Math 317: Linear Algebra Homework 6 Solutions Turn in the following problems. 1. For each of the following sets V , determine if V forms a subspace in Rn . Justify your answer in each case by providing a proof or a counterexample where appropriate. ( ) n X (a) V = x ∈ Rn : xi = 0 i=1 Proof : We claim that 0= Pn is V is a subspace. We note that 0 ∈ VPsince n (0, 0,P . . . , 0) =⇒ = 0. Suppose Then i=1 xi = 0 i=1 0P P that x, y ∈ V .P n and ni=1 yi = 0. Thus ni=1 xi + ni=1 yi = 0 =⇒ i=1 (xi + yi ) = 0. P Thus Suppose that c ∈ R and x ∈ V . Then ni=1 xi = 0 =⇒ Pn x + y ∈ V . P c i=1 xi = c0 = ni=1 cxi = 0. Thus cx ∈ V . Since the three properties of a subspace have been satisfied we may say that V is a subspace of Rn . (b) V = x ∈ Rn : Ax = b, A 6= 0, b 6= 0, b ∈ Rm , A ∈ Rm×n V is not a subspace because 0 is not an element of V . This is true since if 0 were to be in V , then A0 = b for some b 6= 0. This of course is impossible. 2. Determine which of the following subsets of Rn×n are in fact subspaces of Rn×n . Justify your answer in each case by providing a proof or a counterexample where appropriate. (a) The symmetric matrices. Let V = A ∈ Rn×n | A = AT . We claim that V is a subspace of Rn . We begin by observing that 0n×n ∈ V , since 0T = 0. That is, the zero matrix is symmetric. Let A, B ∈ V . Then A = AT and B = B T (wts. A + B is symmetric). Then (A + B)T = AT + B T = A + B so A + B is symmetric and hence is an element of V . Let c ∈ R and suppose that A ∈ V . Then (cA)T = cAT = cA and so cA is symmetric and hence an element of V . Since all three conditions of a subspace were met, we may conclude that V is a subspace of Rn×n . (b) The nonsingular matrices. The set of all nonsingular matrices do not form a subspace of Rn×n . This is true since 0n×n is a singular matrix (0x = 0 has infinitely many solutions.). Since the zero matrix is not in the set of all nonsingular matrices, the set does not form a subspace of Rn×n . (c) The singular matrices. The set of singular matrices do not form a subspace in Rn×n . This is true because the set is not closed under matrix addition. To see why, consider the singular matrices (all of which have a row of zeros in its echelon form) 1 Math 317: Linear Algebra Homework 6 Solutions 1 0 0 0 1 0 A= and B = . Then A + B = which is the 2 × 2 0 0 0 1 0 1 identity matrix and is nonsingular. So A+B is not in the set, even though A and B are. 3. Section 3.1, Problem 6 Proof: Suppose that U and V are subspaces of Rn . We prove that U ∩ V (the intersection of U and V ) is a subspace of Rn . Since U and V are subspaces of mathbbRn , we know that 0 ∈ U and 0 ∈ V . Thus 0 ∈ U ∩ V . Suppose that x, y ∈ U ∩ V . Then x, y ∈ U and x, y ∈ V . Since U, V are subspaces, we know that x + y ∈ U and x + y ∈ V . Thus x + y ∈ U ∩ V . Now suppose that c ∈ R and let x ∈ U ∩ V . Then U, V being subspaces =⇒ cx ∈ U, cx ∈ V . Thus, cx ∈ U ∩ V and so U ∩ V is a subspace of Rn . For examples of this consider U = Rn and V = {0}, which