MATH1030 Linear Algebra Assignment 5 Solution

2015-16 MATH1030 Linear Algebra Assignment 5 Solution (Question No. referring to the 8 th edition of the textbook) Section 3.4 4 Note that x 2 = x 1, x 3 = 2x 1 and x 1 0, hence span {x 1, x 2, x 3 } has dimension 1. 7 For any a, b, c R, (a + b, a b + 2c, b, c) T = a(1, 1, 0, 0) T + b(1, 1, 1, 0) T + c(0, 2, 0, 1) T Note that X := { (1, 1, 0, 0) T, (1, 1, 1, 0) T, (0, 2, 0, 1) T } is a linear independent set. As spav(x) = S, it follows that X is a basis for S. 8 Since the dimension of R 3 is 3, it takes at least three vectors to span R 3. Therefore x 1 and x 2 cannot span R 3. The matrix X must be nonsingular, that is, det(x) 0. If x 3 = (a, b, c) T and X = (x 1, x 2, x 3 ), then det(x) = 1 3 a 1 1 b 1 4 c = 5a b 4c If one chooses a, b, and c so that 5a b 4c 0, then {x 1, x 2, x 3 } will be a basis for R 3. 10 We must find a subset of three vectors that are linearly independent. independent, but x 3 = x 2 x 1 Clearly x 1 and x 2 are linearly so x 1, x 2 and x 3 are linearly dependent. Next, we consider {x 1, x 2, x 4 }. Let X = (x 1, x 2, x 4 ), then 1 2 2 det(x) = 2 5 7 2 4 4 = 0 1

so these three vectors are also linearly dependent. If we pick Y = (x 1, x 2, x 5 ), then 1 2 1 det(y ) = 2 5 1 2 4 0 = 2 so the vectors x 1, x 2 and x 5 are linearly independent and hence form a basis for R 3. 11 We claim that X = {x 2 + 2, x + 3} is a basis for S. To prove it, firstly note that if there exists a, b R such that a(x 2 + 2) + b(x + 3) = 0 Then a = b = 0, so X is linearly independent. Next, for any element p S, there exists a p, b p R such that p = a p x 2 + b p x + 2a p + 3b p Since we can express p as p = a p (x 2 + 2) + b p (x + 3) so p is spanned by X, i.e., p span(x). So, S span(x). And by reversing the above argument we see that span(x) S. Therefore, S = span(x). It follows that X is a basis for S. 14 Note that the dimension of a finite-dimensional space equals to the number of elements of its basis. So we just need to count the number of linearly independent elements of the given set. Note that x 2 1 = 2x + 2(x 1) + (x 2 + 1) So the four vectors are linearly dependent. Hence, the dimension of the span of the four vectors is at most 3 (i.e. co-dimension 1 of P 3 ). Next, suppose there exists a, b, c R such that Then a(x) + b(x 1) + c(x 2 + 1) = 0 c = 0 a + b = 0 c b = 0 It implies a = b = c = 0. Hence, X = {x, x 1, x 2 + 1} is linearly independent. So, span(x {x 2 1}) = span(x) has dimension 3. Clearly x 2 x 1 = x 2 (x + 1) And it is also clear that {x 2, x + 1} is linearly independent. Therefore, span({x 2, x 2 x 1, x + 1}) = span({x 2, x + 1}) has dimension 2. (d) Clearly {2x, x 2} is linearly independent. Therefore, span({2x, x 2}) has dimension 2. 15 2

Note that S = {ax 3 + bx 2 + cx a, b, c R}. Hence, a basis of S can be given by X = {x 3, x 2, x}. Note that T = {ax 3 + bx 2 + cx + d a, b, c, d R and a + b + c + d = 0}. Since a + b + c + d = 0 if and only if d = a b c, so we can also express T as T = { ax 3 + bx 2 + cx + ( a b c) a, b, c R } Therefore, a basis of T can be given by Y = {x 3 1, x 2 1, x 1}. Note that S T = { ax 3 + bx 2 + cx a, b, c R } { ax 3 + bx 2 + cx + ( a b c) a, b, c R } = { ax 3 + bx 2 + cx a, b, c R and a + b + c = 0 } = { ax 3 + bx 2 + ( a b)x a, b, c R } So a basis of S T can be given by {x 3 x, x 2 x}. 18 Let U and V be subspaces of R n with the property that U \V = 0. If either U = 0 or V = 0, then the result is obvious. So, assume that both subspaces are nontrivial with dim(u) = k > 0 and dim(v ) = r > 0. Let {u 1,..., u k } be a basis for U and let {v 1,..., v r } be a basis for V. The vectors u 1,..., u k, v 1,..., v r span U + V. We claim that these vectors form a basis for U + V and hence that dim(u) + dim(v ) = k + r. To show this we must show that the vectors are linearly independent. Thus we must show that if then c 1 = = c k+r = 0. Now, if we set then the previous equation becomes c 1 u 1 + + c k u k + c k+1 v 1 + + c k+r v r = 0 u = c 1 u 1 + + c k u k and v = c k+1 v 1 + + c k+r v r u + v = 0 This implies u = v and hence that both u and v are in both in U V = 0. Thus we have u = c 1 u 1 + + c k u k = 0 v = c k+1 v 1 + + c k+r v r = 0 So, by the independence of u 1,..., u k and the independence of v 1,..., v r it follows that c 1 = = c k+r = 0 Section 3.6 1 The reduced row echelon form of the matrix is 1 0 2 0 1 0 0 0 0 3

