MATB24 – Midterm –Fall 2019 Professors: Camelia Karimianpour, Christopher Kennedy Solutions MATB24 Midterm , Page 1 of 11 1. Finish each of following sentences by giving complete, precise definitions for, or precise mathematical characterizations of, the italicized terms. (a) (3 points) A subspace W of a real vector space V . Solution: W is a subspace of real vector space V i W V and rw1+w2 2 W, 8w1, w2 2 W, r 2 R. (b) (3 points) A linear transformation T : V ! W , where V and W are real vector spaces. Solution: T : V ! W is a linear transformation i T (V ) W and T (rv1 + v2) = rT (v1) + T (v2), 8v1, v2 2 V, r 2 R. (c) (3 points) A linearly dependent subset S (finite or infinite) of a vector space V . Solution: S is a linearly dependent subset of V i there exist {s1, . . . , sn} S and a1, . . . , an 2 R such that Pn j=1 ajsj = 0 and aj 6= 0 for some j 2 {1, . . . , n}. (d) (3 points) An isomorphism between two vector spaces V and W . Solution: T is an isomorphism between vector spaces V and W i T : V ! W is an invertible (one-to-one and onto) linear transformation. MATB24 Midterm , Page 2 of 11 2. State whether each statement is true or false and provide a short justification for your claim (a short proof is you think the statement is true or a counter example if you think it is false). (a) (3 points) The orthogonal complement of Span 8<: 2412 3 359=; R3 is the plane x+2y+ 3z = 0 in R3. Solution: True, h(1, 2, 3)T , (x, y, z)T i = (x+2y+3z) = 0, 8(x, y, z) 2 {x+2y+3z = 0}. (b) (3 points) Let U and W be non-trivial subspaces of V . Then U [W is a subspace of V . Solution: False, for let U = {(x, 0)|x 2 R} and V = {(0, y)|y 2 R} be non-trivial sub- spaces of R2. Then U [W is not closed under addition since U [W 3 (1, 1) = (1, 0) + (0, 1) 2 U +W . (c) (3 points) If V is not finitely generated then every non-trivial subspace of V is also not finitely generated. Solution: False, for let P0 = span{1} be a non-trivial subspace of P = 1n=0Pn = span{1, x, x2, . . . }. MATB24 Midterm , Page 3 of 11 (d) (3 points) The set {ex, ex + e2x, ex + e2x + e3x} is a basis for Span{ex, e2x, e3x}. Solution: True, c1ex+c2e2x+c3e3x = (c1c2c3)ex+(c2c3)(ex+e2x)+c3(ex+e2x+e3x). (e) (3 points) The space of all upper triangular 3 3 matrices is isomorphic to P5. Solution: True, with isomorphism 0@a b c0 d e 0 0 f 1A 7! a+ bx+ cx2 + dx3 + ex4 + fx5. (f) (3 points) Let V and W be vector spaces, and suppose that T : V ! W is an injective linear transformation. If there are vectors ~v1,~v2, . . . ,~vk in V such that the vectors T (~v1), T (~v2), . . . , T (~vk) span W , then the vectors ~v1,~v2, . . . ,~vk span V . Solution: True, 8v 2 V, T (v) = w =Pkj=1 rjT (vj) = T (Pkj=1 rjvj) =) v =Pkj=1 rjvj. MATB24 Midterm , Page 4 of 11 3. In each part, give an explicit example of the mathematical object described or explain why such object does not exist. (a) (3 points) A 4-dimensional subspace of F , the vector space of all functions from R to R. Solution: Take P3 = span{1, x, x2, x3} F . (b) (3 points) A surjective and non-injective map from a 