this is a basis for .

(The results of this subsection can be used to simplify this job.)

Answer

In a four-dimensional space a set of four vectors is linearly independent if and only if it spans the space. The form of these vectors makes linear independence easy to show (look at the equation of fourth components, then at the equation of third components, etc.).

Problem 10

Refer to Example 2.9.

  1. Sketch a similar subspace diagram for .
  2. Sketch one for .
Answer
  1. The diagram for has four levels. The top level has the only three-dimensional subspace, itself. The next level contains the two-dimensional subspaces (not just the linear polynomials; any two-dimensional subspace, like those polynomials of the form ). Below that are the one-dimensional subspaces. Finally, of course, is the only zero-dimensional subspace, the trivial subspace.
  2. For , the diagram has five levels, including subspaces of dimension four through zero.
This exercise is recommended for all readers.
Problem 11
Where is a set, the functions form a vector space under the natural operations: the sum is the function given by and the scalar product is given by . What is the dimension of the space resulting for each domain?
Answer
  1. One
  2. Two
Problem 12

(See Problem 11.) Prove that this is an infinite-dimensional space: the set of all functions under the natural operations.

Answer

We need only produce an infinite linearly independent set. One is where is

the function that has value only at .

Problem 13

(See Problem 11.) What is the dimension of the vector space of functions , under the natural operations, where the domain is the empty set?

Answer

Considering a function to be a set, specifically, a set of ordered pairs , then the only function with an empty domain is the empty set. Thus this is a trivial vector space, and has dimension zero.

Problem 14

Show that any set of four vectors in is linearly dependent.

Answer

Apply Corollary 2.8.

Problem 15

Show that the set is a basis if and only if there is no plane through the origin containing all three vectors.

Answer

A plane has the form . (The first chapter also calls this a "-flat", and contains a discussion of why this is equivalent to the description often taken in Calculus as the set of points subject to a condition of the form ). When the plane passes through the origin we can take the particular vector to be . Thus, in the language we have developed in this chapter, a plane through the origin is the span of a set of two vectors.

Now for the statement. Asserting that the three are not coplanar is the same as asserting that no vector lies in the span of the other two— no vector is a linear combination of the other two. That's simply an assertion that the three-element set is linearly independent. By Corollary 2.12, that's equivalent to an assertion that the set is a basis for .

Problem 16
  1. Prove that any subspace of a finite dimensional space has a basis.
  2. Prove that any subspace of a finite dimensional space is finite dimensional.
Answer

Let the space be finite dimensional. Let be a subspace of .

  1. The empty set is a linearly independent subset of . By Corollary 2.10, it can be expanded to a basis for the vector space .
  2. Any basis for the subspace is a linearly independent set in the superspace . Hence it can be expanded to a basis for the superspace, which is finite dimensional. Therefore it has only finitely many members.
Problem 17

Where is the finiteness of used in Theorem 2.3?

Answer

It ensures that we exhaust the 's. That is, it justifies the first sentence of the last paragraph.

This exercise is recommended for all readers.
Problem 18

Prove that if and are both three-dimensional subspaces of then is non-trivial. Generalize.

Answer

Let be a basis for and let be a basis for . The set is linearly dependent as it is a six member subset of the five-dimensional space . Thus some member of is in the span of , and thus is more than just the trivial space .

Generalization: if are subspaces of a vector space of dimension and if then they have a nontrivial intersection.

Problem 19

Because a basis for a space is a subset of that space, we are naturally led to how the property "is a basis" interacts with set operations.

  1. Consider first how bases might be related by "subset". Assume that are subspaces of some vector space and that . Can there exist bases for and for such that ? Must such bases exist? For any basis for , must there be a basis for such that ? For any basis for , must there be a basis for such that ? For any bases for and , must be a subset of ?
  2. Is the intersection of bases a basis? For what space?
  3. Is the union of bases a basis? For what space?
  4. What about complement?

(Hint. Test any conjectures against some subspaces of .)

Answer

First, note that a set is a basis for some space if and only if it is linearly independent, because in that case it is a basis for its own span.

  1. The answer to the question in the second paragraph is "yes" (implying "yes" answers for both questions in the first paragraph). If is a basis for then is a linearly independent subset of . Apply Corollary 2.10 to expand it to a basis for . That is the desired . The answer to the question in the third paragraph is "no", which implies a "no" answer to the question of the fourth paragraph. Here is an example of a basis for a superspace with no sub-basis forming a basis for a subspace: in , consider the standard basis . No sub-basis of forms a basis for the subspace of that is the line .
  2. It is a basis (for its span) because the intersection of linearly independent sets is linearly independent (the intersection is a subset of each of the linearly independent sets). It is not, however, a basis for the intersection of the spaces. For instance, these are bases for :
    and , but is empty. All we can say is that the intersection of the bases is a basis for a subset of the intersection of the spaces.
  3. The union of bases need not be a basis: in
    have a union that is not linearly independent. A necessary and sufficient condition for a union of two bases to be a basis
    it is easy enough to prove (but perhaps hard to apply).
  4. The complement of a basis cannot be a basis because it contains the zero vector.
This exercise is recommended for all readers.
Problem 20

Consider how "dimension" interacts with "subset". Assume and are both subspaces of some vector space, and that .

  1. Prove that .
  2. Prove that equality of dimension holds if and only if .
  3. Show that the prior item does not hold if they are infinite-dimensional.
Answer
  1. A basis for is a linearly independent set in and so can be expanded via Corollary 2.10 to a basis for . The second basis has at least as many members as the first.
  2. One direction is clear: if then they have the same dimension. For the converse, let be a basis for . It is a linearly independent subset of and so can be expanded to a basis for . If then this basis for has no more members than does and so equals . Since and have the same bases, they are equal.
  3. Let be the space of finite-degree polynomials and let be the subspace of polynomails that have only even-powered terms . Both spaces have infinite dimension, but is a proper subspace.
? Problem 21

For any vector in and any permutation of the numbers , , ..., (that is, is a rearrangement of those numbers into a new order), define to be the vector whose components are , , ..., and (where is the first number in the rearrangement, etc.). Now fix and let be the span of . What are the possibilities for the dimension of ? (Gilbert, Krusemeyer & Larson 1993, Problem 47)

Answer

The possibilities for the dimension of are , , , and .

To see this, first consider the case when all the coordinates of are equal.

Then for every permutation , so is just the span of , which has dimension or according to whether is or not.

Now suppose not all the coordinates of are equal; let and with be among the coordinates of . Then we can find permutations and such that

for some . Therefore,

is in . That is, , where , , ..., is the standard basis for . Similarly, , ..., are all in . It is easy to see that the vectors , , ..., are linearly independent (that is, form a linearly independent set), so .

Finally, we can write

This shows that if then is in the span of , ..., (that is, is in the span of the set of those vectors); similarly, each will be in this span, so will equal this span and . On the other hand, if then the above equation shows that and thus , so and .

References[edit]

  • Gilbert, George T.; Krusemeyer, Mark; Larson, Loren C. (1993), The Wohascum County Problem Book, The Mathematical Association of America .