1. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 22. It is almost the same.
2. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 25.
3. Solution: If there exists an invertible operator such that for every , then is injective since is injective.
If is injective. Assume , , is a basis of , we can extend it to a basis of as , , , , , . Since is injective, by Problem 9 of Exercises 3B, we have , , is linearly independent in . Hence we can extend it to a basis of as , , , , , . Define as below The existence of is guaranteed by 3.5(unique). Then for any , , we have Moreover, is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.
Compare this problem with Problem 11 of Exercises 3A.
4. Solution: If we assume . Since is finite-dimensional, so is range . Let , , be a basis of , then there exist , , such that Now we will show that . For any , we for some , , . Hence namely this implies . Moreover, if , then we have Note that , , is linearly independent, it follows . Thus we have Similarly, , , is linearly independent. For if , we have Note that , it follows that Thus . Now extend , , to a basis of as , , , , , and , , to a basis of as , , , , , . Define by Note that any can be expressed as where and , , . Hence we have namely . Moreover, is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.
If there exists an invertible operator such that , then for any , we have As is invertible, we have . Hence , it follows that . Similarly, consider , . Thus we conclude null = null .
Compare this problem with Problem 24 of Exercises 3B.
5. Solution: If we assume . Let , , be a basis of null , then we can extend it to a basis of as , , , , , . Then range is and is linearly independent. There exist , , such that for since . As is linearly independent, it follows that , , is linearly independent by Problem 4 of Exercises 3A. Note that range range implies null and null have the same dimension. Let , , be a basis of null , then , , , , , is a basis of by the proof of 3.22. Define by and , then we have and hence by uniqueness in 3.5. Moreover, is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.
If there exists an invertible operator such that , then for any , we have Hence . As is invertible, we have . Similarly, we conclude . Thus range range .
Compare this problem with Problem 25 of Exercises 3B.
6. Solution: If there exist invertible operators and such that , then . Hence null = null by Problem 4. Note that we have range = range by Problem 5, it follows that Hence .
Conversely, if null null . Let , , be a basis of null , then we can extend it to a basis of as , , , , , . Let , , be a basis of null , then we can extend it to a basis of as , , , , , . Then , , is linearly independent in , hence we can extend it to a basis of as , , , , , . Similarly, , , is linearly independent in , hence we can extend it to a basis of as , , , , , . Define by Define by Since and map basis to basis, hence and are invertible(surjective). Moreover, it is easy to check and hence .
Compare this problem with Problem 4 and Problem 5.
7. Solution: a) Let , then hence , namely is closed under addition. For every , hence , namely is closed under scalar multiplication. Therefore is a subspace of .
(b) Since , we can extend it to a basis of , namely , , , . Let , , be a basis of . Under these bases, we have a isometric between and by 3.60. Moreover, if and only if the first column vector of is zero. Hence is exactly all matrices in such the first column vector is zero. Now , it is easily seen thatOr it is equivalent to the dimension all linear map from to . Then we can use 3.61.
8. Solution: Let , , be a basis of . Since is surjective, there exist , , such that . Moreover, by Problem 4 of Exercises 3A, it follows that , , is linearly independent. Consider , then maps a basis of to a basis of . Hence is an isomorphism of onto .
9. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 22.
10. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 23.
11. Solution: By Problem 9, we have is invertible. Again by Problem 9, we have is invertible. Multiply both side of by on left, we get . Multiply both side of by on right, we have . Multiply both side of by on left, we get .
12. Solution: Consider . Define , , by and . Then . However is not surjective.
13. Solution: Since is finite-dimensional, the surjectivity of implies is invertible. Hence is invertible by Problem 11.
14. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 20.
15. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 21.
16. Suppose is finite-dimensional and . Prove that is a scalar multiple of the identity if and only if for every .
Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 24.
17. Solution: Let , , be a basis of . Suppose , then there is nonzero . This implies there is some such that . Let Since , there is some such that . Define by , where if and for every .
By definition, for every , we have By assumption, . Note that is a subspace of . It follows that and . This implies , hence . Therefore for any , , namely .
18. Solution: For given , define by . Then (check it). Hence we can define by . It suffices to show is an isomorphic from to . First, we should check is linear. For every and , we have Hence is linear. Then imples for every , thus . We conclude injectivity. For any , if then . Hence every can be express as , namely is surjective.
19. Solution:
(a) Since for every nonzero polynomial , we have for any . Note that is injective, hence is injective. Note that is finite-dimensional, we conclude is surjective by 3.56. Hence is surjective as any polynomial must be contained in some .
(b) We argue it by induction on . It is true for . Suppose it is true for . As is surjective, every polynomial with degree can be attained by for some . Moreover, is surjective, every polynomial with degree can be attained by for some . If there exist with degree such that . Then there exist some such that since is surjective. Hence . Note that is injective and (different degrees), we get a contradiction. Hence for every nonzero with degree . The proof is complete.
20. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 26.