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APPM 3310 — Problem Set 4 Solutions 1. Problem 2.1.2 – Note: Since these are nonstandard definitions of addition and scalar multiplication, be sure to show that they satisfy all of the vector space axioms. Solution: The space Q is defined by Q = {(x, y) | x, y > 0} ⊂ R2 with (x1 , y1 ) + (x2 , y2 ) = (x1 x2 , y1 y2 ) and c (x, y) = (xc , y c ). We need to show that Q satisfies the two closure conditions and the seven axioms of a vector space outlined in Definition 2.1 of the textbook. (i) Closure under addition: Let (x1 , y1 ) , (x2 , y2 ) ∈ Q, i.e. x1 , y1 , x2 , y2 > 0. Then (x1 , y1 ) + (x2 , y2 ) = (x1 x2 , y1 , y2 ) ∈ Q because x1 , x2 > 0 ⇒ x1 x2 > 0 and y1 , y2 > 0 ⇒ y1 y2 > 0. (ii) Closure under scalar multiplication: Let (x, y) ∈ Q. Then c (x, y) = (xc , y c ) ∈ Q because a positive number to raised to any exponent is positive. (a) Commutativity of addition: Since multiplication of real numbers is commutative we have (x1 , y1 ) + (x2 , y2 ) = (x1 x2 , y1 y2 ) = (x2 x1 , y2 y1 ) = (x2 , y2 ) + (x1 , y1 ) (b) Associativity of addition: Since multiplication of real numbers is associative we have (x1 , y1 )+[(x2 , y2 ) + (x3 , y3 )] = [x1 (x2 x3 ) , y1 (y2 y3 )] = [(x1 x2 ) x3 , (y1 y2 ) y3 ] = [(x1 , y1 ) + (x2 , y2 )]+(x3 , y3 ) (c) Additive identity: The zero element is the element (1, 1). Note that since 1 > 0 the zero element is in the space and we have (x, y) + (1, 1) = (x · 1, y · 1) = (x, y) = (1 · x, 1 · y) = (1, 1) + (x, y) (d) Additive Inverse: The additive inverse of (x, y) ∈ Q is (1/x, 1/y). Note that since x, y > 0 we have that 1/x, 1/y > 0 and so the additive inverse is also in Q. Remembering that the zero element in Q is (1, 1) we have (x, y) + (1/x, 1/y) = (x/x, y/y) = (1, 1) = (1/x · x, 1/y · y) = (1/x, 1/y) + (x, y) (e) Distributivity: Let c, d ∈ R and (x1 , y1 ) , (x2 , y2 ) ∈ Q. Then c+d (c + d) (x1 , y1 ) = xc+d = xc1 xd1 , y1c y1d = (xc1 , y1c ) + xd1 , y1d = c (x1 , y1 ) + d (x1 , y1 ) 1 , y1 and also c [(x1 , y1 ) + (x2 , y2 )] = c (x1 x2 , y1 y2 ) = ((x1 x2 )c , (y1 y2 )c ) = (xc1 xc2 , y1c y2c ) = (xc1 , y1c ) + (xc2 , y2c ) = c (x1 , y1 ) + c (x2 , y2 ) (f) Associativity of scalar multiplication: We have c [d (x, y)] = c xd , y d = xd c , yd c = xcd , y cd = (cd) (x, y) (g) Unit for scalar multiplication We have 1 (x, y) = x1 , y 1 = (x, y) 2. Problem 2.2.2 Solution (a) Not a subspace because it does not contain the zero element: 0 − 0 + 4 (0) + 1 = 1 6= 0 (b) The set of vectors of the form (t, −t, 0)T for t ∈ R is a subspace. To see this we need to show that it contains the zero element, and is closed under vector addition and scalar multiplication. zero element: Taking t = 0 we have (0, −0, 0)T = (0, 0, 0)T which is in the space. closure: We want to show that since (t, −t, 0)T and (s, −s, 0)T are in the subspace, so is a (t, −t, 0)T + b (s, −s, 0)T where a, b ∈ R. We have t s at + bs r a −t + b −s = −at − bs = −r where r = at + bs 0 0 0 0 (c) The set of vectors of the form (r − s, r + 2s, −s)T is a subspace. zero element: Let r = s = 0, then (r − s, r + 2s, −s)T = (0, 0, 0)T closure: Consider the two vectors (r1 − s1 , r1 + 2s1 , −s1 )T and (r2 − s2 , r2 − 2s2 , −s2 )T from the space. Then we have (ar1 + br2 ) − (as1 + bs2 ) r1 − s1 r2 − s2 ar1 − as1 + br2 − bs2 a r1 + 2s1 +b r2 + 2s2 = ar1 + 2as1 + br2 + 2bs2 = (ar1 + br2 ) + 2 (as1 + bs2 ) − (as1 + bs2 ) −s1 −s2 −as1 − bs2 r−s which has the form r + 2s with r = ar1 + br2 and s = as1 + bs2 . −s (d) The set of vectors whose first component is 0 is a subspace. zero element: The zero vector (0, 0, 0)T is in the space. closure: Consider (0, x, y)T and (0, u, v)T from the space. We want to show that for scalars b, c ∈ R the vector b (0, x, y)T + c (0, u, v)T is in the space. We have 0 0 0 b x + c u = bx + cu y v by + cv which has first element 0 so it’s in the space. (e) The space of vectors with last element 1 is not a subspace. To see this note that the zero element (0, 0, 0)T is not in the space. It’s also does not satisfy either of the closure conditions since, for example, the vectors 0 0 0 0 + 0 = 0 1 1 2 0 0 and 3 0 = 0 1 3 are not in the space. (f) The set of all vectors (x, y, z)T