Linear Algebra/Definition and Examples of Vector Spaces/Solutions
Solutions[edit]
 Problem 1
Name the zero vector for each of these vector spaces.
 The space of degree three polynomials under the natural operations
 The space of matrices
 The space
 The space of realvalued functions of one natural number variable
 Answer
 The constant function
 The constant function
 This exercise is recommended for all readers.
 Problem 2
Find the additive inverse, in the vector space, of the vector.
 In , the vector .
 In the space ,
 In , the space of functions of the real variable under the natural operations, the vector .
 Answer
 This exercise is recommended for all readers.
 Problem 3
Show that each of these is a vector space.
 The set of linear polynomials under the usual polynomial addition and scalar multiplication operations.
 The set of matrices with real entries under the usual matrix operations.
 The set of threecomponent row vectors with their usual operations.
 The set
 Answer
Most of the conditions are easy to check; use Example 1.3 as a guide. Here are some comments.
 This is just like Example 1.3; the zero element is .
 The zero element of this space is the matrix of zeroes.
 The zero element is the vector of zeroes.
 Closure of addition involves noting that the sum
 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.)
 Under the operations inherited from , this set
 Under the operations inherited from , this set
 Under the usual matrix operations,
 Under the usual polynomial operations,
 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.
 It is not closed under addition; it fails to meet condition 1.
 It is not closed under addition.
 It is not closed under addition.
 It is not closed under scalar multiplication.
 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 twotall 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.
 Answer
 It is not a vector space.
 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.
 The diagonal matrices
 This set of matrices
 This set
 The set of functions
 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.
 Yes.
 Yes.
 No, it is not closed under addition. The vector of all 's, when added to itself, makes a nonmember.
 Yes.
 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 realvalued 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?
 and
 and
 Answer
 No: .
 No; the same calculation as the prior answer shows a contition 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 .
 The matrices under the usual operations
 The matrices (under their usual operations)
 This set of matrices
 This set of matrices
 Answer
 To see that the answer is , rewrite it as
 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 ,
 their vector sum is an element of . If then
 and
 .
 There is a zero vector such that for all .
 Each has an additive inverse such that . If are scalars, that is, members of ), and then
 each scalar multiple is in . If and then
 , and
 , and
 , and
 .
 This exercise is recommended for all readers.
 Problem 17
Prove these.
 Any vector is the additive inverse of the additive inverse of itself.
 Vector addition leftcancels: if then implies that .
 Answer
 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 .
 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
 Prove that every point, line, or plane thru the origin in is a vector space under the inherited operations.
 What if it doesn't contain the origin?
 Answer
 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.
 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 .
 Prove that if and only if .
 Prove that if and only if .
 Prove that any nontrivial vector space is infinite.
 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 .
 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.
 Where are scalars, holds if and only if . By the prior item, then .
 A nontrivial space has a vector . Consider the set . By the prior item this set is infinite.
 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 realvalued 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 .)
 The set of degree two polynomials with complex coefficients
 This set
 Answer
The check is routine. Note that "1" is and the zero elements are these.
 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

 Prove that a sum of four vectors can be associated in any way without changing the result.
 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
 A small rearrangement does the trick.
 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.
 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.
 Show that is a subspace of the vector space of degree two polynomials.
 Show that this is a subspace of the matrices.
 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
 We outline the check of the conditions from Definition 1.1. Additive closure holds because if and then
 This is similar to the prior answer.
 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.