3-dimensional vector space V other than R3 to a 3-dimensional vector space W other than R3. Solution: False, by the Rank-Nullity Theorem. Let T : V ! W be a surjective linear map. Then dim(kerT ) = dim(V ) dim(ImT ) = dim(V ) dim(W ) = 3 3 = 0 =) kerT = {0V }, whence T is injective. It is true, however, if T need not be linear. (c) (3 points) Two di erent bases for M2 2(R), the vector space of all 2 2 matrices with entries in R. Solution: Take B1 = 1 0 0 0 ◆ , 0 1 0 0 ◆ , 0 0 1 0 ◆ , 0 0 0 1 ◆ and B2 = 1 0 0 1 ◆ , 0 1 0 0 ◆ , 0 0 1 0 ◆ , 1 0 0 1 ◆ . MATB24 Midterm , Page 5 of 11 (d) (3 points) An isomorphism between M2 2(R), the vector space of all 2 2 matrices with entries in R, and P3, the vector space of all cubic polynomials with coecients in R. Solution: Take isomorphism T : a b c d ◆ 7! a+ bx+ cx2 + dx3. (e) (3 points) Two di erent bases and for R2 and a change of basis matrix C that changes -coordinate of a vector ~v 2 R2 into -coordinates of ~v. Solution: Take = {(1, 1), (1,1)} and = {(1, 0), (0, 1)}. Then C! = 1 1 1 1 ◆ . (f) (3 points) An isomorphism between R3 and V = {A 2 M2 2(R)| tr(A) = 0}, that is the vector space of all 2 2 matrices with entries in R that have zero trace1. Solution: Take isomorphism T : (a, b, c) 7! a b c a ◆ . 1trace of a matrix is the sum ot its diagonal entries MATB24 Midterm , Page 6 of 11 4. Let Pn be the vector space of all polynomials with degree less than or equal to n. Consider the map D : Pn ! Pn, given by D(p(x)) = p0(x). (a) (4 points) Show that D is a linear transformation. Solution: Check D(p(x) + q(x)) = (p+ q)0(x) = p0(x) + q0(x) = D(p(x)) +D(q(x)). (b) (4 points) Prove that {a1x+ a2x2 + · · ·+ an1xn1 | ai 2 R} is a subspace of Pn. Solution: Let p(x) 2 S := {a1x+ a2x2 + · · ·+ an1xn1 | ai 2 R}. Then p(x) = 0 + a1x+ a2x2+· · ·+an1xn1+0xn 2 Pn. Moreover, rp(x)+q(x) = Pn1 j=1 (raj+bj)x n1 2 S, 8p(x) :=Pn1j=1 ajxj, q(x) :=Pn1j=1 bjxj 2 S, r 2 R, whence S is a subspace. (c) (5 points) Suppose n = 2. Let A = {1, x, x2} and B = {1 + x+ x2, 1 x+ x2, 1 + x x2} be bases for P2 (you don’t need to prove this). Suppose B is a matrix such that B[p(x)]B = [p0(x)]B. Find B. Solution: Observe that 1 + 2x = (1 + x+ x2) 12(1 x+ x2) + 12(1 + x x2), 1 + 2x = (1 + x + x2) 32(1 x + x2) 12(1 + x x2) and 1 2x = 1(1 + x + x2) + 3 2(1 x+ x2) + 12(1 + x x2) and conclude B = [D]B = [D(1 + x+ x2)]B | [D(1 x+ x2)]B | [D(1 + x x2)]B = [1 + 2x]B | [1 + 2x]B | [1 2x]B = 0@ 1 1 11 2 3 2 3 2 1 2 1 2 1 2 1A . MATB24 Midterm , Page 7 of 11 (d) (6 points) Let C be a matrix such that (C1BC)[p(x)]A = [p0(x)]A. Find C1. Solution: Recall that [D]A[p(x)]A = [p0(x)]A =) C1BC = [D]A =) C = CA!B and conclude C1 = CB!A = [1 + x+ x2]A | [1 x+ x2]A | [1 + x x2]A = 0@1 1 11 1 1 1 1 1 1A . (e) (6 points) Suppose p(x) = a + bx + cx2 2 P2 is such that [p0(x)]B = 2423 5 35. Find b and c. Solution: Calculate 0@ b2c 0 1A = [p0(x)]A = CB!A[p0(x)]B = 0@1 1 11 1 1 1 1 1 1A0@23 5 1A = 0@104 0 1A , whence (b, c) = (10, 2). Verify that 10 + 4x = 2(1 + x + x2) + 3(1 x + x2) + 5(1 + x x2). MATB24 Midterm , Page 8 of 11 