Abstract Algebra Theory and Applications - Computer Science ...

314 CHAPTER 18 VECTOR SPACES5. −(αv) = (−α)v = α(−v) for all α ∈ F and all v ∈ V .Proof. To prove (1), observe that0v = (0 + 0)v = 0v + 0v;consequently, 0 + 0v = 0v + 0v. Since V is an abelian group, 0 = 0v.The proof of (2) is almost identical to the proof of (1). For (3), we aredone if α = 0. Suppose that α ≠ 0. Multiplying both sides of αv = 0 by1/α, we have v = 0.To show (4), observe thatv + (−1)v = 1v + (−1)v = (1 − 1)v = 0v = 0,and so −v = (−1)v. We will leave the proof of (5) as an exercise.□18.2 SubspacesJust as groups have subgroups and rings have subrings, vector spaces alsohave substructures. Let V be a vector space over a field F , and W a subsetof V . Then W is a subspace of V if it is closed under vector addition andscalar multiplication; that is, if u, v ∈ W and α ∈ F , it will always be thecase that u + v and αv are also in W .Example 5. Let W be the subspace of R 3 defined by W = {(x 1 , 2x 1 +x 2 , x 1 − x 2 ) : x 1 , x 2 ∈ R}. We claim that W is a subspace of R 3 . Sinceα(x 1 , 2x 1 + x 2 , x 1 − x 2 ) = (αx 1 , α(2x 1 + x 2 ), α(x 1 − x 2 ))= (αx 1 , 2(αx 1 ) + αx 2 , αx 1 − αx 2 ),W is closed under scalar multiplication. To show that W is closed undervector addition, let u = (x 1 , 2x 1 + x 2 , x 1 − x 2 ) and v = (y 1 , 2y 1 + y 2 , y 1 − y 2 )be vectors in W . Thenu + v = (x 1 + y 1 , 2(x 1 + y 1 ) + (x 2 + y 2 ), (x 1 + y 1 ) − (x 2 + y 2 )).Example 6. Let W be the subset of polynomials of F [x] with no oddpowerterms. If p(x) and q(x) have no odd-power terms, then neither willp(x) + q(x). Also, αp(x) ∈ W for α ∈ F and p(x) ∈ W .