Linear Algebra/Definition and Examples of Vector Spaces/Solutions

Solutions edit

Problem 1

Name the zero vector for each of these vector spaces.

  1. The space of degree three polynomials under the natural operations
  2. The space of   matrices
  3. The space  
  4. The space of real-valued functions of one natural number variable
Answer
  1.  
  2.  
  3. The constant function  
  4. The constant function  
This exercise is recommended for all readers.
Problem 2

Find the additive inverse, in the vector space, of the vector.

  1. In  , the vector  .
  2. In the space  ,
     
  3. In  , the space of functions of the real variable   under the natural operations, the vector  .
Answer
  1.  
  2.  
  3.  
This exercise is recommended for all readers.
Problem 3

Show that each of these is a vector space.

  1. The set of linear polynomials   under the usual polynomial addition and scalar multiplication operations.
  2. The set of   matrices with real entries under the usual matrix operations.
  3. The set of three-component row vectors with their usual operations.
  4. The set
     
    under the operations inherited from  .
Answer

Most of the conditions are easy to check; use Example 1.3 as a guide. Here are some comments.

  1. This is just like Example 1.3; the zero element is  .
  2. The zero element of this space is the   matrix of zeroes.
  3. The zero element is the vector of zeroes.
  4. Closure of addition involves noting that the sum
     
    is in   because  . Closure of scalar multiplication is similar. Note that the zero element, the vector of zeroes, is in  .
This exercise is recommended for all readers.
Problem 4

Show that each of these is not a vector space. (Hint. Start by listing two members of each set.)

  1. Under the operations inherited from  , this set
     
  2. Under the operations inherited from  , this set
     
  3. Under the usual matrix operations,
     
  4. Under the usual polynomial operations,
     
    where   is the set of reals greater than zero
  5. Under the inherited operations,
     
Answer

In each item the set is called  . For some items, there are other correct ways to show that   is not a vector space.

  1. It is not closed under addition; it fails to meet condition 1.
     
  2. It is not closed under addition.
     
  3. It is not closed under addition.
     
  4. It is not closed under scalar multiplication.
     
  5. It is empty, violating condition 4.
Problem 5

Define addition and scalar multiplication operations to make the complex numbers a vector space over  .

Answer

The usual operations   and   suffice. The check is easy.

This exercise is recommended for all readers.
Problem 6

Is the set of rational numbers a vector space over   under the usual addition and scalar multiplication operations?

Answer

No, it is not closed under scalar multiplication since, e.g.,   is not a rational number.

Problem 7

Show that the set of linear combinations of the variables   is a vector space under the natural addition and scalar multiplication operations.

Answer

The natural operations are   and  . The check that this is a vector space is easy; use Example 1.3 as a guide.

Problem 8

Prove that this is not a vector space: the set of two-tall column vectors with real entries subject to these operations.

 
Answer

The " " operation is not commutative (that is, condition 2 is not met); producing two members of the set witnessing this assertion is easy.

Problem 9

Prove or disprove that   is a vector space under these operations.

  1.  
  2.  
Answer
  1. It is not a vector space.
     
  2. It is not a vector space.
     
This exercise is recommended for all readers.
Problem 10

For each, decide if it is a vector space; the intended operations are the natural ones.

  1. The diagonal   matrices
     
  2. This set of   matrices
     
  3. This set
     
  4. The set of functions  
  5. The set of functions  
Answer

For each "yes" answer, you must give a check of all the conditions given in the definition of a vector space. For each "no" answer, give a specific example of the failure of one of the conditions.

  1. Yes.
  2. Yes.
  3. No, it is not closed under addition. The vector of all  's, when added to itself, makes a nonmember.
  4. Yes.
  5. No,   is in the set but   is not (that is, condition 6 fails).
This exercise is recommended for all readers.
Problem 11

Prove or disprove that this is a vector space: the real-valued functions   of one real variable such that  .

Answer

It is a vector space. Most conditions of the definition of vector space are routine; we here check only closure. For addition,  . For scalar multiplication,  .

