MTH5102 Spring 2017 HW Assignment 1: Prob. Set; Sec. 1.2, #7, 8, 12, 13, 20, 21 The due date for this assignment is 1/18/17. 7. Let S = {0, 1} and F = R. In F (S, R), show that f = g and f + g = h, where f (t) = 2t + 1, g (t) = 1 + 4t 2t 2, and h (t) = 5 t + 1. Proof. First, to show that f = g we must show that f (t) = g (t) for all t S. This is true though, f (1) = 2(1) + 1 = 3 = 1 + 4(1) 2(1) 2 = g (1), f (0) = 2 (0) + 1 = 1 = 1 + 4 (0) 2 (0) 2 = g (0), and S = {0, 1}. Second, to show that f + g = h we must show that f (t) + g (t) = h (t) for all t S. But this is true f (1) + g (1) = 3 + 3 = 6 = 5 1 + 1 = h (1), f (0) + g (0) = 1 + 1 = 2 = 5 0 + 1 = h (0), and S = {0, 1}. This completes the proof. 8. In any vector space V over a field F, show that (a + b) (x + y) = ax + by + bx + by for any x, y V and any a, b F. Proof. Let V be a vector space over a field F. Then for any x, y V and any a, b F we have (a + b) (x + y) (V S 1,2) = ax + by + bx + by. This completes the proof. (V S 8) (V S 7) = a (x + y) + b (x + y) = (ax + ay) + (bx + by) 12. A real-valued function f defined on the real line is called an even function if f ( t) = f (t) Prove that the set of even functions defined on the real line with the operations of addition and scalar multiplication defined in Example 3 is a vector space. Proof. First, let W = {f f : R R is an even function} denote the set of all even functions on R, in particular, W F (R, R) where the latter set has addition and scalar multiplcation well-defined, see Example 3. We now will show that if f, g W and c R then f + g, cf W so that addition and scalar multiplication are well-defined on W over the field R. Thus, if f, g W and c R then (f + g) ( t) = f ( t) + g ( t) = f (t) + g (t) = (f + g) (t), (cf) ( t) = c [f ( t)] = c [f (t)] = (cf) (t) for all t R implying that f + g and cf are even functions from R into R, that is, f +g, cf W. Hence, we have proven that addition and scalar multiplication on W is well-defined. Thus, in order to prove that W is a vector space over the field R with these operations of addition and scalar multiplication we need to prove that statement (V S 1) (V S 8) are true. This is proven in the following: (V S 1) If f, g W then addition in R is commutative we have (f + g) (t) = f (t) + g (t) = g (t) + f (t) = (g + f) (t) for all t R implying f + g = g + f. 1
(V S 2) If f, g, h W then addition in R is associative we have [(f + g) + h] (t) = (f + g) (t) + h (t) = (f (t) + g (t)) + h (t) for all t R, that is, = f (t) + (g (t) + h (t)) = [f + (g + h)] (t) (f + g) + h = f + (g + h). (V S 3) Let 0 denote the constant function 0 : R R defined by Then 0 W 0 (t) = 0 0 ( t) = 0 = 0 (t) This function has the property that if f W then that is (f + 0) (t) = f (t) + 0 (t) = f (t) + 0 = f (t), f + 0 = f. (V S 4) If f W then the function g : R R defined by is in W f W and for all t R, and, moreover, g (t) = f (t) for all t R g ( t) = f ( t) = f (t) f + g = 0 (f + g) (t) = f (t) + g (t) = f (t) f (t) = 0 = 0 (t) (V S 5) For each f W, we have 1f = f (1f) (t) = 1 [f (t)] = f (t) (V S 6) For each a, b R and each f W we have (ab) f = a (bf) [(ab) f] (t) = (ab) [f (t)] = a [bf (t)] = a [(bf) (t)] 2
(V S 7) For each a R and each f, g W we have a (f + g) = af + ag [a (f + g)] (t) = a [(f + g) (t)] = a (f (t) + g (t)) = a [f (t)] + a [g (t)] = (af) (t) + (ag) (t) = (af + ag) (t) (V S 8) For each a, b R and each f W we have (a + b) f = af + bf [(a + b) f] (t) = (a + b) [f (t)] = a [f (t)] + b [f (t)] = (af) (t) + (bf) (t) = (af + bf) (t) Therefore, W is a vector space over the field R. 13. Let V = R 2 denote the set of order pairs of real numbers. If (a 1, a 2 ), (b 1, b 2 ) V and c R, define (a 1, a 2 ) + (b 1, b 2 ) = (a 1 + b 1, a 2 b 2 ) and c (a 1, a 2 ) = (ca 1, a 2 ). Is V a vector space over R with these operations? Justify your answer. Solution. No. Because axiom (V S 3) cannot be true. To see this, if there exists a vector 0 V such that x + 0 = x for all x V then 0 = (c 1, c 2 ) and x = (a 1, a 2 ) and so implies that (a 1 + c 1, a 2 c 2 ) = (a 1, a 2 ) + (c 1, c 2 ) = x + 0 = x = (a 1, a 2 ) a 1 + c 1 = a 1, a 2 c 2 = a 2 implies that this is to be true for all a 1, a 2 R then we must have so that c 1 = 0, c 2 = 1 0 = (c 1, c 2 ) = (0, 1). But this contradicts Theorem 1.2.(a) if a 2 1 then 0x 0 0x = 0 (a 1, a 2 ) = (0a 1, a 2 ) = (0, a 2 ) (0, 1) = 0. 3
