MA1101R Linear Algebra 1 AY 2008/2009 Sem 2 NATIONAL UNIVERSITY OF SINGAPORE MATHEMATICS SOCIETY PAST YEAR PAPER SOLUTIONS with credits to Wu Jiawei, Shi Xiaojie, Terry Lau Shue Chien MA1101R Linear Algebra 1 AY 2008/2009 Sem 2 Question 1 (a) (i) (w, x, y, z) = (w, x, w − x, 2w + x), (w, x ∈ R). (ii) (w, x, w − x, 2w + x) = x(0, 1, −1, 1) + w(1, 0, 1, 2) So V = span{(0, 1, −1, 1), (1, 0, 1, 2)}. So V is a subspace of R4 . (iii) For any a in V , a = x(0, 1, −1, 1) + w(1, 0, 1, 2). (0, 1, −1, 1) and (1, 0, 1, 2) are linearly independent. {(0, 1, −1, 1), (1, 0, 1, 2)} is a basis for V . So dim V = 2. (iv) Because dim V = 2, it is not a linearly independent set as v3 could be expressed as linear combinations of v1 and v2 since dim V = 2. (b) (i) Since u1 , u2 , u3 are linearly independent. Therefore for the equality to hold c1 u1 + c2 u2 + c3 u3 = 0 the only solution is c1 = c2 = c3 = 0. To prove the linear independency u1 + u2 , u1 − u2 , u3 , d1 (u1 + u2 ) + d2 (u1 − u2 ) + d3 u3 = 0 (1) Rearranging the above term, we have (d1 + d2 )u1 + (d1 − d2 )u2 + d3 u3 = 0 Comparing the coefficients of ui with (1), d1 + d2 = 0, d1 − d2 = 0 and d3 = 0. Therefore d1 = d2 = d3 = 0 which implies that u1 + u2 , u1 − u2 , u3 are linearly independent. Now let x ∈ R3 x = c1 u1 + c2 u2 + c3 u3 c1 − c2 c1 + c2 (u1 + u2 ) + (u1 − u2 ) + c3 u3 = 2 2 = d1 (u1 + u2 ) + d2 (u1 − u2 ) + d3 u3 Therefore T is a basis for R3 . (ii) Let v1 = u1 + u2 , v2 = u1 − u2 , v3 = u3 . So u1 transition matrix from S to T is 0.5 0.5 0.5 −0.5 0 0 NUS Math LaTeXify Proj Team Page: 1 of 5 = v1 +v2 2 and u2 = v1 −v2 2 , u3 = v3 . The 0 0 1 NUS Mathematics Society MA1101R Linear Algebra 1 AY 2008/2009 Sem 2 (c) Let ai = (0, 0 . . . , 0, 1, 0, . . . , 0), with the ith entry being 1. Let Vi = span{a1 , a2 , . . . , ai }. Because a1 , a2 , . . . , ai are linearly independent, dim Vi = i. {a1 , a2 , . . . , ai } is a subset of {a1 , a2 , . . . , ai+1 }. So it satisfies the condition. Question 2 (a) (i) The reduced row-echelon form of A is 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 (2) A basis for the row space of A is {(1, 0, 1, 0, 1), (0, 1, 0, 1, 0)}. 2 1 2 1 (ii) A basis for the column space of A is { 3 , 3 } 1 0 (iii) Let Ax = 0. Referring to (2) We get x = s 1 0 −1 0 0 + t 1 0 0 0 −1 +r 0 1 0 −1 0 . with r, s, t ∈ R. A basis for the null space is { (iv) Add two vectors (b) 1 0 0 0 0 and 1 1 0 0 −1 , 0 0 0 −1 0 0 1 0 into it. 0 0 , 0 1 0 −1 0 }. (i) if rank(C)=1, x−2=0 & x2 − x − 2 = 0 & x+1=0 There is no real x that satisfy the simultaneous solution. (ii) If rank(C) =2, either x − 2 6= 0 or x2 − x − 2 6= 0 or x2 − x − 2 = 0 and x + 1 = 0. Case(i) x 6= 2, then x2 − x − 2 must be zero. Then x = −1. Case(ii) x = 2, then x + 1 = 0. Therefore x = 2 or −1. (iii) if x 6= 2 and x 6= −1, rank(C)=3. β1 (c) let B be β2 , where β1 , β2 , β3 are the row vectors. β3 NUS Math LaTeXify Proj Team Page: 2 of 5 NUS Mathematics Society MA1101R Linear Algebra 1 AY 2008/2009 Sem 2 For any α = (x, y, z) in the null space of B, Bα = 0. So β1 α = 0, β2 α = 0, β3 α = 0, (a, b, c) belongs to the row space of B. x (a, b, c) = x1 β1 + x2 β2 + x3 β3 . So (a, b, c) y = 0. z So the nullspace of B is a subset of the plane ax + by + cz = 0. Question 3 (a) (i) r2 = r1 + r3 , ie. r2 is a redundant vector as a basis. r1 · r3 = 0. So S is an orthogonal basis for the vector space V. (ii) Let u = x1 r1 + x3 r3 . Solving, x1 = 5, x3 = 4. (u)s = (5, 4). (iii) The projection of v onto V is v · r3 1 4 4 10 v · r1 r1 + r3 = r1 + r3 = ( , , ) 2 2 |r1 | |r3 | 3 3 3 3 (b) 2 0 −2 | 1 (i) Consider 1 2 1 | 1 . 