1. Solution: (a) is a subspace of . By 1.34, to show a subset is a subspace, we just need to check Additive identity, Closed under addition and Closed under scalar multiplication.
Additive identity: it is clear that the additive identity of is contained in .
Closed under addition: if , thenHencethis means is also in .
Closed under scalar multiplication: if , then . For any , we haveThis means
(b) is not a subspace of since is not in it.
(c) is not a subspace of since and are in it, but the sum is not in it.
(d) is a subspace of .
Additive identity: it is clear that the additive identity of is contained in .
Closed under addition: if , then and . Hencethis means is also in .
Closed under scalar multiplication: if , then . For any , we have . This means
2. Solution: (a) If this set is a subspace of , then , then . Hence . Follow the same steps of Problem 1 (d), we will see the set is a subspace of if .
(b) (c) and (d) is similar to Problem 3 and 4.
Now let us consider (e). Denote the set of all sequences of complex numbers with limit by .
Additive identity: it is clear that .
Closed under addition: if , thenIt is easy to seeThis means .
Closed under scalar multiplication: if , thenFor any , it is easy to seeThis means .
3. Solution: Denote the set of differentiable real-valued functions on the interval such that by .
Additive identity: it is clear that the constant function is contained in .
Closed under addition: if , then and are differentiable real-valued functions. So is . Moreover,This concludes is closed under addition.
Closed under scalar multiplication: if , for any , then is differentiable real-valued functions. So is . Moreover,This deduces is closed under scalar multiplication.
4. Solution: Denote the set of continuous real-valued functions on the interval such that by .
If is a subspace of , then for any , we have . Because is a subspace of , it follows that for any . Hencethis happens if and only if .
Now if , then for any and . We have thatand is continuous real-valued functions since and are. This deduces , i.e. is closed under addition. Similarly,and is continuous real-valued functions since is. This implies , i.e. is closed under scalar multiplication. On the other hand, the constant function , which is also the additive identity in . Hence is a subspace of by 1.34.
5. Solution: Note that we consider complex vector space, so if is a subspace of the complex vector space , thenwe get a contradiction. Hence is not a subspace of the complex vector space .
6. Solution: (a) Because if and only if in , hence is obviously a subspace of by the similar arguments in Problem 1 and Problem 2.
(b) Note thatandHowever,This implies is not closed under addition, hence not a subspace of .
7. Solution: Denote by , then is not empty. If and , then . Hence and are integers. This means , i.e. is closed under addition. Similarly, since , it follows that is closed under additive inverses. However, is not closed under scalar multiplication since while . Hence is not a subspace of .
8. Solution: Denote by , then is not empty. If , then for any , we haveSimilarly, , hence is closed under scalar multiplication. However, while . This implies is not closed under addition, hence not a subspace of .
9. Solution: Denote the set of periodic functions from to by . Then is not a subspace of . Otherwise, we have since both and are periodic functions from to . Assume there exists a positive number such that for all , then . It is equivalent to this implies and . deduces that where . However implies where . Hence which is impossible. Therefore we get the conclusion.
10. Solution: Additive identity: by definition and , hence .
Closed under addition: if and , then and , hence for is closed under addition. Similarly, . Therefore .
Closed under scalar multiplication: if , then . Then for any , we have since is closed under scalar multiplication. Similarly, . Therefore .
11. Solution: Assume are subspaces of , where . Now we will show is a subspace of .
Additive identity: by definition for every , hence .
Closed under addition: if and , then for any given , we have and , hence for is closed under addition. Therefore .
Closed under scalar multiplication: if , then for any given . Then for any , we have since is closed under scalar multiplication. Therefore .
12. Solution: Suppose and are two subspaces of . We argue it by contradiction. If is a subspace of , moreover and . Consider and , then since is a subspace of . Hence or . If , then . We get a contradiction. If , then . We also get a contradiction. Hence if is a subspace of , we must have or .
If or . Without loss of generality, we can assume . Then is obviously a subspace of .
13. See If a field is such that why is a vector space over not equal to the union of proper subspaces of or A question on vector space over an infinite field.
14. Solution: It is clear that and are subspaces of . Now assume that and , then Hence . For any , we have and . However, hence . Combining this with previous argument, it follows that .
15. Solution: Because is a subspace of , hence closed under addition.Therefore for any , we have , i.e. . Note that if , then , hence . Combining with these arguments, it follows that .
16. Solution: For and , because addition in V is commutative, we have . This implies . Similarly, we have . Hence .
17. Solution: Note that in V, we have . Hence this is similar to Problem 16. Let , , then Since every element in can be expressed as the form , it follows that . Similarly, we also have . Hence .
18. Solution: If is a additive identity, then for any subspace of , we have . This means by the similar arguments in Problem 15 and 16. Hence the only possibility is , in fact this is the case. Hence is the additive identity. Suppose subspace of has additive inverses, then there exists a subspace of such that . This can only happen when since .
19. Solution: Here is a counterexample. Let , and . Then it is easy to see that , but .
See comments for the example from Neven Sajko.
20. Solution: Take . For any , we havesince and . We have .
Moreover, if , then we must have and since .
Similarly, since , we have and . Therefore, and , hence . It follows that . Hence by 1.45.
21. Solution: Since we can only choose the first two coordinates arbitrarily in , we can take in which the first two coordinates are zero and the last three are variables.
Let . If , then since . Moreover, we have since . Therefore , hence .
Again, for any , we havesince and . Therefore .
Since , it follows from 1.45 that .
22. Solution: Let , and . By the same argument as in Problem 21, we have
23. Solution: Here is a counter example. Let , , and . By the same argument as in Problem 21, one hashowever .
This is exactly the example from Neven Sajko.
24. Solution: Given any , define to be and define to be for all . Then . Moreover, for all , we haveandHence and . Note that we also havehence . Since we can choose arbitrarily, one has .
By 1.45, to show , it suffices to prove that . Let , then since and since for all . Sum up and , we have for all . Hence , which implies .