Solutions to Homework 2 - Math 3410 - uleth.ca

1 Solutions to Homework 2 - Math 3410. 1. (Page 156: # ) Let V be the set of ordered pairs (a, b) of real numbers with addition in V and scalar multiplication on V defined by (a, b) + (c, d) = (a + c, b + d) and k(a, b) = (ka, 0). Show that V satisfies all the axioms of a vector space except [M4 ], that is, except 1~u = ~u. Hence [M4 ] is not a consequence of the other axioms. Solution The proofs of [A1 ], [A2 ], [A3 ], and [A4 ] are straightforward. Note that ~0 = (0, 0) and (a, b) = ( a, b). We have k((a1 , b1 ) + (a2 , b2 )) = k(a1 + a2 , b1 + b2 ) = (k(a1 + a2 ), 0) = (ka1 +.)

2 Ka2 , 0) = (ka1 , 0) + (ka2 , 0) = k(a1 , b1 ) + k(a2 , b2 ), which proves [M1 ]. We have (k1 + k2 )(a, b) = ((k1 + k2 )a, 0) = (k1 a + k2 a, 0) = (k1 a, 0) + (k2 a, 0) =. k1 (a, b) + k2 (a, b), which proves [M2 ]. We have k1 (k2 (a, b)) = k1 (k2 a, 0) = (k1 (k2 a), 0) = ((k1 k2 )a, 0) = (k1 k2 )(a, 0), which proves [M3 ]. Note that 1(1, 1) = (1, 0) 6= (1, 1), so [M4 ] does not hold. 2. (Page 156: # ) Show that Axiom [A4 ] of a vector space V , that is, that ~u + ~v = ~v + ~u, can be derived from the other axioms for V . Hint: Expand (1 + 1)(~u + ~v ) in two different ways.

3 Solution We calculate (1 + 1)(~u + ~v ) in two different ways. We have (1 + 1)(~u + ~v ) = (1 + 1)~u + (1 + 1)~v [M1 ]. = (1~u + 1~u) + (1~v + 1~v ) [M2 ]. = ~u + ~u + ~v + ~v [M4 ], [A1 ]. Also (1 + 1)(~u + ~v ) = 1(~u + ~v ) + 1(~u + ~v ) [M2 ]. = (1~u + 1~v ) + (1~u + 1~v ) [M1 ]. = ~u + ~v + ~u + ~v [M4 ], [A1 ]. So ~u + ~u + ~v + ~v = ~u + ~v + ~u + ~v ~u + (~u + ~u + ~v + ~v ) = ~u + (~u + ~v + ~u + ~v ). ( ~u + ~u) + ~u + ~v + ~v = ( ~u + ~u) + ~v + ~u + ~v [A1 ]. ~0 + ~u + ~v + ~v = ~0 + ~v + ~u + ~v [A3 ]. ~u + ~v + ~v = ~v + ~u + ~v [A2 ].

4 (~u + ~v + ~v ) + ( ~v ) = (~v + ~u + ~v ) + ( ~v ). ~u + ~v + (~v + ( ~v )) = ~v + ~u + (~v + ( ~v )) [A1 ]. ~u + ~v + ~0 = ~v + ~u + ~0 [A3 ]. ~u + ~v = ~v + ~u [A2 ]. So [M4 ] is a consequence of [A1 ], [A2 ], [A3 ], [M1 ], [M2 ], and [M4 ]. 1. 3. (Page 156: # ) Let V be the set of ordered pairs (a, b) of real numbers. Show that V is not a vector space over R with addition and scalar multiplication defined by: (i) (a, b) + (c, d) = (a + d, b + c) and k(a, b) = (ka, kb), (ii) (a, b) + (c, d) = (a + c, b + d) and k(a, b) = (a, b), (iii) (a, b) + (c, d) = (0, 0) and k(a, b) = (ka, kb), (iv) (a, b) + (c, d) = (ac, bd) and k(a, b) = (ka, kb).

5 Solution (i) We have (1 + 2)(3, 4) = (9, 12) however 1(3, 4) + 2(3, 4) = (3, 4) +. (6, 8) = (11, 10) so [M2 ] is not satisfied and thus V is not a real vector space. (ii) Note that (1 + 1)(1, 1) = 2(1, 1) = (1, 1) but on the other hand 1(1, 1) +. 1(1, 1) = (1, 1) + (1, 1) = (2, 2) and these two expressions are unequal. Therefore [M 2] does not hold. (iii) Note that (1 + 1)(1, 1) = 2(1, 1) = (2, 2) but on the other hand 1(1, 1) +. 1(1, 1) = (1, 1)+(1, 1) = (0, 0). Thus these two expressions are [M 2] does not hold. (iv) Since (1, 1) + (a, b) = (a, b) + (1, 1) = (a, b) so 0 = (1, 1).

