MTH5102 Spring 2017 HW Assignment 2: Sec. 1.3, #8, 16, 28, 31; Sec. 1.4, #5 (a), (e), (g), 10 The due date for this assignment is 1/25/17.

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1 MTH510 Spring 017 HW Assignment : Sec. 1. # ; Sec. 1.4 #5 (a (e (g 10 The due date for this assignment is 1/5/17. Sec. 1. #8. Determine whether the following sets are subspaces of the vector space R (over the field R under the operations of addition and scalar multiplication defined on R. Justify your answers. (a W 1 = { (a 1 a a R } : a 1 = a and a = a (b W = { (a 1 a a R : a 1 = a + } (c W = { (a 1 a a R : a 1 7a + a = 0 } (d W 4 = { (a 1 a a R : a 1 4a a = 0 } (e W 5 = { (a 1 a a R : a 1 + a a = 1 } (f W 6 = { (a 1 a a R : 5a 1 a + 6a = 0 } Solution. (a First W 1 R and (0 0 0 W 1 where (0 0 0 is the zero element of the vector space R. Second if (b 1 b b (c 1 c c W 1 and d R then d (b 1 b b + (c 1 c c = (db 1 + c 1 db + c db + c and since b 1 = b b = b and c 1 = c c = c we have db 1 + c 1 = db + c = (db + c db + c = d ( b + ( c = (db + c which implies that d (b 1 b b + (c 1 c c W 1. Therefore by Theorem 1. W 1 is a subspace of R. (b As (0 0 0 W then by Theorem 1. W cannot be a subspace of R. (c First W R and (0 0 0 W where (0 0 0 is the zero element of the vector space R. Second if (b 1 b b (c 1 c c W and d R then d (b 1 b b + (c 1 c c = (db 1 + c 1 db + c db + c and since b 1 7b + b = 0 and c 1 7c + c = 0 we have (db 1 + c 1 7 (db + c + (db + c = d (b 1 7b + b + (c 1 7c + c = d0 + 0 = 0 which implies that d (b 1 b b + (c 1 c c W. Therefore by Theorem 1. W is a subspace of R. (d First W 4 R and (0 0 0 W 4 where (0 0 0 is the zero element of the vector space R. Second if (b 1 b b (c 1 c c W 4 and d R then d (b 1 b b + (c 1 c c = (db 1 + c 1 db + c db + c 1

2 and since b 1 4b b = 0 and c 1 4c c = 0 we have (db 1 + c 1 4 (db + c (db + c = d (b 1 4b b + (c 1 4c c = d0 + 0 = 0 which implies that d (b 1 b b + (c 1 c c W 4. Therefore by Theorem 1. W 4 is a subspace of R. (e As (0 0 0 W 5 then by Theorem 1. W 5 cannot be a subspace of R. (f As ( 0 1 ( W 6 but ( ( ( = W 6 since 5 ( ( 1 + ( ( = ( ( = ( ( = = < 0 1 ( + then by Theorem 1. W 6 cannot be a subspace of R. Sec. 1. #16. Let C n (R denote the set of all real-valued functions defined on the real line that have a continuous nth derivative. Prove that C n (R is a subspace of the vector space F (R R over the field R. Proof. First C n (R F (R R and the zero function 0 is the zero vector of F (R R with 0 C n (R since constant functions are infinitely differentiable. Second if f g C n (R and c R then by calculus we know that cf + g has a continuous nth derivative since f and g and (cf + g (n = cf (n + g (n so that cf + g C n (R. Therefore by Theorem 1. C n (R is a subspace of the vector space F (R R over the field R. Sec. 1. #8. A matrix M is called skew-symmetric if M t = M. Clearly a skew-symmetric matrix is square. Let F be a field. Prove that the set W 1 of all skew-symmetric n n matrices with entries from F is a subspace of M n n (F. Now assume that F is not of characteristic and let W be the subspace of M n n (F consisting of all symmetric n n matrices. Prove that M n n (F = W 1 W. Proof. First W 1 M n n (F and the zero matrix 0 is the zero vector of M n n (F satisfies 0 t = 0 = 0 so that 0 W 1. Second if A B W 1 and c F then (ca + B t = ca t + B t = ca B = (ca + B so that ca + B W 1. Therefore by Theorem 1. W 1 is a subspace of M n n (F. Now suppose that F is not of characteristic i.e. for the identity element 1 of F we assume that Then if A W 1 W then A = A t = A implies A = 0 and since 1

