Solution to Set 7, Math 2568

Transcription

1 Solution to Set 7, Math 568 S 5.: No. 18: Let Q be the set of all nonsingular matrices with the usual definition of addition and scalar multiplication. Show that Q is not a vector space. In particular, show that the property (C1) is not valid; also show that properties (C) and (a3) are also violated. Solution: Now, with respect to property (C1), we construct a counter example by taking matrices A and B as below, which of which is clearly nonsingular, and addition resulting in the zero matrix which is singular. ( ) ( ) A =, B = A and B are nonsingular since the only solution of AX = 0 is X = 0 since AX = X. This is true for BX = 0 as well since BX = X. Yet, A + B = 0 matrix which is clearly singular. So, property (C1) does not hold. Also, note A is nonsingular, yet if the scalar a = 0, we have aa = 0 which is singular. Therefore, (C) does not hold. Also, clearly the zero element is the matrix with all elements identically zero. But the zero matrix is singular, and therefore not a member of the set Q. Therefore property (a3) does not hold. S 5.: No. 4. Prove that Z = {0} is itself a vector space, where 0 is the zero element of some vector space V. Solution: Since Z V, we only need to show closure properties (C1)-(C) since all other properties (a1)-(a4), (m1)-(m4) are automatically inherited from the vector space V. However, since Z only contains one element 0, showing (C1) amounts to showing = 0, which is true from (a3). Also, for any scalar a, since for any v V we have av + a0) = a (v + 0) = av implying a0 = 0 and therefore property (C) holds for subset Z. Thus Z is a vector space itself. S 5.: No. 30. The set C [a, b] is defined to be the set of all real valued functions for which f(x), f (x), f (x) are continuous for x [a, b]. Verify that C [a, b] is a vector space by using theorems from calculus. Solution: We know from calculus that sum of continuous functions is continous that that sum of differentiable functions are continuous. Thus, if f, g C [a, b], d then f(x)+g(x), dx [f(x)+g(x)] = f (x)+g (x) d dx [f(x)+g(x)] = f (x)+g (x) are each continuous when its argument x [a, b]. This implies f + g C [a, b]. Property (c1) holds. property c If a is any scalar, we know from calculus that the scalar multiple of a continuous/differentiable function is also continuous/differentiable and that d dx [af(x)] = af (x) d dx [af(x)] = af (x) are each continuous when its argument x [a, b]. This implies af C [a, b]. a1, a. We have for f, g, h C [a, b] at any x [a, b] from commutativity and 1

2 associativity of addition of numbers, f(x) + g(x) = g(x) + f(x), [f(x) + g(x)] + h(x) = f(x) + [g(x) + h(x)] and thus (a1)-(a) hold. a3. We also have a candidate for θ C [a, b] to be the identically zero function, implying θ(x) = 0 for all x [a, b]. Now, this is indeed the zero element of C [a, b] since it is differentiable any number of times and also has the property that for any f C [a, b], f(x) + θ(x) = f(x) + 0 = f(x) as required for the zero element of a vector space. a4. We also note that for any f C [a, b] we can define its additive inverse f with the property that ( f)(x) = f(x) and it is indeed an additive inverse with this definition since for any x [a, b], we have f(x) + ( f)(x) = f(x) f(x) = 0 and therefore (a4) is satisfied. m1. is satisfied since for any two scalars a, b and any f C, we have from associative properties of ordinary number multiplication a[bf(x)] = (ab)f(x) and therefore a(bf) = (ab)f. m. For any scalar a and f, g C [a, b], we have at any x [a, b] the following functional evaluation using distributive properties of ordinary multiplication of numbers: a(f + g)(x) = a[f(x) + g(x)] = af(x) + bf(x) = (af)(x) + (bf)(x), and therefore as function property a(f +g) = af +ag and property (m) follows. m3. We also have for two scalars a, b and a function f C [a, b] at any x [a, b] from the functional evaluation and use of distribute properties of ordinary multiplication of numbers: (a + b)f(x) = af(x) + bf(x) = (af + bf)(x) implying that as functions, we have (a + b)f = af + bf. m4. Take any f C [a, b]. Then at any x [a, b], we have 1 f(x) = (1)f(x) = f(x) Since this is true for any x [a, b], it follows that 1 f = f, and property (m4) holds as well. Since we verified all ten properties, C [a, b] is indeed a vector space. S 5.: No. 34. Let { ( ) } x1 V = x : x =, where x x 1, x R

