HW 07-solutions.pdf - berana(sb54896 u2013 HW 07 u2013...

This preview shows page 1 out of 12 pages.

Unformatted text preview: berana (sb54896) – HW 07 – gilbert – (53435) This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A = A is not diagonalizable . 10.0 points If the matrix  2 1 −1 4  A = is diagonalizable, i.e., A = P DP −1 with P invertible and D diagonal, which of the following is a choice for D?   −3 0 1. D = 0 2   −3 0 2. D = 0 −3   3 0 3. D = 0 3   3 0 4. D = 0 2 5. A is not diagonalizable correct  1 1 4 −2  is diagonalizable, i.e., A = P DP −1 with P invertible and D diagonal, which of the following is a choice for P ?   2 4 1. P = −1 1   −1 4 correct 2. P = 1 1   −1 4 3. P = 2 1   −1 1 4. P = 1 4 5. A is not diagonalizable Explanation: Since det [A − λI] = Consequently, 002 If the matrix 1  2−λ 1 −1 4−λ Explanation: Since  det [A − λI] = = 1 + (2 − λ)(4 − λ) = 9 − 6λ + λ2 , 4 −2 − λ  the eigenvalues of A are the solutions of 9 − 6λ + λ2 = (3 − λ)2 = 0 , λ2 + λ − 6 = (λ + 3)(λ − 2) = 0 , i.e., λ = 3, 3. 1 0  , so x2 is the only free variable. Thus the eigenspace Nul(A − 3I) has dimension 1. But then, when λ = 3, geo multA (λ) < alg multA (λ) . 1−λ 1 = −4 − (1 − λ)(2 + λ) = λ2 + λ − 6 , the eigenvalues of A are the solutions of On the other hand, when λ = 3,    1 −1 −1 = rref(A − λI) = rref 0 1 1  i.e., λ = −3, 2. Thus A is diagonalizable because the eigenvalues of A are distinct, and A = P DP −1 with P = [v1 v2 ] where v1 and v2 are eigenvectors corresponding to λ1 and λ2 respectively. To determine v1 and v2 we solve the equation Ax = λx. berana (sb54896) – HW 07 – gilbert – (53435) Ax = λx, λ = −3:        x x1 + 4x2 x1 1 4 = −3 1 , = x2 x1 − 2x2 x2 1 −2 which can be written as x1 + 4x2 = −3x1 , x1 − 2x2 = −3x2 , Ax = λx, λ = 2:        x x1 + 4x2 x1 1 4 = 2 1 , = x2 x1 − 2x2 x2 1 −2 which can be written as x1 − 2x2 = 2x2 , i.e., x1 = 4x2 . So one choice of v2 is   4 . v2 = 1 Consequently, A = P DP −1 with   −1 4 P = . 1 1 003 1 0 2 3. P = 1 1 0 0 0 1 4. A is not diagonalizable i.e., x1 = −x2 . So one choice of v1 is   −1 . v1 = 1 x1 + 4x2 = 2x1 , 2 10.0 points The eigenvalues of the matrix 2 0 −2 A = 1 3 2 0 0 3 are λ = 2, 3, 3. If A is diagonalizable, i.e., A = P DP −1 with P invertible and D diagonal, which of the following is a choice for P ? −1 0 −2 1. P = 1 1 0 correct 0 0 1 1 0 −2 2. P = 1 1 0 0 0 1 −1 5. P = 1 0 0 2 1 0 0 1 Explanation: We first determine the eigenspaces corresponding to λ = 2, 3: λ = 2 : since 0 0 −2 rref(A − 2I) = rref 1 1 2 0 0 1 1 1 0 = 0 0 1, 0 0 0 there is only free variable and −1 Nul(A − 2I) = s 1 : s in R . 0 Thus Nul(A − 2I) has dimension 1, and −1 v1 = 1 0 is a basis for Nul(A − 2I). λ = 3 : since −1 0 rref(A − 3I) = rref 1 0 0 0 1 0 2 = 0 0 0, 0 0 0 −2 2 0 there are two free variables and −2 0 Nul(A−3I) = s 1 + t 0 : s, t in R . 