3 exists in F (because F is not of characteristic then 0 = 1 (A = ( 1 A = 1A = A. This proves that V := W 1 + W = {A + B : A W 1 B W } is a direct sum i.e. V = W 1 W. The complete the proof since V M n n (F we need only prove that M n n (F V. Let B M n n (F. Then B = 1 (B B t + 1 (B + B t V since [ 1 ( B + B t] t = 1 ( B t + ( B t t = 1 ( B + B t [ 1 ( B B t] t = 1 ( B t ( B t t = 1 ( B B t implies 1 (B B t W 1 and 1 (B + B t W. Therefore M n n (F V. This completes the proof. Sec. 1. #1. Let W be a subspace of a vector space V over a field F. For any v V the set {v} + W = {v + w : w W } is called the coset of W containing v. It is customary to denote this coset by v + W rather than {v} + W. Addition and scalar multiplication by scalars of F can be defined in the collection S = {v + W : v V } of all cosets of W as follows: for all v 1 v V and (v 1 + W + (v + W = (v 1 + v + W a (v + W = av + W for all v V and all a F. (a Prove that v + W is a subspace of V if and only if v W. (b Prove that v 1 + W = v + W if and only if v 1 v W. (c Prove that the preceeding operations are well-defined. (d Prove that the set S is a vector space with those operations. Note: This vector space is called the quotient space of V modulo W and is denoted by V/W. Proof. (a: Suppose that v + W is a subspace of V. Then by this and the definition of v + W there exists a vector w W such that v + w = 0. In particular by the uniqueness of the additive inverse we have w = v. And since v + W is a subspace containing w and 1 F and then by Theorem 1. v = ( 1 w v + W. Conversely suppose that v W then by Theorem 1. v + w W for all w W implying v + W W. Moreover if w W then since W is a vector space containing w and v then w v W and hence w = v + (w v v + W. Therefore v + W = W which is a subspace of V. (b Suppose that v 1 + W = v + W for some v 1 v V. Then since 0 W then v = v + 0 v + W = v 1 + W implying there exists w W such that v 1 + w = v so that this and Theorem 1. v 1 v = w = ( 1 w W. Conversely suppose that v 1 v W. Then by Theorem 1. v v 1 = ( 1 (v 1 v W implying that v 1 = v + (v 1 v v + W and v = v 1 + (v v 1 v 1 + W. Thus by Theorem 1. if w W then (v 1 v + w (v v 1 + w W implying that v 1 + w = v + [(v 1 v + w] v + W and v + w = v 1 + [(v v 1 + w] v 1 + W. This proves that v 1 + W = v + W. (c We begin by proving that the operation of addition defined above is well-defined. Suppose that v 1 + W = v 1 + W and v + W = v + W then we must prove that (v 1 + v + W = (v 1 + v + W. To prove this note that

4 by part (b we must have v 1 v 1 v v W so that by Theorem 1. (v 1 + v (v 1 + v = v 1 v 1 + v v W. By part (b this implies that (v 1 + v + W = (v 1 + v + W. This proves that the operation of addition defined above is well-defined. Now we will prove that the operation of scalar multiplication defined above is well-defined. Suppose that v + W = v + W and a F. Then we must prove that av + W = av + W. By part (b it suffi ces to prove that av av W. But since v + W = v + W then by part (b we have v v W implying by Theorem 1. that av av = a (v v W. Therefore the operation of scalar multiplication defined above is well-defined. (d We have already proven in part (c that the definitions of addition and scalar multiplication defined above are well-defined on V/W so that to prove it is a vector space we need only prove that V/W together with these operations over the field F satisfy axioms (V S 1 (V S 8. We do this now: (V S 1 For all x + W y + W V/W x + W + y + W = x + y + W = y + x + W = y + W + x + W. (V S For all x + W y + W z + W V/W (x + W + y + W + z + W = x + y + W + z + W = (x + y + z + W = x + (y + z + W = x + W + y + z + W = x + W + (y + W + z + W. (V S The element 0 + W = W has the property that for each x + W V/W x + W W = (x W = x + W. (V S 4 For each x + W V/W we have x + W V/W and x + W + x + W = (x x + W = 0 + W. (V S 5 For each element x + W V/W 1 (x + W = 1x + W = x + W. (V S 6 For each pair of element a b F and each element x + W V/W (ab (x + W = (ab x + W = a (bx + W = a (bx + W = a [b (x + W ]. (V S 7 For each element a F and each pair of elements x + W y + W V/W a (x + W + y + W = a (x + y + W = a (x + y + W = ax + ay + W = ax + W + ay + W = a (x + W + a (y + W. 4

5 (V S 8 For each pair of element a b F and each element x + W V/W (a + b (x + W = (a + b x + W = ax + bx + W = ax + W + bx + W = a (x + W + b (x + W. Therefore V/W together with these operations over the field F satisfy axioms (V S 1 (V S 8 and hence V/W is a vector space over the field F. Sec. 1.4 #5. In each part determine whether the given vector is in the space of S. (a ( 1 1 S = {(1 0 ( 1 1 1}; (e x + x + x + S = { x + x + x + 1 x + x + 1 x + 1 } ; (g ( 1 4 S = Solution. (a We must find scalars a b R such that i.e. Solving these equations we find that ( 1 1 = a (1 0 + b ( = (a b b a + b a b = b = 1 a + b = 1. b = 1 a = 1. Therefore ( 1 1 span (S. (e We must find scalars a b c R such that i.e. {( ( x + x + x + = a ( x + x + x b ( x + x c (x + 1 Solving these equations we find that = ax + (a + b x + (a + b + c x + (a + b + c a = 1 a + b = a + b + c = a = 1 b = c = 1. Therefore x + x + x + span (S. (g We must find scalars a b c R such that ( ( ( ( = a + b + c ( a + c b + c = a b ( }. i.e. a + c = 1 b + c = a = b = 4 5

6 Solving these equations we find that a = b = 4 c =. ( 1 Therefore span (S. 4 Sec. 1.4 #10. Show that if ( 1 0 M 1 = M 0 0 = ( 0 0 M 0 1 = ( then the span ( of {M 1 M M } is the set of all symmetric matrices. a b Proof. First M c d (F is a symmetric matrix if and only if if and only if ( ( t ( a b a b a c = = c d c d b d b = c. Therefore the set of all symmetric matrices with entries in a field F is given by {( } a b : a b d F = {am b d 1 + dm + bm : a b d F } This completes the proof. = span ({M 1 M M }. 6

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.

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