such that x ≥ y ≥ z is not a subspace because it is not closed under scalar multiplication. Consider that the vector (3, 2, 1)T is in the space but 3 −3 −1 2 = −2 1 −1 has −3 < −2 < −1. (g) The set of all vectors (x, y, z)T such that z = x − y forms a subspace. It’s easier to see this if we note that all vectors from the space have entries that satisfy z − x + y = 0. zero element: The zero vector (0, 0, 0)T is in the space since z−x+y = 0−0+0 = 0. closure: Let (x1 , y1 , z1 ) be such that z1 − x1 + y1 = 0 and similarly (x2 , y2 , z2 ) be such that z2 − x2 + y2 = 0. Then for any b, c ∈ R we have x1 x2 bx1 + cx2 b y1 + c y2 = by1 + cy2 z1 z2 bz1 + cz2 The resulting vector is in the space because (bz1 + cz2 )−(bx1 + cx2 )+(by1 + cy2 ) = b (z1 − x1 + y1 )+c (z2 − x2 + y2 ) = b·0+c·0 = 0 (h) The set of all solutions to z = xy does not form a subspace because it does not satisfy either of the closure conditions. Note that (1, 1, 1)T and (1, 2, 2)T are in the space, but the following are not 1 2 2 1 = 2 since 2 6= 2 (2) = 4 1 2 1 1 2 1 + 2 = 3 since 3 6= 2 (3) = 6 1 2 3 (i) The set of all solutions of the equation x2 + y 2 + z 2 = 0 is a subspace because the only elements of the space is the zero element (0, 0, 0)T . Recall that the trivial subspace {0} is always a subspace. (j) The set of all solutions to the system xy = yz = xz is a subspace. To see this note that the only solutions to this system are constant vectors of the form (t, t, t)T . zero element: Clearly (0, 0, 0)T is a solution to the system. closure: Let b, c ∈ R and consider the two constant vectors (r, r, r)T and (s, s, s)T . Then we have r s br + cs t b r + c s = br + cs = t r s br + cs t if we let t = br + cs 3. Problem 2.2.8 The claim that we wish to prove is an if-and-only-if so we need to prove the implication in both directions. In other words we need to prove the following two conditionals: If the set of all solutions x of Ax = b is a subspace then the system is homogeneous. If the system is homogeneous then the set of all solutions x of Ax = b is a subspace. Proof (⇒) Assume that the set of all solutions x of Ax = b is a subspace. Since it is a subspace it must contain the zero element 0. But if 0 is a solution we have A0 = 0 so the right-hand side must be 0 and the system is homogeneous. (⇐) Assume that the system is homogeneous, i.e. Ax = 0. To prove that the set of all solutions x to this system is a subspace we need to show that it contains the zero element and satisfies the closure conditions. Clearly the zero element is in the space since A0 = 0. For the closure conditions we assume x and y satisfy Ax = 0 and Ay = 0. Then, for c, d ∈ R we have A (cx + dy) = cAx + dAy = c0 + d0 = 0 showing that (cx + dy) is in the subspace. 4. Problem 2.2.10 Proof: We wish to prove that the set of all n × n traceless matrices form a subspace of Mn×n . We need to show that the space contains the zero element and satisfies the closure conditions. zero element: The zero element of Mn×n is the n × n zero matrix. Since every entry of the zero matrix is zero it’s diagonal entries are all zero. Since the sum of n zeros is zero we have that the zero matrix is traceless. closure: Let A and B be n × n traceless matrices. Then we have tr (A) = a11 + a22 + · · · ann = 0 and tr (B) = b11 + b22 + · · · bnn = 0 Let c, d ∈ R then tr (cA + dB) = = = = (ca11 + db11 ) + (ca22 + db22 ) + · · · + (cann + dbnn ) c (a11 + a22 + · · · + ann ) + d (b11 + b22 + · · · + bnn ) c (0) + d (0) 0 5. Problem 2.3.22 Solution: I’m going to save time by doing parts (a) and (c) simultaneously since the result of (c) makes part (b) trivial. (a) and (c): To show that a set of vectors is linearly independent we stack the vectors side-by-side in a matrix A and show that Ax = 0 has only the trivial solution by performing Gaussian Elimination and showing that the reduced system has all nonzero pivots. We also augment the system with a general right-hand side vector to determine conditions on the range of A. 1 −2 2 a 1 −2 2 a 0 0 b b 3 −2 3 −2 ∼ 2 −1 1 c 0 3 −3 c − 2a 1 1 −1 d 0 3 −3 d − a 1 −2 2 a 3 −2 b ∼ 0 0 0 −1 c − 2a − b 0 0 −1 d − a − b 1 −2 2 a 0 3 −2 b ∼ 0 0 −1 c − 2a − b 0 0 0 d−c+a Since we’ve reduced to