0 1 berana (sb54896) – HW 07 – gilbert – (53435) Thus Nul(A − 3I) has dimension 2, and −2 0 v2 = 1 , v3 = 0 , 0 1 form a basis for Nul(A − 3I). Consequently, A is diagonalizable because v1 , v2 , v3 are linearly independent, and A = P DP −1 with −1 0 −2 P = [v1 v2 v2 ] = 1 1 0 . 0 0 1 004 10.0 points 2. f (A) = Explanation: An n × n matrix A can be diagonalized, i.e. written as A = P DP −1 for some invertible matrix P and diagonal matrix D, if and only if A has linearly independent eigenvectors v1 , v2 , . . . , vn . f (A) = P f (D)P −1   f (d1 ) 0 P −1 . = P 0 f (d2 ) Now A can be diagonalized if we can find an eigenbasis of R2 of eigenvectors v1 , v2 of A corresponding to eigenvalues λ1 , λ2 , for then:   λ1 0 P −1 , P = [v1 v2 ] . A = P 0 λ2 But 3 − λ det[A − λI] = 2 By diagonalizing the matrix   3 −4 , A = 2 −3 compute f (A) for the polynomial f (x) = 3x3 − x2 − 2x − 2 . 1. f (A) =  0 −2 4 6  −4 −3 − λ = 8 − (3 − λ)(3 + λ) = λ2 − 1 = 0 , FALSE . 10.0 points 0 −2 then Consequently, the statement is 005  Explanation: If A can be diagonalized by   d1 0 −1 P −1 , A = P DP = P 0 d2 True or False? 2. TRUE 0 4 2 −6  −4 3. f (A) = 6   0 −4 correct 4. f (A) = 2 −6  Every n × n matrix A having eigenvectors v1 , v2 , . . . , vn can be diagonalized. 1. FALSE correct  3 i.e., λ1 = 1 and λ2 = −1. Corresponding eigenvectors are     1 2 , , v2 = v1 = 1 1 so P =  2 1  2 1 1 1 1 1  , P −1 =  1 −1 −1 2  . Thus f (A) =  f (1) 0 0 f (−1)  1 −1 −1 2  . berana (sb54896) – HW 07 – gilbert – (53435) Now f (1) = 3x3 − x2 − 2x − 2 x=1 = −2 , while f (−1) = 3x − x − 2x − 2 3 2 = −4 . x=−1 −1 2 10.0 points 1 2 1 x + . . . + xn + . . . , 2! n! tA compute e when 1. etA = as a matrix-valued function of t   −7 8 . A = −4 5 # " t 2e + et 2(e−3t − et ) et − e−3t t 2. etA = " 2e − e 3. etA = " 2e−3t + et tA 4. e = t 2e−3t + et 2(e −3t −3t e −e " 2e −3t e −3t −3t t −e ) t −e 2(et − e−3t ) t t −e 2e + e −3t # 2(et − e−3t ) e−3t − et 2et − e−3t # cor- e = Pe P = P  etd1 0 0 etd2  P P =  2 1 1 1  , −1 P −1 =  0 et  Consequently,    −3t 2 1 e tA e = 0 1 1 " 1 −1 . . 1 −1 −1 2  2(et − e−3t ) e−3t − et 2et − e−3t 10.0 points Find a matrix P so that   d1 0 P −1 , P 0 d2  1 4   1 −4  0 2. P = 1  −1 2 2e−3t − et −4 1. P = 1 then −1 so d1 ≥ d2 is a diagonalization of the matrix   2 8 A= 0 0 Explanation: If A can be diagonalized by   d1 0 −1 P −1 , A = P DP = P 0 d2 tD i.e., λ1 = −3 and λ2 = 1. Corresponding eigenvectors are     1 2 , , v2 = v1 = 1 1 007 rect tA = λ2 − λ − 3 = (λ + 3)(λ − 1) = 0 , = # 2e−3t − et −7 − λ 8 det[A − λI] = −4 5 − λ  Using the fact that ex = 1 + x + Now A can be diagonalized if we can find an eigenbasis of R2 of eigenvectors v1 , v2 of A corresponding to eigenvalues λ1 , λ2 , for then:   λ1 0 P −1 , P = [v1 v2 ] . A = P 0 λ2 But Consequently,    1 −2 0 2 1 f (A) = −1 0 −4 1 1   0 −4 . = 2 −6 006 4 # . berana (sb54896) – HW 07 – gilbert – (53435)  1 −4  4 1   1 0  −4 3. P = −1 −1 4. P = 0 −4 5. P = 1  1 6. P = 0 So, P = [u1 u2 ] and  A = P DP −1 is a diagonalization of A. Consequently,   1 2 0 , P = D= 0 0 0   −4 correct 1 008 Explanation: To begin, we must find the eigenvectors and eigenvalues of A. To do this, we will use the characteristic equation, det(A − λI) = 0. That is, we will look for the zeros of the characteristic polynomial. λ1 D= 0 0 λ2   2 = 0  0 . 0 Now to find the eigenvectors of A, we will solve for the nontrivial solution of the characteristic equation by row reducing the related augmented matrices:   2−2 8 0 [ A − λ1 I 0 ] = 0 0−2 0     0 −1 0 0 8 0 ∼ = 0 0 0 0 −2 0   1 , =⇒ u1 = 0 while [ A − λ2 I =  0] 2 8 0 0 =⇒  2+0 8 0 = 0 0+0 0    1 4 0 0 ∼ 0 0 0 0   −4 . u2 = 1  . 10.0 points The eigenvalues of an n × n matrix A are the entries on the main diagonal of A. True or False? 1. TRUE Explanation: The eigenvalues of a triangular n × n matrix A are the entries on the main diagonal of A. This is not necessarily true for matrices that are not triangular. = λ2 − 2λ = (λ − 2)(λ + 0) = 0.  −4 1 2. FALSE correct det(A − λI) = (2 − λ)(0 − λ) So 5  Consequently, the statement is FALSE . 009 10.0 points When  −5 A= −1 0 −4  find matrices D and P in a diagonalization of A given that λ1 > λ2 .     −3 −1 −4 0 , P = 1. D = −1 1 0 −5  −4 2. D = 0   −1 0 , P = 0 −5   0 5 0 , P = 3. D = −1 0 4  1 1 1 1   berana (sb54896) – HW 07 – gilbert – (53435) 4. D = rect  −4 0   0 0 , P = −1 −5   −3 5 0 , P = 5. D = −1 0 4 −1 1   −1 5 0 , P = 6. D = 0 0 4 1 1   1 1  6 So, P = [u1 u2 ] and cor- A = P DP −1 is a diagonalization of A. Consequently,    0 −4 0 , P = D= −1 0 −5   010 Explanation: To begin, we must find the eigenvectors and eigenvalues of A. To do this, we will use the characteristic equation, det(A − λI) = 0. That is, we will look for the zeros of the characteristic polynomial. det(A − λI) = (−5 − λ)(−4 − λ) 1 1  . 10.0 points For n × n matrix A is said to be diagonalizable when A = P DP −1 for some matrix D and invertible matrix P . True or False? 1. TRUE 2. FALSE correct Explanation: An n × n matrix A is said to be diagonalizable when A can be factored as = λ2 + 9λ + 20 = (λ + 4)(λ + 5) = 0. A = P DP −1 So  λ1 D= 0 0 λ2   −4 = 0  0 . −5 with D a DIAGONAL matrix and P an invertible matrix. Now to find the eigenvectors of A, we will solve for the nontrivial solution of the characteristic equation by row reducing the related augmented matrices:   −5 + 4 0 0 [ A − λ1 I 0 ] = −1 −4 + 4 0     −1 0 0 −1 0 0 ∼ = 0 0 0 −1 0 0   0 , =⇒ u1 = −1 while  −5 + 5 0 0 −1 −4 + 5 0     1 −1 0 0 0 0 ∼ = 0 0 0 −1 1 0   1 . =⇒ u2 = 1 [ A − λ2 I 0] =  Consequently, the statement is FALSE . 