3 with addition and scalar multiplication defined as follows: ( ) ( ) ( ) ( ) u1 v1 u1 + v u + v = + = cu, cu = 1 u + v 1 cu + v 1 u v a. Show that the V under the operations defined above satisfy (c1),(c), (a1)- (a4), (m1), and (m4). b. Give examples to illustrate that property (m), (m3) are not satisfied. Solution to part a: (Note to distinguish between addition of vectors in the space V and addition of mere numbers, we use + in bold for the former). c1. If u = [u 1, u ] T, v = [v 1, v ] T V, we have from above definition of addition u + v = [u 1 + v 1 + 1, u + v 1] T, a ordered pair of real numbers, and therefore in V. c. If u = [u 1, u ] T V, then from above definition of scalar multiplication cu = [cu 1, cu ] T, we also an a ordered pair of real numbers, and therefore in cu V for any scalar c. a1. From definition we have u + v = [u 1 + v 1 + 1, u + v 1] T = [v 1 + u 1 + 1, v + u 1] T = v + u, where used used commutativity of addition of mere numbers. a. From definition we have (u+v)+w = [u 1 +v 1 +1, u +v 1] T +[w 1, w ] T = [u 1 +v 1 +w 1 +, u +v +w ] T, where as u+(v+w) = [u 1, v 1 ] T +[v 1 +w 1 +1, v +w 1] T = [u 1 +v 1 +w 1 +, u +v +w ] T and therefore property (a): associate property of addition, follows. a3. Claim θ = [ 1, 1] T since for any v V, we have from definition of the addition in this space v + θ = [v 1, v ] T + [ 1, 1] T = [v , v + 1 1] T = [v 1, v ] T = v a4. For any v = [v 1, v ] T V claim that the additive inverse v = [ v 1, v + ] since with the definition of addition in the space V, we have v+( v) = [v 1, v 1 ] T +[ v 1 1, v +1] T = [v 1 + ( v 1 ) + 1, v + ( v + ) 1] T = [ 1, 1] T = θ m1. We have any scalars a, b and vector v = [v 1, v ] T V, we have using scalar multiplication definition, a(bv) = a[bv 1, bv ] T = [abv 1, abv ] T = (ab)v m4. Indeed by definition of scalar multiplication, we have for any v = [v 1, v ] T V, 1.v = [1v 1, 1v ] T = [v 1, v ] T = v. Part b. 3

4 Consider property (m). Take for example a =, u = [1, 1] T = v V. Then, from definition of the addition in the vector space V where as ( [1, 1] T + [1, 1] T ) = [ , ] T = [6, ] T, [1, 1] T + [1, 1] T = [, ] T + [, ] T = [ + + 1, + 1] T = [5, 3] T [6, ] T Therefore (m) is violated. Consider property (m3). Take a = 1, b = 1 and v = [1, 1] T. Then we notice (a + b)v = 0[1, 1] T = [0, 0] T, where as av+bv = v+( 1)v = [v 1, v ] T +[ v 1, v ] T = [v 1 v 1 +1, v v 1] T = [1, 1] T [0, 0] T and so property (m3) does not hold. S 5.3: No. 10. Determine if the subset F = {f C[ 1, 1] : f(x) 0 for all x [ 1, 1]} is a subspace of C[ 1, 1]. Solution: It does not satisfy (c) since if we take f(x) = x F, and take scalar a = 1 clearly af(x) = x is not in F since it is negative in some places in the interval [ 1, 1]. Therefore F is not a subspace of vector space C[ 1, 1]. S 5.3: No. 14. Determine if F defined below is a subset of C [ 1, 1] F = { f : f C [ 1, 1] : f (x) e x f (x) + xf(x) = 0 } Solution: Checking c1. Take g, h F. Then, we know g + h C [ 1, 1] since C [ 1, 1] is a vector space. Also, we have (g+h) (x) e x (g+h) (x)+x(g+h)(x) = g (x)+h (x) e x g (x) e x h (x)+xg(x)+xh(x) = [g (x) e x g (x) + xg(x)] + [h (x) e x h (x) + xh(x)] = = 0, where each of the terms within each of the square parentheses [ ] above vaniish because g, h F. Thus g + h F and property (c1) holds. Checking c. Take g F. Then, we know for any scalar a, ag C [ 1, 1] since C [ 1, 1] is a vector space. Also, we have (ag) (x) e x (ag) (x) + x(ag)(x) = ag (x) e x ag (x) + xag(x) = a [g (x) e x g (x) + x] = 0 where the square parentheses [ ] = 0 because g F. Hence ag F for any scalar a and the property (c) holds. S 5.3: No. 16. Same question as (14) but with F now given as F = { f : f C [ 1, 1] : f (x) = 0 for x [ 1, 1] } 4