Thus, (1, 0, 2) and (0, 1, 0) form a basis for the row space. The first and second columns of the original matrix form a basis for the column space: a 1 = (1, 2, 4) T and a 2 = (3, 1, 7) T The reduced row echelon form involves one free variable and hence the nullspace will have dimension 1. Setting x 3 = 1, we get x 1 = 2 and x 2 = 0. Thus (2, 0, 1) T is a basis for the nullspace. The reduced row echelon form of the matrix is 1 0 0 10/7 0 1 0 2/7 0 0 1 0 Clearly then, the set {(1, 0, 0, 10/7), (0, 1, 0, 2/7), (0, 0, 1, 0)} is a basis for the row space. Since the reduced row echelon form of the matrix involves one free variable the nullspace will have dimension 1.Setting the free variable x 4 = 1 we get x 1 = 10/7, x 2 = 2/7, x 3 = 0 Thus { (10/7, 2/7, 0, 1) } T is a basis for the nullspace. The dimension of the column space equals the rank of the matrix which is 3. Thus the column space must be R 3 and we can take as our basis the standard basis {e 1, e 2, e 3 }. The reduced row echelon form of the matrix is 1 0 0 0.65 0 1 0 1.05 0 0 1 0.75 The set {(1, 0, 0, 0.65), (0, 1, 0, 1.05), (0, 0, 1, 0.75)} is a basis for the row space. The set { (0.65, 1.05, 0.75, 1) T } is a basis for the nullspace. As in part the column space is R 3 and we can take {e 1, e 2, e 3 } as our basis. 2 Note that 2 1 2 = 2 1 4 2 and 3 1 2 2 = 1 + 1 5 2 3 Also, it is obvious that { (1, 1, 2) T, (2, 1, 3) T } is a linearly independent set. Therefore, the dimension of the space spanned by the given four vectors is 2. 3 The reduced row echelon form of A is given by 1 2 0 5 3 0 U = 0 0 1 1 2 0 0 0 0 0 0 1 4

So a 1, a 3 and a 6 corresponds to the free variables. And we have a 1 = a 1 /2 a 3 = 5a 2 /2 a 4 a 6 = 0 The lead variables correspond to columns 1, 3, and 6. Thus, a 1, a 3 and a 6 form a basis for the column space of A. The remaining column vectors satisfy the following dependency relationships: a 2 = 2a 1 a 4 = 5a 1 a 3 a 5 = 3a 1 + 2a 3 8 Since null(a) = 0, so by the dimension theorem, rank(a) = n. Hence, the column vectors of A are linearly independent. However, since m > n so the column vectors of A cannot span R m. If b is not in the column space of A, then {a 1, a 2,..., a n, b} is a linearly independent set. Therefore, there is no solution for the system Ax = b. If b is in the column space of A, then there exists a set of parameters {x 1, x 2,..., x n } with not all x i are zero, such that x 1 a 1 + + x n a n = b Therefore, there exists one solution for the system Ax = b. 10 If Ac = Ad, then A(c d) = 0. Suppose rank(a) = n, then by the dimension theorem, null(a) = n n = 0. Therefore, the null space of A is N(A) = {0}. Therefore, c d = 0 and hence c = d. Suppose rank(a) < n, by the dimension theorem, null(a) = n rank(a) > 0. Therefore, there exists at least one non-zero vector v 1 such that span({v 1 }) N(A). Therefore, we cannot conclude that c d = 0 and hence c and d may not be equivalent. 12 Since A and B differ only by a product of elementary matrices E = E 1... E n (i.e. B = EA), so the system of linear equations Ax = 0 is equivalent to Bx = EAx = E0 = 0. Therefore, the null spaces of A and B are equivalent, and hence null(a) = null(b). By the dimension theorem, rank(a) = rank(b). Therefore, the dimension of the column space of A equals to that of the column space of B. Let A = [ ] 1 1 1 1 and B = [ ] 1 1 0 0 Then clearly A and B are row equivalent. Note that the column space of A is span((1, 1) T ), but the column space of B is span((1, 0) T ). Therefore, the column spaces of A and B may not be the same. 14 5

