Chapter 2 Solutions

Here are the solutions to the exercises in Chapter 2 of Mathematical Physics: A Modern Introduction to Its Foundations by Sadri Hassani.

• 2.1
• 2.2
• 2.3
• 2.4
• 2.5
• 2.6
• 2.7
• 2.8

2.1

Let denote the set of positive real numbers. Define the “sum” of two elements of to be their usual product, and define scalar multiplication by elements of as being given by where and . With these operations, show that is a vector space over .

To show that is a vector space over with the given operations, we need to verify that it satisfies the vector space axioms. That is, we need to check that it is an abelian group under the defined addition, and that scalar multiplication satisfies the necessary properties.

Let and . To avoid confusion, we will denote the newly defined addition by , so that .

As is a field, the usual multiplication operation forms an abelian group. The identity element is , since for any , we have . The inverse of any is , since . Therefore, the newly defined addition satisfies the axioms for an abelian group.

Next, we need to verify the properties of scalar multiplication: .

1. As , for any (closure),
2. (compatibility), and
3. (identity).

Finally, we need to check the distributive properties:

1. , and
2. .

Since all the vector space axioms are satisfied, we conclude that is indeed a vector space over with the given operations.

2.2

Show that the intersection of two subspaces is also a subspace.

Let and be two subspaces of a vector space . Let be their intersection.

Let be two arbitrary vectors in the intersection, and let be two scalars from the underlying field.

As the vectors are in the intersection, we have and . Therefore, , since is a subspace, and similarly, . As is in both and , it must also be in their intersection . Thus, we have shown that for any and any , . Therefore, is cloesd under vector addition and scalar multiplication, and is thus a subspace of .

2.3

For each of the following subsets of , determine whether it is a subspace of :

Let and be two arbitrary vectors in , and let be two scalars. Let's call the sets in the question , , and respectively.

1. Let . Then, we have and . We have

   Does this vector belong to ?

   Therefore, , and is a subspace of .

2. Let . Then, we have and .

   For this to be equal to , we need . However, this is not true for arbitrary . Therefore, is not closed under vector addition and scalar multiplication, and is not a subspace of .

3. Let . Then, we have and .

   This expression is not necessarily equal to for arbitrary . Therefore, is not closed under vector addition and scalar multiplication, and is not a subspace of .

   We can also show this with a simple counterexample. Let and . Both vectors are in , but their sum is not in since .

2.4

Prove that the components of a vector in a given basis are unique.

Let be a vector space over a field , and let be a basis for . Assume that a vector can be expressed in two different ways in this basis:

where for .

Subtracting the second equation from the first gives

As the basis is linearly independent by definition, the only solution to this equation is for all . This implies that for all . Therefore, the components of the vector in the given basis are unique.

2.5

Show that the following vectors form a basis for (or ):

2.6

Prove Theorem 2.1.6.

Here is the proof. The TL;DR is as follows:

Let be a set of vectors in a vector space over a field . Trivially, as . And , so is a subspace of .

2.7

Let be a subspace of defined by

Find a basis for .

Notice that choosing determines , and choosing determines . And and determine . Therefore, the only free variables are and . Letting and , we have

Thus, a basis for is given by the set

2.8

Let and be subspaces of . Show that

1. . Hint: Let be a basis of . Extend this to , a basis for , and to , a basis for . Now show that is a basis for .
2. If and , then .
3. If , then .

1. Let be an arbitrary vector in the sum of the two subspaces. Then, we can write where and . Using the bases given in the hint, we can express and as follows:

   where for and respectively. Therefore, we have

   This shows that any vector in can be expressed as a linear combination of the vectors in the set . Next, we need to show that this set is linearly independent. Suppose . Then,

   where for , , and respectively. Rearranging gives

   The left-hand side is in , while the right-hand side is in . Since the basis vectors and are not in , the only solution is for all , for all , and for all . Therefore, the set is linearly independent.

   As such, it forms a basis for , and we have

   This completes the proof.

2. Given that and , using the result from the first part, we have . This implies that the only vector in the intersection is the zero vector, i.e., . Therefore, by definition, .

3. From the first part, we have . If , then . This implies that there exists at least one non-zero vector in the intersection, i.e., .