1. Solution: Suppose for some basis , , of and some basis , , of , the matrix of has at most nonzero entries. Then there are at most nonzero vectors in , , . Note that , it follows that We get a contradiction, hence completing the proof.
2. Solution: The basis of is , , , . The corresponding basis of is , , (with the indicated order).(Check it!)
3. Solution: We use the notation and the proof of 3.22. Extend , , to a basis of as , , , , , . Then with respect to the basis , , , , , of and the basis , , , , , of , all entries of are 0 except that the entries in row , column , equal for .
4. Solution: If , then any basis , , of will satisfy the desired conditions. If , then any basis , , of such that will satisfy the desired conditions.
5. Solution: Let , , be a basis of , denote the first row of with respect to the bases , , and , , by . If , then we can choose , . If , suppose . Then let for , . Then you can check that , , satisfies the desired conditions.
6. Solution: Suppose there exist a basis , , of and a basis , , of such that with respect to these bases, all entries of equal . Then Hence , it follows that .
Conversely, if , then . Let , , , be a basis of such that , , . Note that , hence we can extend it to a basis of as , , , . Let and , for . It is obvious that , , is a basis of and , , is a basis of . Now we can directly check that all entries of with respect to these bases equal .
7. Solution: Given a basis , , of and a basis , , of , denote and with respect to these bases by and , respectively. Then we have and Hence it follows that the entries in row , column of with respect to these bases are . By 3.35, we deduce that .
8. Verify 3.38.
Solution: It is almost the same as the previous Exercise.
9. Prove 3.52.
Solution: It is almost the same as Problem 11. Just consider the entries.
10. Suppose is an -by- matrix and is an -by- matrix. Prove that for . In other words, show that row of equals (row of ) times .
Solution: It is almost the same as Problem 11. Just consider the entries.
These exercises are tedious. I prefer solving other interesting exercises… If you have problems regarding to them, please make a comment.
11. Solution: By 3.41, we have It is obvious that . Hence by 3,35. Thus we deduce that .
12. Solution: Let and , then we have andHence .
13. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 17.
14. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 18.
15. Solution: Note that , denote , then by definition (3.41) Similarly, the entry in row , column , of is Hence by , we have