2 a 3 = 2a 1 + a 2 = 7 11 1 13 a 4 = a 1 + 4a 2 = 7 30 3 16 If A is 5 8 with rank 5, then the column space of A will be R 5. So by the Consistency Theorem, the system Ax = b will be consistent for any b in R 5. Since A has 8 columns, its reduced row echelon form will involve 3 free variables. A consistent system with free variables must have infinitely many solutions. 18 Since A is 5 3 with rank 3, its nullity is 0. Therefore N(A) = {0}. If c 1 y 1 + c 2 y 2 + c 3 y 3 = 0 then c 1 Ax 1 + c 2 Ax 2 + c 3 Ax 3 = 0 A(c 1 x 1 + c 2 x 2 + c 3 x 3 ) = 0 and it follows that c 1 x 1 + c 2 x 2 + c 3 x 3 is in N(A). However, we know from part that N(A) = {0}. Therefore c 1 x 1 + c 2 x 2 + c 3 x 3 = 0 Since x 1, x 2, x 3 are linearly independent it follows that c 1 = c 2 = c 3 = 0 and hence y 1, y 2, y 3 are linearly independent. Since dim(r 5 ) = 5 it takes 5 linearly independent vectors to span the vector space. The vectors y 1, y 2, y 3 do not span R 5 and hence cannot form a basis for R 5. 19 Given A is m n with rank n and y = Ax where x 0. If y = 0 then x 1 a 1 + x 2 a 2 + + x n a n = 0 But this would imply that the columns vectors of A are linearly dependent. Since A has rank n we know that its column vectors must be linearly independent. Therefore y cannot be equal to 0. 20 If the system Ax = b is consistent, then b is in the column space of A. Therefore the column space of (A b) will equal the column space of A. Since the rank of a matrix is equal to the dimension of the column space it follows that the rank of (A b) equals the rank of A. 6

Conversely, if (A b) and A have the same rank, then b must be in the column space of A. If b were not in the column space of A, then the rank of (A b) would equal rank(a) + 1. 22 If x N(A), then BAx = B0 = 0 and hence x N(BA). Thus N(A) is a subspace of N(BA). On the other hand, if x N(BA) then B(Ax) = BAx = 0 and hence Ax N(B). But N(B) = {0} since B is nonsingular. Therefore Ax = 0 and hence x N(A). Thus BA and A have the same nullspace. It follows from the dimension theorem that rank(a) = n dim(n(a)) = n dim(n(ba)) = rank(ba) By part, left multiplication by a nonsingular matrix does not alter the rank. Thus rank(a) = rank(a T ) = rank(c T A T ) = rank((ac) T ) = rank(ac) 24 If N(A B) = R n then the nullity of A B is n and consequently the rank of A B must be 0. Therefore A B = O A = B Section 4.1 3 For any α 1 and a 0, L(αx) = αx + a αl(x) = αx + αa Hence, the translation is not a linear operation. Remark: Students can also prove that the addition also fails to be linear for translation. 4 Let u 1 = [ ] [ ] 1 1, u 2 2 =, x = 1 [ ] 7 5 7