011 10.0 points If an n × n matrix A is diagonalizable, then A has n distinct eigenvalues. True or False? 1. TRUE 2. FALSE correct Explanation: If an n×n matrix has n distinct eigenvalues, it is diagonalizable, but the converse does not necessarily have to be true. For example, when 1 A = −3 3 3 −5 3 3 −3 , 1 berana (sb54896) – HW 07 – gilbert – (53435) then then A need not be diagonalizable. det[A − λI] = −(λ − 1)(λ + 2)2 = 0. Thus the eigenvalues of A are λ = 1, −2, −2. Now 1 rref(A − I) = 0 0 0 −1 1 1 , 0 0 1 Nul(A − I) = s −1 : s in R . 1 On the other hand, so True or False? 1. FALSE 2. TRUE correct so 7 1 1 rref(A + 2I) = 0 0 0 0 1 0, 0 Nul(A + 2I) −1 −1 = s 1 + t 0 : s, t in R . 0 1 But then A is diagonalizable because it has 3 linearly independent eigenvectors 1 −1 −1 −1 , 1 , 0 . 1 0 1 Since the eigenvalue λ = −2 is repeated, however, A does not have distinct eigenvalues. Consequently, the statement is FALSE . 012 10.0 points If A is a 7 × 7 matrix having eigenvalues λ1 , λ2 , and λ3 such that (i) the eigenspace corresponding to λ1 is two-dimensional, (ii) the eigenspace corresponding to λ2 is three-dimensional, Explanation: The eigenspace corresponding to λ3 must be at least one-dimensional because all eigenspaces must have a dimension at least one. But if this eigenspace is one-dimensional, then the sum of the dimensions of the three eigenspaces is six, which is less than seven, in which case A will not be diagonalizable. On the other hand, if the eigenspace corresponding to λ3 is two-dimensional, then the sum of the dimensions of the three eigenspaces is seven, in which case A will be diagonalizable. Consequently, the statement is TRUE . 013 10.0 points Find the solution of the differential equation   du 5 = Au(t), u(0) = −2 dt when A is a 2 × 2 matrix with eigenvalues 5 and 4 and corresponding eigenvectors     3 −1 . , v2 = v1 = −2 1 1. u(t) = 8e5t v1 + 3e4t v2 2. u(t) = 8e5t v1 − 3e4t v2 3. u(t) = 4e5t v1 − 6e4t v2 4. u(t) = 4e5t v1 + 3e4t v2 correct 5. u(t) = 8e5t v1 + 6e4t v2 6. u(t) = 4e5t v1 + 6e4t v2 berana (sb54896) – HW 07 – gilbert – (53435) Explanation: Since v1 and v2 are eigenvectors corresponding to distinct eigenvalues of A, they form an eigenbasis for R2 . Thus u(0) = c1 v1 + c2 v2 To compute c1 and c2 we apply row reduction to the augmented matrix   −1 3 5 [ v1 v2 u(0) ] = 1 −2 −2   1 0 4 . ∼ 0 1 3 This shows that c1 = 4, c2 = 3 and u(0) = 4v1 + 3v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues 5 and 4 respectively, u(t) = 4e5t v1 + 3e4t v2 . Then u(0) is the given initial value and Au(t) = 4e5t Av1 + 3e4t Av2   du(t) = 4 5e5t v1 + 3 4e4t v2 = . dt Thus u(t) = 4e5t v1 + 3e4t v2 solves the given differential equation. 