This exercise is recommended for all readers.
Problem 12

Show that the set   of positive reals is a vector space when " " is interpreted to mean the product of   and   (so that   is  ), and " " is interpreted as the  -th power of  .

Answer

We check Definition 1.1.

First, closure under " " holds because the product of two positive reals is a positive real. The second condition is satisfied because real multiplication commutes. Similarly, as real multiplication associates, the third checks. For the fourth condition, observe that multiplying a number by   won't change the number. Fifth, any positive real has a reciprocal that is a positive real.

The sixth, closure under " ", holds because any power of a positive real is a positive real. The seventh condition is just the rule that   equals the product of   and  . The eight condition says that  . The ninth condition asserts that  . The final condition says that  .

Problem 13

Is   a vector space under these operations?

  1.   and  
  2.   and  
Answer
  1. No:  .
  2. No; the same calculation as the prior answer shows a condition in the definition of a vector space that is violated. Another example of a violation of the conditions for a vector space is that  .
Problem 14

Prove or disprove that this is a vector space: the set of polynomials of degree greater than or equal to two, along with the zero polynomial.

Answer

It is not a vector space since it is not closed under addition, as   is not in the set.

Problem 15

At this point "the same" is only an intuition, but nonetheless for each vector space identify the   for which the space is "the same" as  .

  1. The   matrices under the usual operations
  2. The   matrices (under their usual operations)
  3. This set of   matrices
     
  4. This set of   matrices
     
Answer
  1.  
  2.  
  3.  
  4. To see that the answer is  , rewrite it as
     
    so that there are two parameters.
This exercise is recommended for all readers.
Problem 16

Using   to represent vector addition and   for scalar multiplication, restate the definition of vector space.

Answer

A vector space (over  ) consists of a set   along with two operations " " and " " subject to these conditions. Where  ,

  1. their vector sum   is an element of  . If   then
  2.   and
  3.  .
  4. There is a zero vector   such that   for all  .
  5. Each   has an additive inverse   such that  . If   are scalars, that is, members of  ), and   then
  6. each scalar multiple   is in  . If   and   then
  7.  , and
  8.  , and
  9.  , and
  10.  .
This exercise is recommended for all readers.
Problem 17

Prove these.

  1. Any vector is the additive inverse of the additive inverse of itself.
  2. Vector addition left-cancels: if   then   implies that  .
Answer
  1. Let   be a vector space, assume that  , and assume that   is the additive inverse of   so that  . Because addition is commutative,  , so therefore   is also the additive inverse of  .
  2. Let   be a vector space and suppose  . The additive inverse of   is   so   gives that  , which says that   and so  .
Problem 18

The definition of vector spaces does not explicitly say that   (it instead says that  ). Show that it must nonetheless hold in any vector space.

Answer

Addition is commutative, so in any vector space, for any vector   we have that  .

This exercise is recommended for all readers.
Problem 19

Prove or disprove that this is a vector space: the set of all matrices, under the usual operations.

Answer

It is not a vector space since addition of two matrices of unequal sizes is not defined, and thus the set fails to satisfy the closure condition.

Problem 20

In a vector space every element has an additive inverse. Can some elements have two or more?

Answer

Each element of a vector space has one and only one additive inverse.

For, let   be a vector space and suppose that  . If   are both additive inverses of   then consider  . On the one hand, we have that it equals  . On the other hand we have that it equals  . Therefore,  .

Problem 21
  1. Prove that every point, line, or plane thru the origin in   is a vector space under the inherited operations.
  2. What if it doesn't contain the origin?
Answer
  1. Every such set has the form   where either or both of   may be  . With the inherited operations, closure of addition   and scalar multiplication   are easy. The other conditions are also routine.
  2. No such set can be a vector space under the inherited operations because it does not have a zero element.
This exercise is recommended for all readers.
Problem 22

Using the idea of a vector space we can easily reprove that the solution set of a homogeneous linear system has either one element or infinitely many elements. Assume that   is not  .

  1. Prove that   if and only if  .
  2. Prove that   if and only if  .
  3. Prove that any nontrivial vector space is infinite.
  4. Use the fact that a nonempty solution set of a homogeneous linear system is a vector space to draw the conclusion.
Answer

Assume that   is not  .