To determine L(x) we must first express x as a linear combination x = c 1 u 1 + c 2 u 2 To do this we must solve the system Uc = x for c. The solution is c = (4, 3) T and it follows that [ ] 7 L(x) = L(4u 1 + 3u 2 ) = 4L(u 1 ) + 3L(u 2 ) = 18 6 For any x 1, x 2, y 1, y 2 R, L((x 1, x 2 ) + (y 1, y 2 )) = L((x 1 + y 1, x 2 + y 2 )) = (x 1 + y 1, x 2, y 2, 1) but L((x 1, x 2 )) + L((y 1, y 2 )) = (x 1, x 2, 1) + (y 1, y 2, 1) = (x 1 + y 1, x 2 + y 2, 2) Therefore, L is not a linear operation. For any x 1, x 2, y 1, y 2 R, L((x 1, x 2 ) + (y 1, y 2 )) = L((x 1 + y 1, x 2 + y 2 )) = (x 1 + y 1, x 2, y 2, (x 1 + y 1 ) + 2(x 2 + y 2 )) L((x 1, x 2 )) + L((y 1, y 2 )) = (x 1, x 2, x 1 + 2x 2 ) + (y 1, y 2, y 1 + 2y 2 ) = (x 1 + y 1, x 2 + y 2, x 1 + y 1 + 2x 2 + 2y 2 ) Also, for any α R, αl((x 1, x 2 )) = α(x 1, x 2, x 1 + 2x 2 ) = (αx 1, αx 2, αx 1 + 2αx 2 ) = L((αx 1, αx 2 )) = L(α(x 1, x 2 )) Therefore, L is a linear operation. For any x 1, x 2, y 1, y 2 R, L((x 1, x 2 ) + (y 1, y 2 )) = (x 1 + y 1, 0, 0) = (x 1, 0, 0) + (y 1, 0, 0) = L((x 1, x 2 )) + L((y 1, y 2 )) Also, for any α R, Therefore, L is a linear operation. (d) For any α 1, x 1, x 2 R \ {0}, αl((x 1, x)2) = α(x 1, 0, 0) = (αx 1, 0, 0) = L(α(x 1, x 2 )) αl((x 1, x 2 )) = α(x 1, x 2, x 2 1, x 2 2) = (αx 1, αx 2, αx 2 1 + αx 2 2) but L(α(x 1, x 2 )) = (αx 1, αx 2, (αx 1 ) 2 + (αx 2 ) 2 ) = (αx 1, αx 2, α 2 x 2 1 + α 2 x 2 2) Hence, L is not a linear operation. 8 For any A, B R n n and α R, L(αA) = C(αA) + (αa)c = α(ca + AC) = αl(a) 8

And Therefore, L is a linear operator. For any A, B R n n and α, β R, L(A + B) = C(A + B) + (A + B)C = CA + CB + AC + BC = (CA + AC) + (CB + BC) = L(A) + L(B) L(αA + βb) = C 2 (αa + βb) = αc 2 A + βc 2 B = αl(a) + βl(b) Therefore, L is a linear operator. If C O, then L is not a linear operator. For example, but L(2I) = (2I) 2 C = 4C 2L(I) = 2C 9 For any p, q P 2 for any x R, L((p + q)(x)) = x(p + q)(x) = xp(x) + xq(x) = L(p(x)) + L(q(x)) Also, for any α R, L(αp(x)) = x(αp)(x) = α x(p(x)) = αl(p(x)) Therefore, L is a linear operator on P 2. Let p(x) = x 2 for any x R, then p P 2, and but L(2p(x)) = x 2 + (2p)(x) 2L(p(x)) = 2(x 2 + p(x)) = 2x 2 + (2p)(x) Therefore, L is not a linear operator. For any p, q P 2 for any x R, Also, for any α R, L((p + q)(x)) = (p + q)(x) + x(p + q)(x) + x 2 (p + q) (x) = p(x) + xp(x) + x 2 p (x) + q(x) + xq(x) + x 2 q (x) = L(p(x)) + L(q(x)) L(αp(x)) = (αp)(x) + x(αp)(x) + x 2 (αp )(x) = α[p(x) + xp(x) + x 2 p (x)] = αl(p(x)) Hence, L is a linear operator. 12 When n = 1, L(α 1 v 1 ) = α 1 L(v 1 ) 9

Assume the result is true for any linear combination of k vectors and apply L to a linear combination of k + 1 vectors. L(α 1 v 1 + + α k v k + α k+1 v k+1 ) = L([α 1 v 1 + + α k v k ] + [α k+1 v k+1 ]) Thus, the result follows then by mathematical induction. = L(α 1 v 1 + + α k v k ) + L(α k+1 v k+1 ) = α 1 L(v 1 ) + + α k L(v k ) + α k+1 L(v k+1 ) 13 If v is any element of V then v = α 1 v 1 + α 2 v 2 + + α n v n Since L 1 (v i ) = L 2 (v i ) for any i = 1,..., n, it follows that L 1 (v) = α 1 L 1 (v 1 ) + α 2 L 1 (v 2 ) + + α n L 1 (v n ) = α 1 L 2 (v 1 ) + α 2 L 2 (v 2 ) + + α n L 2 (v n ) = L 2 (α 1 v 1 + α 2 v 2 + + α n v n ) = L 2 (v) 16 If v 1, v 2 V, then Therefore, L is a linear transformation. L(αv 1 + βv 2 ) = L 2 (L 1 (αv 1 + βv 2 )) = L 2 (αl 1 (v) + βl 1 (v 2 )) = αl 2 (L 1 (v 1 )) + βl 2 (L 1 (v 2 )) = αl(v 1 ) + βl(v 2 ) 17 For any x = (x 1, x 2, x 3 ) R 3, L(x) = (x 3, x 2, x 1 ) = (0, 0, 0) = 0 if any only if x 1 = x 2 = x 3 = 0. Hence, ker(l) = {0}. By the dimension theorem, null(l) = 0 rank(l) = dim(r 3 ) = 3. Therefore, range(l) = R 3. For any x = (x 1, x 2, x 3 ) R 3, L(x) = (x 1, x 2, 0) = (0, 0, 0) = 0 if and only if x 1 = x 2 = 0 and x 3 R. Therefore, ker(l) = span({e 3 }). Then, for any (x 1, x 2, 0) R 3, we have (x 1, x 2, 1) R 3 and L(x 1, x 2, 1) = (x 1, x 2, 0). Hence, range(l) = span({e 1, e 2 }). For any x = (x 1, x 2, x 3 ) R 3, L(x) = (x 1, x 1, x 1 ) = (0, 0, 0) = 0 if and only if x 1 = 0 and x 2, x 3 R. Therefore, ker(l) = span({e 2, e 3 }). For any (x 1, x 1, x 1 ) R 3, we have (x 1, 0, 0) R 3 and L(x 1, 0, 0) = (x 1, x 1, x 1 ). Hence, range(l) = span((1, 1, 1) T ). 10