014 10.0 points Find the solution of the differential equation   du 13 = Au(t), u(0) = −14 dt when A is the matrix  5 A= −6 1. u(t) =  3 −4 12e2t − e−t −12e2t + 2e−t   8 2. u(t) =  12e2t − e−t 12e2t − 2e−t 3. u(t) =  −12e2t + e−t −12e2t + 2e−t 4. u(t) =  −12e2t + e−t 12e2t − 2e−t 5. u(t) =  12e2t + e−t −12e2t − 2e−t 6. u(t) =  −12e2t − e−t 12e2t + 2e−t     correct  Explanation: Since 5 − λ 3 det[A − λI] = −6 −4 − λ = (5 − λ)(−4 − λ) + 18 = λ2 − λ − 2 = (λ − 2)(λ + 1), the eigenvalues of A are λ1 = 2, λ2 = −1 and corresponding eigenvectors     −1 −3 , v2 = v1 = 2 3 form a basis for R2 because λ1 6= λ2 . Thus u(0) = c1 v1 + c2 v2 . To compute c1 and c2 we apply row reduction to the augmented matrix   −3 −1 13 [ v1 v2 u(0) ] = 3 2 −14   1 0 −4 . ∼ 0 1 −1 This shows that c1 = −4, c2 = −1 and u(0) = −4v1 − v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues 2 and −1 respectively, set u(t) = −4e2t v1 − e−t v2 . berana (sb54896) – HW 07 – gilbert – (53435) Then u(0) is the given initial value and 6. u(t) = Au(t) = −4e2t Av1 − e−t Av2   du(t) = −4 2e2t v1 − −e−t v2 = . dt Thus u(t) is a solution of the differential equation. But     −t −1 2t −3 + (−)e u(t) = −4e 2 3   12e2t + e−t = . −12e2t − 2e−t Consequently, u(t) =  12e2t + e−t −12e2t − 2e−t  solves the given differential equation. 015 10.0 points Find the solution of the differential equation   du −10 = Au(t), u(0) = −12 dt when A is the matrix  1 −5 A= 3 0 2 −3 5  5  1. u(t) =  12e−t + 2e− 3 t 12e−t 2. u(t) =  12e−t + 2e− 3 t −12e−t 3. u(t) =  12e−t − 2e− 3 t 12e−t 4. u(t) =  −12e − 2e 12e−t 5. u(t) =  −12e−t − 2e− 3 t −12e−t 5 −t  5 5 −12e−t + 2e− 3 t −12e−t  correct Explanation: Since 5 − − λ 3 det[A − λI] = 0 2 3 −1 − λ = (− 53 − λ)(−1 − λ) + 0 5 8 = λ2 + λ + = (λ + 1)(λ+ 53 ), 3 3 the eigenvalues of A are λ1 = −1, λ2 = − 53 and corresponding eigenvectors     −1 −3 , v2 = v1 = 0 −3 form a basis for R2 because λ1 6= λ2 . Thus u(0) = c1 v1 + c2 v2 . To compute c1 and c2 we apply row reduction to the augmented matrix   −3 −1 −10 [ v1 v2 u(0) ] = −3 0 −12   1 0 4 . ∼ 0 1 −2 This shows that c1 = 4, c2 = −2 and u(0) = 4v1 − 2v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues −1 and − 35 respectively, set 5 u(t) = 4e−t v1 − 2e− 3 t v2 . Then u(0) is the given initial value and 5 Au(t) = 4e−t Av1 − 2e− 3 t Av2    du(t) −t 5 − 53 t v2 = = 4 −e v1 − 2 − 3 e dt  − 35 t  9   Thusu(t) solves the differential equation. But     − 53 t −1 −t −3 − 2e u(t) = 4e −3 0   −t − 35 t −12e + 2e = −12e−t berana (sb54896) – HW 07 – gilbert – (53435) 10 This shows that c1 = −5, c2 = −1 and Consequently, u(t) =  5 −12e−t + 2e− 3 t −12e−t  solves the given differential equation. 