Therefore, axiom (V S 3) cannot be true so that V under the above definition of addition and scalar multiplication cannot be a vector space over R. 20. Let V be the set of sequences {a n } of real numbers. For {a n }, {b n } V and any real number t, define {a n } + {b n } = {a n + b n } and t {a n } = {ta n }. Prove that, with these operations, V is a vector space over R. Proof. Recall, that Example 5 (p. 11) we define a sequence {f n } is the notation for a (unique) function f : N R (where N = {1, 2, 3,...} denotes the set of natural numbers) which satisfies f (n) = f n for all n N. Thus, according to the notation of Example 3 (p. 9), the set of sequences of real numbers is just the set F (N, R) (i.e., the set of all functions from the set N to the vector space R over the field R). Also, by Example 3 (p. 9), the operations of addition and scalar multiplication defined for f, g F (N, R) and c R defined by (f + g) (n) = f (n) + g (n) and (cf) (n) = c [f (n)] for each n N, make F (N, R) a vector space over the field R. Thus, with this notation, the sequences {a n }, {b n } F (N, R) is notation for the functions a : N R and b : N R defined by a (n) = a n and b (n) = b n, for all n N. Thus, for any t R, the sequences {a n + b n } and {ta n } just denote the functions a + b and ta (a + b) (n) = a (n) + b (n) = a n + b n, (ta) (n) = t [a (n)] = ta n. Therefore, we have proved that the set of sequences of real numbers, i.e., V = F (N, R), is a vector space over R with the operations of addition and scalar multiplication {a n } + {b n } = {a n + b n } and t {a n } = {ta n } which is none other than the operations of addition of functions and scalar multiplication of a function as defined in Example 3 (p. 9) for F (S, F ), but for the special case that S = N and F = R. This completes the proof. 21. Let V and W be vector spaces over a field F. Let Z = {(v, w) : v V and w W }. Prove that Z is a vector space over F with the operations of addition and scalar multiplication defined by (v 1, w 1 ) + (v 2, w 2 ) = (v 1 + v 2, w 1 + w 2 ) and c (v 1, w 1 ) = (cv 1, cw 1 ), respectively. 4
Proof. First, addition and scalar multiplication on Z are well-defined for any (v 1, w 1 ), (v 2, w 2 ) Z and c F we have v 1 + v 2, cv 1 V and w 1 + w 2, cw 1 W because V and W are vector spaces of F which implies that (v 1 + v 2, w 1 + w 2 ) Z and (cv 1, cw 1 ) Z. Thus, in order to prove that Z is a vector space over the field F with these operations of addition and scalar multiplication we need to prove that statement (V S 1) (V S 8) are true. This is proven in the following: (V S 1) If (v 1, w 1 ), (v 2, w 2 ) Z then addition in V and W are commutative we have (v 1, w 1 ) + (v 2, w 2 ) = (v 1 + v 2, w 1 + w 2 ) = (v 2 + v 1, w 2 + w 1 ) = (v 2, w 2 ) + (v 1, w 1 ). (V S 2) If (v 1, w 1 ), (v 2, w 2 ), (v 3, w 3 ) Z then addition in V and W is associative we have [(v 1, w 1 ) + (v 2, w 2 )] + (v 3, w 3 ) = (v 1 + v 2, w 1 + w 2 ) + (v 3, w 3 ) = ((v 1 + v 2 ) + v 3, (w 1 + w 2 ) + w 3 ) = (v 1 + (v 2 + v 3 ), w 1 + (w 2 + w 3 )) = (v 1, w 1 ) + (v 2 + v 3, w 2 + w 3 ) = (v 1, w 1 ) + [(v 2, w 2 ) + (v 3, w 3 )]. (V S 3) Let 0 = (0 V, 0 W ), where 0 V and 0 W are the zero vectors in V and W, respectively. Then (0 V, 0 W ) Z and for each (v, w) Z we have (v, w) + (0 V, 0 W ) = (v + 0 V, w + 0 W ) = (v, w). (V S 4) If (v, w) Z then ( v, w) Z and (v, w) + ( v, w) = (v v, w w) = (0 V, 0 W ). (V S 5) If (v, w) Z then 1 (v, w) = (1v, 1w) = (v, w). (V S 6) For each a, b F and each (v, w) Z we have (ab) (v, w) = ((ab) v, (ab) w) = (a (bv), a (bw)) = a [b (v, w)]. (V S 7) For each a F and each (v 1, w 1 ), (v 2, w 2 ) Z we have a ((v 1, w 1 ) + (v 2, w 2 )) = a (v 1 + v 2, w 1 + w 2 ) = (a (v 1 + v 2 ), a (w 1 + w 2 )) = (av 1 + av 2, aw 1 + aw 2 ) = (av 1, aw 1 ) + (av 2, aw 2 ) = a (v 1, w 1 ) + a (v 2, w 2 ). (V S 8) For each a, b F and each (v, w) Z we have (a + b) (v, w) = ((a + b) v, (a + b) w) = (av + bv, aw + bw) = (av, aw) + (bv, bw) = a (v, w) + b (v, w). Therefore, W is a vector space over the field R. 5