  1. One direction of the if and only if is clear: if   then  . For the other way, let   be a nonzero scalar. If   then   shows that  , contrary to the assumption.
  2. Where   are scalars,   holds if and only if  . By the prior item, then  .
  3. A nontrivial space has a vector  . Consider the set  . By the prior item this set is infinite.
  4. The solution set is either trivial, or nontrivial. In the second case, it is infinite.
Problem 23

Is this a vector space under the natural operations: the real-valued functions of one real variable that are differentiable?

Answer

Yes. A theorem of first semester calculus says that a sum of differentiable functions is differentiable and that  , and that a multiple of a differentiable function is differentiable and that  .

Problem 24

A vector space over the complex numbers   has the same definition as a vector space over the reals except that scalars are drawn from   instead of from  . Show that each of these is a vector space over the complex numbers. (Recall how complex numbers add and multiply:   and  .)

  1. The set of degree two polynomials with complex coefficients
  2. This set
     
Answer

The check is routine. Note that "1" is   and the zero elements are these.

  1.  
  2.  
Problem 25

Name a property shared by all of the  's but not listed as a requirement for a vector space.

Answer

Notably absent from the definition of a vector space is a distance measure.

This exercise is recommended for all readers.
Problem 26
  1. Prove that a sum of four vectors   can be associated in any way without changing the result.
     
    This allows us to simply write " " without ambiguity.
  2. Prove that any two ways of associating a sum of any number of vectors give the same sum. (Hint. Use induction on the number of vectors.)
Answer
  1. A small rearrangement does the trick.
     
    Each equality above follows from the associativity of three vectors that is given as a condition in the definition of a vector space. For instance, the second " " applies the rule   by taking   to be  , taking   to be  , and taking   to be  .
  2. The base case for induction is the three vector case. This case   is required of any triple of vectors by the definition of a vector space. For the inductive step, assume that any two sums of three vectors, any two sums of four vectors, ..., any two sums of   vectors are equal no matter how the sums are parenthesized. We will show that any sum of   vectors equals this one  . Any parenthesized sum has an outermost " ". Assume that it lies between   and   so the sum looks like this.
     
    The second half involves fewer than   additions, so by the inductive hypothesis we can re-parenthesize it so that it reads left to right from the inside out, and in particular, so that its outermost " " occurs right before  .
     
    Apply the associativity of the sum of three things
     
    and finish by applying the inductive hypothesis inside these outermost parenthesis.
Problem 27

For any vector space, a subset that is itself a vector space under the inherited operations (e.g., a plane through the origin inside of  ) is a subspace.

  1. Show that   is a subspace of the vector space of degree two polynomials.
  2. Show that this is a subspace of the   matrices.
     
  3. Show that a nonempty subset   of a real vector space is a subspace if and only if it is closed under linear combinations of pairs of vectors: whenever   and   then the combination   is in  .
Answer
  1. We outline the check of the conditions from Definition 1.1. Additive closure holds because if   and   then
     
    is in the set since   is zero. The second through fifth conditions are easy. Closure under scalar multiplication holds because if   then
     
    is in the set as   is zero. The remaining conditions here are also easy.
  2. This is similar to the prior answer.
  3. Call the vector space  . We have two implications: left to right, if   is a subspace then it is closed under linear combinations of pairs of vectors and, right to left, if a nonempty subset is closed under linear combinations of pairs of vectors then it is a subspace. The left to right implication is easy; we here sketch the other one by assuming   is nonempty and closed, and checking the conditions of Definition 1.1. First, to show closure under addition, if   then   as  . Second, for any  , because addition is inherited from  , the sum   in   equals the sum   in   and that equals the sum   in   and that in turn equals the sum   in  . The argument for the third condition is similar to that for the second. For the fourth, suppose that   is in the nonempty set   and note that  ; showing that the   of   acts under the inherited operations as the additive identity of   is easy. The fifth condition is satisfied because for any   closure under linear combinations shows that the vector   is in  ; showing that it is the additive inverse of   under the inherited operations is routine. The proofs for the remaining conditions are similar.