201y03fin - B U Department of Mathematics Math 201 Matrix...

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Unformatted text preview: B U Department of Mathematics Math 201 Matrix Theory Summer 2003 Final Exam This archive is a property of Bo˘azi¸i University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. gc This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 1. For what values of a does the system : ax + y = 1 4x + ay = 2 have (i) a unique solution (ii) infinitely many solutions (iii) no solution? Find also the rank of the coefficient matrix in each case. Solution: To have a unique solution the coefficient matrix A= a1 4a must be non-singular since the system is square. detA = a2 − 4 = 0 ⇒ a = 2, a = −2 (i) unique solution if a = ±2 , which is : x y = A− 1 1 2 In this case rankA = 2 (ii) if a = 2 : 2x + 1 = 1 4x + 2y = 2 Aug = 21|1 42|2 −−−> 21|1 00|0 2x + y = 1. One equation two unknowns ⇒infinitely many solutions. In this case rankA = 1 −2 1 | 1 4 −2 | 2 sistent and hence no solution. (iii) if a = −2 : Aug = In this case 00 3 2 2. Let A= 41 −−− > −2 1 | 1 0 0|4 rankA = 1 1 2 . 3 (a) Find QR-decomposition of A, where Q is an orthogonal matrix. Solution: . Incon- Q is the orthogonal matrix obtained from A by Gram-Schmidt process α1 =< 0, 3, 4 >, α2 =< 0, 2, 1 > , α3 =< 1, 2, 3 > labelling the columns x1 = α1 αT x 2 x2 = α2 − ||x1 ||1 x1 =< 0, 4 , −3 > 2 55 αT x αT x 3 3 x3 = α3 − ||x1 ||1 x1 − ||x2 ||2 x2 =< 1, 0, 0 > 2 2 x1 , x2 , x3 is an orthogonal set. Then q1 = q2 = 34 x1 =< 0, , > ||x1 || 55 x2 4 −3 =< 0, , > ||x2 || 55 q3 = x3 =< 1, 0, 0 > ||x3 || form an orthonormal set. 0 0 1 Q = [q1 |q2 |q3 ] = 3/5 4/5 0 . 4/5 −3 0 /5 T T T 5 2 18/5 q1 α1 q1 α2 q1 α3 T T q2 α2 q2 α3 = 0 1 −1/5 . R= T q3 α3 00 1 So that A = QR. (b) Find the inverse of Q, if it exists. Solution: Since Q is orthogonal, we have QT Q = QQT = I . Hence Q−1 = QT . 3. (a) Show that if A is similar to B , then Ak is similar to B k Solution: Given that there exists an invertible M such that M −1 AM = B or A = M BM −1 , compute Ak Ak = (M BM −1 )(M BM −1 )...(M BM −1 )[ktimes] ⇒ Ak = M B k M −1 . Hence Ak is similar to B k , via the same matrix M . (b) Let A be an m × n matrix. Prove that if tr(AT A) = 0 then A = 0. Solution: Let A = [q1 |q2 |...|q3 ]mxn T T T q1 q1 q1 q2 ... q1 qn T q T q q T q ... q2 qn Hence, AT A = 2 1 2 2 : : T T T qn q1 qn q2 ... qn qn T ⇒ (AT A)ii = qi qi = ||qi ||2 T T T tr(AT A) = q1 q1 + q2 q2 + ... + qn qn = ||q1 ||2 + ... + ||qn ||2 tr(AT A) = 0 ⇒ qi = 0 for every i = 1, 2, ..., n So each column of A is zero. Therefore A = 0. 2 −1 3 −1 6 has an eigenvector v1 = −2 . It is also known that 4. The matrix A = 0 0 0 3 −7 3 λ = 2 is an eigenvalue of A. (a) Using the information, diagonalize A. Solution: If v1 is an eigenvector, then there is an eigenvalue λ such that 2 −1 3 −1 −1 9 −1 6 −2 = λ −2 ⇒ 18 = λ −2 Av1 = λv1 ⇒ 0 0 0 3 −7 3 3 −27 3 Therefore another eigenvalue is λ = −9, with an eigenvector v1 λ = 2 is an eigenvalue(given). Let us find 0 −1 3 0 A − 2I = 0 −2 6 − −− > 0 0 3 −9 0 the eigenvector(s) for λ = 2 : −1 3 0 0 ⇒ −x2 + 3x3 = 0 00 x1 1 0 3x3 = x1 0 + x3 3 x1 is free.