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Midterm Exam I, Math 121 A 27 October, 2006 Name: Solutions ID: Instructions: Put away any electronic devices, papers, notes, and books. Use only a pencil and eraser (or a pen if you are daring). Write neatly and only on the paper provided. Use the back of the page if necessary. 1. (25 pts.) Let T : R3 → R2 be defined via T ((a1 , a2 , a3 )) = (5a1 , a3 − 2a2 ). Show that T is linear. Find bases for N(T) and R(T). Confirm the Dimension Theorem for this example. Is T one-to-one? Is T onto? Solution: To show that T is linear let x = (a1 , a2 , a3 ) and y = (b1 , b2 , b3 ) and let c ∈ F . Then T(x + cy) = = = = = T ((a1 + cb1 , a2 + cb2 , a3 + cb3 )) (5(a1 + cb1 ), a3 + cb3 − 2(a2 + cb2 )) (5a1 + c5b1 , a3 − 2a2 + c(b3 − 2b2 )) (5a1 , a3 − 2a2 ) + c (5b1 , b3 − 2b2 ) T(x) + cT(y), (1) which shows that T is linear. N(T) is the set of all x = (a1 , a2 , a3 ) such that T(x) = (0, 0). Equivalently, x ∈ N(T) if and only if 5a1 = 0 a3 − 2a2 = 0. (2) Thus N(T) = {(0, s, 2s) | s ∈ R}. The set SN = {(0, 1, 2)} spans N(T). Since it is also linearly independent, SN is a basis for N(T). R(T) = R2 . To see this we will show that R2 ⊆ R(T). Let y = (b1 , b2 ) ∈ R2 be arbitrary. If we can find some x = (a1 , a2 , a3 ) ∈ R3 such that y = T(x), then y ∈ R(T). Now, y = T(x) if and only if b1 = 5a1 b2 = a3 − 2a2 . 1 (3) A solution to this system exists; one solution is x = (a1 , a2 , a3 ) = (b1 /5, 0, b2 ). Thus y ∈ R(T), which implies that R2 ⊆ R(T). Since R(T) ⊆ R2 always, it must be that R(T) = R2 . A basis for R2 is SR = {(1, 0), (0, 1)}; thus SR is a basis for R(T). By definition dim(N(T)) = 1 and dim(R(T)) = 2. The Dimension Theorem states that dim(N(T))+dim(R(T)) = dim(V), where V = R3 . Thus the Dimension Theorem is confirmed. T is not one-to-one because N(T) 6= {(0, 0, 0)}. Alternatively, T ((1, 1, 1)) = (5, −1) and T ((1, 0, 1/2)) = (5, −1), which shows directly that T is not one-to-one. T is onto because R(T) = R2 ./// 2 2. (25 pts.) Let V and W be vector spaces and let T : V → W be linear. If V1 is a vector subspace of V show that T(V1 ) = {w ∈ R(T) | T(v) = w for some v ∈ V1 } is a vector subspace of W. Solution: We must show that (a) 0W ∈ T(V1 ), (b) w1 + w2 ∈ T(V1 ), for all w1 , w2 ∈ T(V1 ), and (c) cw1 ∈ T(V1 ), for all w1 ∈ T(V1 ) and c ∈ F . (a) Since T is linear T(0V ) = 0W . It must be true that 0V ∈ V1 , because V1 is a vector subspace of V. Thus 0W ∈ T(V1 ), because there is some element v ∈ V1 , namely v = 0V , such that T(v) = 0W . (b) Let w1 , w2 ∈ T(V1 ). Then there exist v1 , v2 ∈ V1 such that T(v1 ) = w1 and T(v2 ) = w2 . By the linearity of T, w1 + w2 = T(v1 + v2 ). Now, it must be true that v1 + v2 ∈ V1 , because v1 , v2 ∈ V1 and V1 is a vector subspace of V. Thus w1 + w2 ∈ T(V1 ), because there is some element v ∈ V1 , namely v = v1 + v2 , such that T(v) = w1 + w2 . (c) Let w1 ∈ T(V1 ). Then there exists some v1 ∈ V1 such that T(v1 ) = w1 . Let c ∈ F . Then cv1 ∈ V1 , because V1 is a vector subspace of V. By the linearity of T, cw1 = T(cv1 ). Thus cw1 ∈ T(V1 ), because there is some element v ∈ V1 , namely v = cv1 , such that T(v) = cw1 ./// 3 3. (25 pts.) Prove that if {A1 , . . . , Ak } is a linearly independent subset of Mm×n (F ), then {At1 , . . . , Atk } is a linearly independent subset of Mn×m (F ). Solution: Suppose that S = {A1 , . . . , Ak } is linearly independent. Let Z be the zero matrix of Mm×n (F ): Zi,j = 0 for 1 ≤ i ≤ m and 1 ≤ j ≤ n. Then Z = 0Mm×n (F ) and Z t = 0Mn×m (F ) . Let a1 , . . . , ak ∈ F be arbitrary. S is linearly independent if and only if k X al Al = Z = 0Mm×n (F ) (4) l=1 implies that al = 0 for all l = 1, . . . , k. Equation (4) holds if and only if the following holds k X al (Al )i,j = 0, 1 ≤ i ≤ m, 1 ≤ j ≤ n. (5) l=1 Now, consider an arbitrary representation of zero as a linear combination of vectors from S t = {At1 , . . . , Atk }: k X al Atl = Z t = 0Mn×m (F ) . (6) l=1 Equation (6) holds if and only if the following holds k X al (Atl )i,j = 0, 1 ≤ i ≤ n, 1 ≤ j ≤ m. (7) l=1 Using (A)i,j = (At )j,i, Eq. (7) holds if and only if the following holds k X al (Al )j,i = 0, 1 ≤ i ≤ n, 1 ≤ j ≤ m. (8) l=1 Equations (8) and (5) are equivalent (only the indices are different) and Eq. (5) implies that al = 0 for all l = 1, . . . , k. Thus Eq. (8) implies that al = 0 for all l = 1, . . . , k. Hence, S t is linearly independent./// 4 4. (25 pts.) Let V and W be vector spaces such that dim(V) = dim(W) < ∞, and let T : V → W be linear. Show that there exist ordered bases β and γ for V and W, respectively, such that [T]γβ is diagonal. Solution: By the Dimension Theorem dim(V) = dim(N(T)) + dim(R(T)). Let dim(V) = n, dim(N(T)) = k, and dim(R(T)) = n − k, where 0 ≤ k ≤ n. Let βN = {vi }ki=1 be an ordered basis for N(T). We may extend βN to β = {vi }ni=1 , where β is an ordered basis for V. In the proof of the Dimension Theorem we showed that γR = {T(vi )}ni=k+1 is an ordered basis for R(T). We may extend γR to γ = {wi }ni=1 , where wi = T(vi ) for i = k + 1, . . . , n and γ is an ordered basis for W. Let A ∈ Mn×n (F ) denote the matrix representation of T in the order bases β and γ: A = [T]γβ . By definition T(vj ) = n X Ai,j wi . (9) i=1 Since vj ∈ N(T) for 1 ≤ j ≤ k, we have T(vj ) = 0W for 1 ≤ j ≤ k. Thus n X Ai,j wi = i=1 ( 0W wj if 1 ≤ j ≤ k . if k + 1 ≤ j ≤ n (10) The unique matrix that satisfies the last equation is given by Ai,j = ( 0 1 if i 6= j . if i = j and k + 1 ≤ j ≤ n Thus A = [T]γβ is diagonal./// 5 (11)