1. Solution: Assume is 5-dimensional vector space with a basis , , . Define by Then , hence . Similarly, , hence .
2. Solution: Since , it follows that for any . Hence for any , i.e. .
3. Solution:
(a) Note that , hence if , , spans , it follows that is surjective.
(b) Note that if and only if since , , is linearly independent. Hence , then is injective.
4. Solution: Let , , be a basis of and , , , be a basis of . Define and by Then it is obvious that . However, and Then you can check that . Hence is not closed under addition, this implies it is not a subspace of .
5. Solution: Let , , , be a basis of . Define by Then , and . Hence .
6. Solution: By 3.22, we know that If , we will get that . This is impossible since dimension is an integer.
7. Solution: Let , , be a basis of and , , be a basis of , then we have . Define by and Then . However, we have Then by Problem 3 (b), it follows that is injective. Hence is not closed under addition, which implies it is not a subspace of .
8. Solution: Let , , be a basis of and , , be a basis of , then we have . Define by and Then . However, we have Then by Problem 3 (a), it follows that is surjective. Hence is not closed under addition, which implies it is not a subspace of .
9. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 5.
10. Solution: Note that , , spans , any can be written as a linear combination of , , . That is there are , , such that Since , it follows that Hence . On the other hand , , are contained in . By the definition of span, we conclude that , , spans .
See a similar problem in Linear Algebra Done Right Solution Manual Chapter 3 Problem 7.
11. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 6.
12. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 8.
13. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 9.
14. Solution: By 3.22, we have Note that and , it follows that Therefore is surjective by Problem 1 of Exercises 2.C.
15. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 10.
16. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 11.
17. Solution: By 3.22, it follows that for any injective , we have Hence there exists an injective linear map from to , then .
If , then let , , and , , be the bases of and , respectively. Define such that Here we use . Similar to Problem 3(b), we can show that is injective.
18. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 12. It is similar to Problem 17.
19. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 13.
20. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 14.
21. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 15.
22. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 16.
23. Solution: It is obvious that , hence . Let be a basis of , then by Problem 10, we have Hence . Thus
24. Solution: Suppose , we show that there exists such that .
Since is finite-dimensional, is finite-dimensional. Let be a basis of , then is linearly independent. Let , then . Hence for any , we have where and . Let such that . Such an exists since is linearly independent.
Since , we have Hence is spanned by .
On the other hand, we also have
Hence for all , i.e. .
Conversely, if , it is clear that .
25. Solution: If we assume . Let , , be a basis of , then we can find , , such that for since . Define by , then we have hence by uniqueness in 3.5.
If there exists such that , then for any , we have Hence .
26. Solution: Consider Problem 10 of Exercise 2C. Note that we have it follows that for any , there is a polynomial with degree . Note that is a subspace, hence Moreover, by Problem 10 of Exercise 2C, we have Hence , namely .
27. Solution: Denote the differentiation map by , then is of the type in Problem 26. Hence is surjective, thus there exists such that . Note that , the proof completes.
28. Solution: For any , there is a uniquely linear combination where are uniquely determined by , as , , is a basis of range . Denote by . Now we check that , . For , we have and Similarly, we also have Since , it follows that As , , is a basis of range , it follows that Similarly, by consider for , it follows that Hence .
29. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 4.
30. Solution: If , then .
If , then there is a is not in null , namely . By problem 29, we haveHence for any , can be written as where . Then , we have i.e. . Let , we complete the proof.
31. Solution: Let , , , , and , be a basis of and , respectively. Define by Define by Then and is not a scalar multiple of . Otherwise, if , then We get . However, from we get .