⇒ (A − 2I )x = 0 is satisfied if x = x3 0 1 There are eigenvectors for λ = 2. Taking these two eigenvectors to be : two 1 0 v2 = 0 and v3 3 , We can diagonalize A 0 1 ||| −1 1 0 −9 2 S −1 AS = Λ where S = v1 v2 v3 = −2 0 3 and = ||| 3 01 2 (b) Find A2003 . (Leave it as a product.) Solution: A = S ΛS −1 ⇒ A2003 = S Λ2003 S −1 − 1 −1 1 0 −92003 −1 1 0 −2 0 3 22003 = −2 0 3 2003 3 01 2 3 01 ⇒ A2003 5. Let T : R2 − > R2 be a linear transformation satisfying T (< 1, 0 >) =< −4, 3 > and T (< 1, 1 >) =< −10, 8 > . Let A be a matrix of T in the standart basis. (a) Find A. Solution: We need T (< 0, 1 >). But < 0, 1 >=< 1, 1 > − < 1, 0 > ⇒ T (< 0, 1 >) = T (< 1, 1 >) − T (< 1, 0 >) ⇒ T (< 0, 1 >) =< −10, 8 > − < −4, 3 >=< −6, 5 > −4 −6 ⇒ A = [T e1 |T e2 ] = 3 5 (b) What is the matrix B representing T in the basis that consist of eigenvectors of A ? Solution: Eigenvalues of A : |A − λI | = −4 − λ −6 3 5−λ = (5 − λ)(−4 − λ) + 18 = 0 ⇒ λ = 2, λ = −1 λ = 2 : A − 2λ = −6 −6 3 3 ⇒ x1 = −x2 ⇒ an eigenvector is p1 = λ = −1 : A + λ = −3 −6 3 6 ⇒ x1 = −2x2 ⇒ an eigenvector is p2 = T (p1 ) = −4 −6 3 5 1 −1 = 2 −2 −4 −6 3 5 −2 1 = 2 −1 −2 1 = 2p1 + 0p2 T (p2 ) = 1 −1 = 0p1 − 1p2 B = [T p1 |T p2 ] = 2 2 −2 −1 (c) Solve the system of differential equations −1 u0 = −2 du dt = Au with the initial condition Solution: du dt = Au has the general solution : u = SeΛt S −1 u0 S= 1 −2 −1 1 e2t e−t (found above, eigenvectors of A.) Λ = 2 −1 ⇒ eΛt = S −1 = −1 −2 −1 −1 , c = S −1 u0 = ⇒ u = c1 eλ1 t p1 + c2 eλ2 t p2 −2 1 u = −e2t − e−t −1 1 −1 −2 −1 −1 1 0 = −1 −1 = c1 c2 u1 u2 . Let u = u1 = −e2t + 2e−t u2 = +e2t − e−t 6. Let A be square matrix with eigenvalues λ1 = 2, λ2 = 1 and λ3 = 5 with multiplicities 3, 2 and 2 respectively. Let Ei be the eigenspace associated with the eigenvalue λi , i = 1, 2, 3. Assume that dimE1 = 2, dimE2 = 1 and dimE3 = 2 (a) Is A diagonalizable? Explain. Solution: Counting the multiplicities, we understand that A is 7x7. dimE1 + dimE2 + dimE3 = 2 + 1 + 2 = 5 = 7. Hence A is not diagonalizable. (b) Give a Jordan form J of A. Solution: λ = (0), (0, 0) 2 eigenvectors. λ = (1, 1) 1 eigenvector. λ = (5), (5) 2 eigenvectors. J contains 5 blocks. 01 0 [0] 11 J = 1 [5] [5] 7 x7 (c) Write down the characteristic polynomial of A. Solution: Using the fact that eigenvalues are roots of the characteristic polynomial p(λ) = −λ3 (λ − 1)2 (λ − 5)2 (the front minus sign comes from the order of A) (d) Is A invertible?Justify your answer fully. Solution: To be invertible, detA = 0 has to be satisfied. But A has λ = 0 as an eigenvalue and detA = λ1 λ2 ...λ7 which yields that detA = 0. The answer is no. (e) Find the trace of A. Solution: trA = λ1 + λ2 + ... + λ7 = 0 + 0 + 0 + 1 + 1 + 5 + 5 = 12 ...
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This note was uploaded on 11/16/2011 for the course MATH 201 taught by Professor Soysal during the Fall '08 term at Boğaziçi University.

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