3 4 1 | 1 1 0 −1 | 12 The reduced row-echelon form is 0 1 1 | 14 0 0 0 | − 38 Ax = b is inconsistent. (ii) Consider AT Ax = AT b. 14 14 0 6 14 20 6 x = 6 0 6 6 0 s + 37 The least squares solutions are −s , (s ∈ R). s (c) Consider B T Bx = 0. rank(B T B)≤rank(B)≤ m < n. So it has infinitely many solutions. So Bx = b has infinitely many least squares solutions. Question 4 (a) λ−4 0 0 0 (i) λI − A = −1 λ − 4 0 0 λ−5 2 det(λI − A) = (λ − 4) (λ − 5) The eigenvalues of A are 4 and 5. NUS Math LaTeXify Proj Team Page: 3 of 5 NUS Mathematics Society MA1101R Linear Algebra 1 AY 2008/2009 Sem 2 (ii) λ1 = 4. Consider (4I − A)x = 0. 0 E4 = span 1 0 0 Similarly, we get E5 = span 0 1 (iii) dim E4 = 1, dim E5 = 1. A is not a diagonalizable matrix. −1 0 0 (iv) Let B = 0 0 0 , 0 0 0 3 0 0 A + B = 1 4 0 , 0 0 5 0 0 A + B has eigenvalues 4 with respect to 0 , 3 with respect to 1 , and 5 with respect 1 0 −1 to 1 Hence. it’s diagonalizable. 0 (b) Let A = 1 2 1 1 ,B = 2 0 0 3 −1 2 −1 . . A = 1 −1 C = ABA−1 . The eigenvalues of C are the same as B. The eigenvalues of C T are the same as C. So the eigenvalues of C T are 2 and 3. Consider (λI − C T )x = 0. 1 −1 , and the eigenvector corresponding to 3 is The eigenvector corresponding to 2 is 2 −1 T −1 T T T T −1 T T Alternatively, C = (A ) B A = (A ) B A , therefore we have AT C T (AT )−1 = B T −1 The eigenvalues are 2 with respect to eigenvectors , and 3 with respect to eigenvectors 2 1 −1 (c) X is diagonalizable matrix with one eigenvalue λ. dim Eλ = n. Consider (λI − X)x = 0, nullity(λI − X) = n. rank(λI − X) = 0. So λI − X = 0, X = λI. Question 5 (a) (i) A = 1 0 1 0 1 −1 . (ii) Consider Ax = 0, ker(T )= {(s, −s, −s)|s ∈ R}. (iii) rank(T )=rank(A)=2. nullity(T ) = nullity(A)=1 NUS Math LaTeXify Proj Team Page: 4 of 5 NUS Mathematics Society MA1101R Linear Algebra 1 AY 2008/2009 Sem 2 1 0 (iv) AT = 0 1 . 1 −1 rank(T )=rank(AT )=2, nullity(T )=nullity(AT )=0. 1 0 (v) R(T )= the column space of A. Both , belong to R(T ). So R(T ) = R2 . 0 1 (vi) Let the standard matrix of T and S be A and B. The standard matrix of S · T is I. I = BA. rank(I)≤rank(A)≤ 2. It contradicts rank(I)=3. So it is not possible. a b Alternatively, one can let B = c d , and compute BA, e f a b a b a−b 1 0 0 1 0 1 BA = c d = c d c−d = 0 1 0 0 1 −1 e f e f e−f 0 0 1 But if a = 1, then b = 1 and b = 0, which does not make sense. Hence it is impossible to find such a S. (b) Let the standard matrix of F be C. Let U = (u1 u2 . . . un ), V = (F (u1 ) F (u2 ) . . . F (un )). U, V are orthogonal matrices. So U U T = In , V V T = In , F (ui ) = Cui for 1 ≤ i ≤ n. V = CU . C = V U −1 and therefore CC T = I. So the standard matrix of F is an orthogonal matrix. NUS Math LaTeXify Proj Team Page: 5 of 5 NUS Mathematics Society

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