are both subspaces of Rn . Then U ∩V = {0} which is still a subspace of Rn . For another example, consider the x-axis and y-axis in R2 which can be interpretated as lines passing through the origin. Thus, these are subspaces of R2 . In set notation, we may write U = {(x1 , 0) | x1 ∈ R} and V = {(0, x2 ) | x2 ∈ R}. Then U ∩ V = {0} which is a subspace of R2 . However, U ∪V is not a subspace in this example. To see why, we note that (1, 0) ∈ U ∪V and (0, 1) ∈ U ∪V but (1, 0)+(0, 1) = (1, 1) which is in neither U nor V , so U ∪V is not closed under vector addition. Hence, it cannot be a subspace. 4. Section 3.1, Problem 8 Proof : As this is an if and only if statement, we must prove both directions of the claim. To begin, we assume that v ∈ span {v1 , v2 , . . . , vk }. We show that span {v1 , v2 , . . . , vk } = span {v1 , v2 , . . . , vk , v}. To show that two sets are equal, we must show containment on both sides. Suppose that x ∈ span {v1 , v2 , . . . , vk } (wts. x ∈ span {v1 , v2 , . . . , vk , v}). Then x = c1 v1 + c2 v2 + . . . + ck vk for some c1 , c2 , . . . ck ∈ R. But x = c1 v1 + c2 v2 + . . . + ck vk + 0v and hence x ∈ span {v1 , v2 , . . . , vk , v}. So span {v1 , v2 , . . . , vk } ⊂ span {v1 , v2 , . . . , vk , v}. We now show containment in the other direction. Suppose that x ∈ span {v1 , v2 , . . . , vk , v}. (wts. x ∈ span {v1 , v2 , . . . , vk }). Then x = c1 v1 + c2 v2 + . . . + ck vk + ck+1 v for c1 , . . . , ck+1 ∈ R. But, v ∈ span {v1 , v2 , . . . , vk }, and thus, v = d1 v1 + . . . + dk vk for some d1 , . . . , dk ∈ R. Thus, we have that x = c1 v1 + c2 v2 + . . . + ck vk + ck+1 v = c1 v1 + c2 v2 + . . . + ck vk + ck+1 (d1 v1 + . . . + dk vk ) = (c1 + ck+1 d1 )v1 + (c2 + ck+1 d2 )v2 + . . . + (ck + ck+1 dk )vk , 2 Math 317: Linear Algebra Homework 6 Solutions and thus x ∈ span {v1 , v2 , . . . , vk }. So, span {v1 , v2 , . . . , vk , v} ⊂ span {v1 , v2 , . . . , vk }. Thus span {v1 , v2 , . . . , vk } = span {v1 , v2 , . . . , vk , v}. To go in the other direction, suppose that span {v1 , v2 , . . . , vk } = span {v1 , v2 , . . . , vk , v}. We show that v ∈ span {v1 , v2 , . . . , vk }. We note that v = 0v1 +0v2 +. . .+0vk +1v ∈ span {v1 , v2 , . . . , vk , v} = span {v1 , v2 , . . . , vk }. Thus, v ∈ span {v1 , v2 , . . . , vk }. 5. Section 3.1, Problem 11 Proof: Suppose that U and V are orthogonal subspaces. That is u · v = 0 for any u ∈ U and v ∈ V . The since U and V are subspaces, we know that U ∩ V is also a subspace. Hence 0 ∈ U ∩ V . In particular, {0} ⊂ U ∩ V . We now show that U ∩ V ⊂ {0}. Suppose that x ∈ U ∩ V . Then x ∈ U and x ∈ V and so x · x = 0. This implies that x = 0, by an earlier proposition (i.e. x · x = x21 + x22 + . . . + x2n = 0 =⇒ x1 = x2 = . . . = xn = 0). Thus U ∩ V ⊂ {0} and hence U ∩ V = {0}. 