19 For any p P 3, L(p(x)) = xp (x) = 0 if and only if p (x) = 0 for any x R. Therefore, ker(l) = P 0, the set of all constant polynomial. For any q P 3 such that q(0) = 0, there exists a, b, c R such that q(x) = ax 3 +bx 2 +cx. Let p(x) = a 3 x3 + b 2 x2 +cx, then p P 3 and L(p) = q. On the other hand, for any q P 3 such that q(0) 0, there exists d R \ {0} and q 0 P 3 such that q 0 (0) = 0 and q(x) = q 0 (x) + d. If there exists p P 3 such that L(p) = q, it implies d = 0, which is contradicting. Therefore, range(l) = {q P 3 q(0) = 0}. If p(x) = ax 2 + bx + c is in ker(l), then L(p) = (ax 2 + bx + c) (2ax + b) = ax 2 + (b 2a)x + (c b) must equal the zero polynomial z(x) = 0x 2 + 0x + 0. Equating coefficients we see that a = b = c = 0 and hence ker(l) = 0. The range of L is all of P 3. To see this note that if p(x) = ax 2 + bx + c is any vector in P 3 and we define q(x) = ax 2 + (b + 2a)x + c + b + 2a then L(q(x)) = (ax 2 + (b + 2a)x + c + b + 2a)(2ax + b + 2a) = ax 2 + bx + c = p(x) Clearly, ker(l) = {p P 3 p(0) = p(1) = 0}. Also, for any a, b R, define p(x) = (b a)x + a, then p P 3, p(0) = a and p(1) = b. Hence, L(p(x)) = p(0)x + p(1) = ax + b. Therefore, range(l) = P 1. 21 Suppose L is one-to-one and v ker(l), then L(v) = 0 W and L(0 V ) = 0 W Since L is one-to-one, it follows that v = 0 V. Therefore ker(l) = 0 V. Conversely, suppose ker(l) = 0 V and L(v 1 ) = L(v 2 ). Then L(v 1 v 2 ) = L(v 1 )L(v 2 ) = 0 W Therefore v 1 v 2 ker(l) and hence So, L is one-to-one. v 1 v 2 = 0 V v 1 = v 2 22 To show that L maps R 3 onto R 3 we must show that for any vector y R 3 there exists a vector x R 3 such that L(x) = y. This is equivalent to showing that the linear system x 1 = y 1 x 1 + x 2 = y 2 x 1 + x 2 + x 3 = y 3 is consistent. And this system is consistent since the coefficient matrix is nonsingular. 25 11

If p = ax 2 + bx + c P 3, then Thus, D(p) = 2ax + b D(P 3 ) = span({1, x}) = P 2 The operator is not one-to-one, for if p 1 (x) = ax 2 + bx + c 1 and p 2 (x) = a 2 + bx + c 2 where c 2 c 1, then D(p 1 ) = D(p 2 ). The subspace S consists of all polynomials of the form ax 2 + bx. If p 1 = a 1 x 2 + b 1 x, p 2 = a 2 x 2 + b 2 x and D(p 1 ) = D(p 2 ), then 2a 1 x + b 1 = 2a 2 x + b 2 and it follows that a 1 = a 2, b 1 = b 2. Thus p 1 = p 2 and hence D is one-to-one. D does not map S onto P 3 since D(S) = P 2. 12