5. Let U = { a b 0 c ◆ |a, b, c 2 R} be the vector space of all upper triangular 2 2 matrices. Let B = 1 0 0 0 , 1 1 0 0 , 0 1 0 1 ◆ be a basis of U . Let T : U ! R2 be the linear transformation defined by T a b 0 c ◆ = a b+ 2c ◆ . (a) (5 points) Find a basis for the kernel of T . Solution: Check that a b 0 c ◆ 2 kerT () T a b 0 c ◆ := a b+ 2c ◆ = 0 0 ◆ () (a, b, c) = (0,2, 1), 2 R. Conclude kerT = span{(0,2, 1)T}. (b) (5 points) Is T surjective Justify. Solution: Yes, fix x y ◆ 2 R2. Then x y 0 0 ◆ 2 U satisfies T x y 0 0 ◆ = x y ◆ . Or, show by Rank-Nullity that dim(ImT ) = dim(U) dim(kerT ) = 3 1 = 2. (c) (5 points) Show that Null([T ]B,E) is 1-dimensional. Solution: First calculate matrix [T ]B,E [T ]B,E = [T (b1)]E | [T (b2)]E | [T (b3)]E = 1 1 0 0 1 1 ◆ and deduce that kerT = {(1, 1,1) | 2 R}. MATB24 Midterm , Page 9 of 11 6. Let V andW be n andm-dimensional vector spaces respectively. Let k be a fixed integer between 1 and n. Suppose T : V ! W is a linear transformation. Let B = {~b1, · · · ,~bn} be a basis for V and {~vk, · · · ,~vn} be a subset of V . Let [T ]B,E be the matrix of T with respect to B. Assume the first k columns of [T ]B,E are zero and all other columns are linearly independent. (a) (6 points) Carefully prove that ker(T ) = Span{~b1, · · · ,~bk}. Solution: Recall that [T ]B,E = [T (b1)]E | · · · | [T (bn)]E and T = [·]1E [T ]B,E [·]B. Next calculate T (bj) = [·]1E [T ]B,E [·]B (bj) = [·]1E ([T ]B,E([bj]B)) = [·]1E ([T ]B,E(ij)1i) = [T (bj)] 1 E . Deduce, for 1 j k, T (bj) = [T (bj)]1E = [(0, . . . , 0)T ]1E = 0W and thus conclude that span{b1, . . . bk} kerT . Next, let b = Pn j=1 rjbj 2 kerT and T (bj) = wj, 8j 2 {k + 1, . . . , n} and calculate T (b) = T nX j=1 rjbj ! = nX j=1 rjT (bj) = kX j=1 rj[T (bj)] 1 E + nX j=k+1 rj[T (bj)] 1 E = kX j=1 0W + nX j=k+1 rj[T (bj)] 1 E = nX j=k+1 rj[T (bj)] 1 E = 0W . Then deduce Pn j=k+1 rj[T (bj)]E = hPn j=k+1 rj[T (bj)] 1 E i E = [0W ]E = (0, . . . , 0)T , whence rj = 0, 8j 2 {k + 1, . . . , n} by linear independence and thus conclude that kerT span{b1, . . . , bk} to complete the proof. MATB24 Midterm , Page 10 of 11 (b) (6 points) Suppose [~vj]B = ~ej + ~ek for k j n, where ~ej’s are standard vectors in Rn. Prove that Img(T ) = Span{T (~vk), · · · , T (~vn)}. Solution: First let w 2 span{T (vk), . . . , T (vn)} and verify that w = Pn j=k+1 rjT (vj) = T Pn j=k+1 rjvj 2 ImgT =) span{T (vk), . . . , T (vn)} ImgT . Next let w 2 ImgT . Then find u =Pnj=1 rjbj such that T (u) = w. Observing [vj]B = ej + ek =) vj = bj + bk, T (u) = T nX j=1 rjbj ! = T k1X j=1 rjbj + rk nX j=k+1 rj ! bk + nX j=k+1 rj(bj + bk) ! = k1X j=1 rjT (bj) + 1 2 rk nX j=k+1 rj ! T (2bk) + nX j=k+1 rjT (bj + bk) = 0W + 1 2 rk nX j=k+1 rj ! T (vk) + nX j=k+1 rjT (vj) = 1 2 rk nX j=k+1 rj ! T (vk) + nX j=k+1 rjT (vj). Then deduce w = T (u) 2 span{T (vk), . . . , T (vn)} and thus conclude ImgT span{T (vk), . . . , T (vn)} to complete the proof. Note that T (vk) = T (2bk) = 2T (bk) = 0W and so ImgT = span{T (vk+1), . . . , T (vn)} as well. MATB24 Midterm , Page 11 of 11 Blank page