6 Now we have (0, 1)(a, b) = (0, b) 6= (1, 1), so (0, 1) does not have an additive inverse and so [A3 ]. does not hold. This shows that V is not a vector space over R. 4. (Page 156, # ) Let U and W be vector spaces over a field K. Let V be the ~ where ~u U and w set of ordered pairs (~u, w) ~ W . Show that V is a vector space over K with addition in V and scalar multiplication on V defined by ~ + (~u0 , w (~u, w) ~ 0 ) = (~u + ~u0 , w ~ 0 ) and k(~u, w). ~ +w ~ = (k~u, k w). ~. ~ (~u0 , w Solution Consider elements (~u, w), ~ 0 ), and (~u00 , w ~ 00 ) of V.

7 [A1]. We have ~ + ((~u0 , w (~u, w) ~ 0 ) + (~u00 , w ~ 00 )). ~ + (~u0 + ~u00 , w = (~u, w) ~0 + w ~ 00 ). = (~u + ~u0 + ~u00 , w ~0 + w ~ +w ~ 00 ). = (~u + ~u0 , w ~ 0 ) + (~u00 , w ~ +w ~ 00 ). ~ + (~u0 , w = ((~u, w) ~ 0 )) + (~u00 , w ~ 00 ). [A2]. We choose the zero element to be (~0U , ~0W ) where ~0U and ~0W are the ~ + (~0U , ~0W ) = (~u + ~0U , w respective zero vectors of U and W . Note that (~u, w) ~+. ~0W ) = (~u, w). ~ and thus [A2] is satisfied. ~ its negative will be ( ~u, w). [A3]. Given (~u, w) ~ Thus ~ + ( ~u, w). (~u, w) ~ = (~u + ( ~u), w ~ = (~0U , ~0W ).)

8 ~ + ( w)). [A4]: We have ~ + (~u0 , w (~u, w) ~ 0 ) = (~u + ~u0 , w ~ 0 ) = (~u0 + ~u, w ~ +w ~ 0 + w). ~ = (~u0 , w ~ 0 ) + (~u, w). ~. [M 1]. Let k K. Then ~ + (~u0 , w k((~u, w) ~ 0 )) = k(~u + ~u0 , w ~ 0 ) = (k(~u + ~u0 ), k(w ~ +w ~ 0 )) = (k~u + k~u0 , k w ~ +w ~ 0). ~ + kw ~ + (k~u0 , k w = (k~u, k w) ~ 0 ) = k(~u, w). ~ + k(~u0 , w ~ 0 ). 2. [M 2]. We have (a + b)(~u, w). ~ = ((a + b)~u, (a + b)w). ~ = (a~u + b~u, aw ~ + bw). ~. = (a~u, aw). ~ + (b~u, bw). ~ = a(~u, w). ~ + b(~u, w). ~. [M 3]. We have (ab)(~u, w). ~ = ((ab)~u, (ab)w) ~ = (a(b~u), a(bw)).

9 ~ = a(b~u, bw). ~ =. a(b(~u, w)). ~. [M 4]. We have 1(~u, w). ~ = (1~u, 1w). ~ = (~u, w). ~. 5. (Page 157, # ) Consider the vectors ~u = (1, 2, 3) and ~v = (2, 3, 1) in R3 . (a). Write w ~ = (1, 3, 8) as linear combination of ~u and ~v . (b). Write w ~ = (2, 4, 5) as linear combination of ~u and ~v . (c). Find k so that w ~ = (1, k, 4) is a linear combination of ~u and ~v . (d). Find conditions on a, b, c so that w ~ = (a, b, c) is a linear combination of ~u and ~v . Solution (a). We want to find x and y such that w ~ = x~u + y~v . As explained in class it suffices to solve A~x = w~ where.

10 1 2 1. x A = (~u ~v ) = 2 3 , ~x =. ,w ~= 3 . y 3 1 8. We must row reduce the following augmented matrix.. 1 2 1. 2 3 3 . 3 1 8. By the row operations R2 R2 2R1 and R3 R3 3R1 we obtain . 1 2 1. 0 1 1 . 0 5 5. We now apply R3 R3 + 5R2 to obtain . 1 2 2. 0 1 1 . 0 0 0. Next apply R1 R1 + 2R2 and R1 ( 1) R2 to obtain . 1 0 3. 0 1 1 . 0 0 0. It follows that x = 3 and y = 1 are Solutions . Thus w~ = 3~u + ( 1)~v . (b). As before it suffices to solve A~x = w~ where . 1 2 2. x A = (~u ~v ) = 2 3 , ~x = ~ = 4 . ,w y 3 1 5. We reduce the augmented matrix.