Problem 1. Prove that for every .
Solution. By definition we have . But . So by uniqueness of the additive inverse (1.27), we conclude that .
Problem 2. Suppose , , and . Prove that or .
Solution. Suppose that . Then admits a multiplicative inverse , and it follows that by (1.31).
Problem 3. Suppose . Explain why there exists a unique such that .
Proof. Let . This vector exists because admits an additive inverse in and is closed under addition and scalar multiplication. It then follows that which establishes the existence of such an . Now, suppose that for . Adding on the left and cancelling yields . Scaling these vector by yields , so this vector is unique.
Problem 4. The empty set is not a vector space. The empty set fails to satisfy only one of the requirements listed in the definition of a vector space (1.20). Which one?
Solution. The empty set contains no elements, and hence it fails to admit an additive identity element. So a vector space , as a set, must be nonempty.
Problem 5. Show that in the definition of a vector space (1.20) , the additive inverse condition can be replaced with the condition that for all . Here the on the left side is the number , and the on the right side is the additive identity of .
Solution. Given the vector-space axioms, one has for every by (1.30). Conversely, suppose that the additive-inverse axiom were replaced with the condition that for every . Fix . Then . But by (1.32), so one has meaning that admits an additive inverse, as required.