6. Section 3.1, Problem 18 Proof: Suppose that U and V are subspaces of Rn and let x ∈ (U + V )⊥ . Then x · y = 0 for every y ∈ U + V . Of course, y ∈ U + V =⇒ y = u + v for some u ∈ U and v ∈ V . Thus x · (u + v) = x · u + x · v = 0 for every u ∈ U and v ∈ V . Now since U and V are subspaces, 0 ∈ U and 0 ∈ V . Let v = 0. Then x · u = 0 for every u ∈ U . Thus x ∈ U ⊥ . Similarly, by letting u = 0, we have that x ∈ V ⊥ . Thus x ∈ U ⊥ ∩ V ⊥ and so (U + V )⊥ ⊂ U ⊥ ∩ V ⊥ . Suppose that x ∈⊂ U ⊥ ∩ V ⊥ . Then x · u = 0 for every u ∈ U and x · v = 0 for every v ∈ V . Thus x · u + x · v = x · (u + v) = x · y = 0 for every y ∈ U + V . Thus x ∈ (U + V )⊥ . So U ⊥ ∩ V ⊥ ⊂ (U + V )⊥ and so (U + V )⊥ = U ⊥ ∩ V ⊥ . 7. Section 3.2, Problem 3b 1 1 0 Let A = 2 1 1. We look for matrices X and Y such that C(A) = N (X) 1 −1 2 and N (A) = C(Y ). To find such matrices, we put [A|b] in row echelon form where b = (b1 , b2 , b3 ) to find the constraint equations for b so that Ax = b is consistent and so that we may solve Ax = 0 to find N (A). Performing Gaussian elimination yields 1 1 0 b1 b1 1 1 0 b2 [A|b] = 2 1 1 b2 → 0 −1 1 1 −1 2 b3 0 0 0 3b1 − 2b2 + b3 . So we may write C(A) = {b ∈ R3 | 3b1 − 2b2 + b3 = 0}. Letting X = [3 − 2 1] will give a matrix whose nullspace is the column space of A. To find the nullspace of A we use the row echelon form of A (noting that x3 is 3 Math 317: Linear Algebra Homework 6 Solutions the free variable) to obtain: x1+x2 = 0 and −x2 + x3 = 0. So, x2 = x3 =⇒ −1 −1 x1 = −x3 and so N (A) = span 1 . Thus letting Y = 1 will ensure 1 1 that the column space of Y is the nullspace of A. 8. Section 3.2, Problem 6a 0 1 1 , 1 Suppose that A is a matrix such that ⊂ C(A) and 1 1 0 1 0 , 1 ⊂ N (A). We show that such a matrix is not possible to con 1 0 struct. Note the the elements in both C(A) and N (A) will force A to be a 3 × 3 matrix. I already have two columns of A, so we just need to find the third one. Using the nullspace as a guide, we have that 1 0 x1 1 0 1 1 x2 0 = 0 =⇒ x1 = x2 = x3 = −1. 1 1 x3 1 0 1 0 −1 0 0 So A = 1 1 −1. However A 1 6= 0. So such an A is impossible to 1 1 −1 0 0 construct. 9. Section 3.2, Problem 10 Suppose that A is an m × n matrix and that B is an n × p matrix. (a) We first show that N (B) ⊂ N (AB). Suppose that x ∈ N (B). Then Bx = 0. So A(Bx) = A0 = 0 =⇒ AB(x) = 0. Thus x ∈ N (AB). (b) Next, we show that C(AB) ⊂ C(A). Suppose that b ∈ C(AB). Then ABx = b for some x ∈ Rp . So A(Bx) = b =⇒ Ay = b for y = Bx. Thus, by definition, b ∈ C(A). (c) We now show that N (B) = N (AB) when A is n × n and nonsingular. It suffices to prove that N (AB) ⊂ N (B) since we have containment on one side from (a). Let x ∈ N (AB). Then ABx = 0 =⇒ A(Bx) = 0. Now since A is nonsingular, Ay = 0 =⇒ y = 0. Thus Bx = 0 =⇒ x ∈ N (B). (d) Finally, we show that C(AB) = C(A) when B is n × n and nonsingular. Once again, it suffices to only show that C(A) ⊂ C(AB) as we have containment on one side from part (b). Suppose that b ∈ C(A). Then Ax = b for some x ∈ Rn . Now B is nonsingular and hence B is invertible and so Ax = ABB −1 x = AB(B −1 x) = b. So b ∈ C(AB). Thus 4 Math 317: Linear Algebra Homework 6 Solutions C(A) = C(AB). 5