the matrix A to a matrix with three nonzero pivots we know that the vectors we started with are linearly independent. By performing Gaussian Elimination with a general right-hand side vector we know a general condition on vectors in the span of the three vectors. A vector b is in the span of the columns of A if there is a solution to the linear system Ax = b. The system will have a solution provided that the right-hand side vector b makes the system compatible, which we can see from the result of Gaussian Elimination will happen if d − c + a = 0. (b) To see if the following vectors are in the span of the given vectors we could solve attempt to solve the system Ax = b for each vector. We conclude that the vector is in the span if the linear system has a solution. Since above we solved the Ax = b with a general right-hand side we need only check that the given vectors satisfy the resulting compatibility condition d − c + a = 0: (i) – For (ii) – For the span. (iii) – For span. 1 1 2 1 1 0 0 0 T T 0 1 0 0 we have d − c + a = 1 − 2 + 1 = 0 so the vector is in the span. we have d − c + a = 0 − 0 + 1 = 1 6= 0 so the vector is not in T we have d − c + a = 0 − 0 + 0 = 0 so the vector is in the T (iv) – For 0 0 0 0 we have d − c + a = 0 − 0 + 0 = 0 so the vector is in the span. Note that the zero vector is in the span of any set of vectors because we the trivial solution is always a solution of Ax = 0. 6. Problem 2.3.28 Proof: It is much easier to prove the following equivalent contrapositive biconditional statement: u and v are linearly dependent if and only if ad − bc = 0. Assume that u and v are linearly dependent. Then there exist nonzero scalars α and β such that αu + βv = 0. Then 0 = αu + βv = α (ax + by) + β (cx + dy) = (αa + βc) x + (αb + βd) y Since x and y are linearly independent by assumption, this equality holds if and only if the coefficients in front of x and y are zero. Since α and β are nonzero this is true if and only if ad − bc = 0. To see this let α = d and β = −b, then 0 = (αa + βc) x + (αb + βd) y = (ad − bc) x Alternatively we could take α = −c and β = a, to get 0 = (αa + βc) x + (αb + βd) y = (−bc + ad) y which confirms the claim. 7. Each of the following statements is either true or false. If the statement is true, please provide a proof or some other sufficient justification. If it is false, explain why it is false or give a counterexample. (a) An interval is a vector space. Solution The statement is False. Consider the interval [a, b]. If 0 is not in the interval we’re done because the zero element is not in the space. If 0 is in the interval [a, b] then we note that b + b = 2b is not in [a, b] so the interval is not closed under addition. Similarly we could note that 3b is not in the interval so [a, b] is not closed under scalar multiplication. (b) The set of all real n × n nonsingular matrices is a subspace of Mn×n . Solution: The statement is False. The set of all nonsingular matrices is not a subspace of Mn×n for multiple reasons. First note that the zero element of Mn×n is the zero matrix, which is singular. Alternatively, we could note that the n × n identity matrix I is nonsingular, and so is it’s negative, −I. But I + (−I) = 0 and the zero matrix is singular, so the space is not closed under additon. (c) The set of all real n × n symmetric matrices is a subspace of Mn×n . Solution: The statement is True. Note that the transpose of the n×n zero matrix is 0T = 0 so the space contains the zero element. To show closure we assume that A and B are symmetric and that c, d ∈ R. Then (cA + dB)T = cAT + dB T = cA + dB (1) where here (1) follows from the assumption that A and B are symmetric. Thus the space of symmetric matrices is closed under addition and scalar multiplication. (d) If v1 , v2 , . . . , vk are elements of a vector space V and do not span V , then v1 , v2 , . . . , vk are linearly independent. Solution The statement is (super duper) False. Consider the following three vectors from R3 : v1 = e1 , v2 = 2e1 , v3 = 3e1 , where here e1 is the first cannoncial basis vector of R3 . Clearly these vectors do not span R3 but they are linearly dependent since they are each scalar multiples of each other.