5 Checking c1. Take g, h F. Then, we know g + h C [ 1, 1] since C [ 1, 1] is a vector space. Also, we have (g + h) (x) = g (x) + h (x) = = 0, where each of the two zeros arise due to g, h F. Thus g + h F and property (c1) holds. Checking c. Take g F. Then, we know for any scalar a, ag C [ 1, 1] since C [ 1, 1] is a vector space. Also, we have (ag) (x) = a [g (x)] = a[0] = 0 where the square parentheses [ ] = 0 because g F. Hence ag F for any scalar a and the property (c) holds. S 5.3: No. 18. Express p(x) = 4x + x as a linear combination of vectors in Q = {p 1, p, p 3, p 3 } where p 1 (x) = 1+x +x 3, p (x) = 1+x+x 3, p 3 (x) = 1 3x+4x 4x 3, p 4 (x) = 1+x x +x 3 Solution: We want to find scalars a 1, a, a 3, a 4 so that which implies that for any x p(x) = a 1 p 1 (x) + a p (x) + a 3 p 3 (x) + a 4 p 4 (x), 4x+x = a 1 (1+x +x 3 )+a (1+x+x 3 )+a 3 ( 1 3x+4x 4x 3 )+a 4 (1+x x +x 3 ) Since the above equation is true for any x, we have to equate coefficients of 1, x, x, x 3 on both sides of the above equation to obtain the following set of linear equations for unknowns (a 1, a, a 3, a 4 ): a 1 + a a 3 + a 4 = a 3a 3 + a 4 = 4 a 1 + 4a 3 a 4 = 1 a 1 + a 4a 3 + a 4 = 0 for which the the corresponding augmented matrix row reduces as shown R4 R3 R 3 R 4 R3 R1 R 4 R R 1 R R+3R3 R3+R R 4 R,R 1 R 1 R3+ 4 R4,R1+ 1 R4 R 5 4 R

6 Therefore, (a 1, a, a 3, a 4 ) = ( 1, 5, 0, 3), which we may directly verify since 4x + x = (1 + x + x 3 ) 5 ( 1 + x + x 3 ) 3(1 + x x + x 3 ) S 5.4: No.. Let W be the set of all real matrices A written in the form ( ) a b A = c d such that a = d,, b = d, c = 3d Exhibit a basis for W. Solution: Clearly with the relations we have we have for any A W, the following form ( ) ( ) d d 1 A = = d 3d d 3 1 Therefore it is clear that W = Sp S, where S = {( 1 )} 3 1 Since S contains one nonzero vector it must be linearly independent (noting that in any vector space V a relation c 1 v 1 = 0 immediately implies c 1 = 0 when v 1 0). Therefore, S is a basis for W as it both spans W and is linearly independent. S 5.4: No. 10. Prove that the set Q of all symmetric matrices is a subspace of the set of all matrices. Solution: Suppose A, B are symmetric matrices. From the way we define matrix addition, their sum A + B is also a matrix, with (A + B) T = A T + B T = A + B, implying A + B is also symmetric. Also, if A is symmetric, i.e. A T = A, then for any constant a, aa is also a matrix from the way we define scalar multiplication of matrices. Also, (ca) T = ca T = ca and so ca is symmetric. So, both properties (c1)-(c) hold and Q is a subspace of M the set of all matrices. For the second part, note that a symmetric matrix has a representation ( ) ( ) ( ) ( ) a b A = = a + b + d b d Hence the set S given by S = {( ) 1 0, 0 0 ( ) 0 1, 1 0 ( )} is a spanning set for Q. Now, we check linear independence by setting ( ) ( ) ( ) ( ) c 1 + c c =

7 By adding all elements on the left side, we get the above reduce to ( ) ( ) c1 c 0 0 = c c implying immediately that c 1, c, c = 0, implying that S is linearly independent. Hence S is a basis for the set of all symmetric matrices. S 5.4: No. 1. Let B = {p 1, p, p 3 } P, where polynomials p 1, p, p 3 are defined as p 1 (x) = 1, p (x) = x and p 3 (x) = x. Find coordinates of (a). x x 1, (b) x + 4x + 1, (c) x + 5. Solution: Since x x + 1 = ( 1)p 1 + ( 1)p + p 3, it follows [x x + 1] B = [ 1, 1, 1] T. Again, because we have a standard basis, we can read off the coordinates on inspection, and we have for (b.) [x + 4x + 1] B = [1, 4, 1] T and for (c.) [x + 5] B = [5,, 0] T. S 5.4: No. 14. Prove that set S below is a linearly independent set in P n, the vector space of n-th degree polynomial, by using derivatives Solution: Set S := { 1, x, x, x 3,, x n} c 1 + c x + c 3 x + + c n x n = 0 for all x (1) If we set x = 0, we have at once c 1 = 0. We now differentiate both sides of equation (1) with respect to x to obtain c + c 3 x + + nc n x n 1 = 0 for all x () Now, we set x = 0 in () to obtain c = 0. Now, differentiate () with respect to x to obtain c 3 + 6c 4 x + n(n 1)c n x n = 0 for all x (3) Then, setting x = 0, we obtain c 3 = 0. In this manner by differentiating upto n times, and setting x = 0, we arrive at the conclusion that every c j = 0 and therefore the set S is linearly independent. 7