016 10.0 points Since v1 and v2 are eigenvectors corresponding to the eigenvalues 4 and 2 respectively, set u(t) = −5e4t v1 − e2t v2 . Then u(0) is the given initial value and Let u(t) satisfy du = Au(t), dt u(0) = −5v1 − v2 . u(0) =   6 . −2 Compute u(2) when A is a 2 × 2 matrix with eigenvalues 4 and 2 and corresponding eigenvectors     4 −2 . , v2 = v1 = −8 2   10e8 + 4e4 1. u(2) = 10e8 + 8e4 2. u(2) =   −10e8 + 4e4 10e8 − 8e4 3. u(2) =  −10e8 − 4e4 10e8 + 8e4  4. u(2) =  10e8 + 4e4 −10e8 − 8e4  5. u(2) =  10e8 − 4e4 −10e8 + 8e4  6. u(2) =  −10e8 − 4e4 −10e8 − 8e4  Au(t) = −5e4t Av1 − e2t Av2   du(t) = −5 4e4t v1 − 2e2t v2 = . dt Thus u(t) = −5e4t v1 − e2t v2 solves the given differential equation. Consequently, u(2) = 017 correct Explanation: Since v1 and v2 are eigenvectors corresponding to distinct eigenvalues of A, they form an eigenbasis for R2 . Thus u(0) = c1 v1 + c2 v2 To compute c1 and c2 we apply row reduction to the augmented matrix   −2 4 6 [ v1 v2 u(0) ] = 2 −8 −2   1 0 −5 . ∼ 0 1 −1  10e8 − 4e4 −10e8 + 8e4  . 10.0 points Find the solution of the differential equation   du −10 = Au(t), u(0) = 6 dt when A is the matrix  −7 A= 9 −12 14 1. u(t) =  −6e5t − 16e2t −6e5t − 12e2t  2. u(t) =  6e5t + 16e2t −6e5t − 12e2t  3. u(t) =  −6e5t − 16e2t 6e5t + 12e2t  4. u(t) =  6e5t + 16e2t 6e5t + 12e2t   berana (sb54896) – HW 07 – gilbert – (53435) 5. u(t) =  −6e5t + 16e2t 6e5t − 12e2t  6. u(t) =  6e5t − 16e2t −6e5t + 12e2t  11 Consequently, u(t) = correct  6e5t − 16e2t −6e5t + 12e2t  solves the given differential equation. Explanation: Since −7 − λ det[A − λI] = 9 −12 14 − λ = (−7 − λ)(14 − λ) + 108 = λ2 − 7λ + 10 = (λ − 5)(λ−2), the eigenvalues of A are λ1 = 5, λ2 = 2 and corresponding eigenvectors     4 −2 , v2 = v1 = −3 2 2 form a basis for R because λ1 6= λ2 . Thus u(0) = c1 v1 + c2 v2 . To compute c1 and c2 we apply row reduction to the augmented matrix   −2 4 −10 [ v1 v2 u(0) ] = 2 −3 6   1 0 −3 . ∼ 0 1 −4 This shows that c1 = −3, c2 = −4 and u(0) = −3v1 − 4v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues 5 and 2 respectively, set u(t) = −3e5t v1 − 4e2t v2 . Then u(0) is the given initial value and Au(t) = −3e5t Av1 − 4e2t Av2   du(t) = −3 5e5t v1 − 4 2e2t v2 = . dt This shows that u(t) solves the differential equation. But     4 2t 5t −2 − 4e u(t) = −3e −3 2   6e5t − 16e2t = . −6e5t + 12e2t 018 10.0 points Let A be a 2 × 2 matrix with eigenvalues −1 and − 34 and corresponding eigenvectors     3 1 . , v2 = v1 = 0 1 Let {xk } be a solution of the difference equation xk+1 = Axk ,  −10 . x0 = −1  Compute x1 .   −1 1. x1 = 13   13 2. x1 = −1   1 3. x1 = 13   −13 4. x1 = −1   −1 5. x1 = −13   13 correct 6. x1 = 1 Explanation: To find x1 we must compute Ax0 . Now, express x0 in terms of v1 and v2 . That is, find c1 and c2 such that x0 = c1 v1 + c2 v2 . This is certainly possible because the eigenvectors v1 and v2 are linearly independent (by inspection and also because they correspond to berana (sb54896) – HW 07 – gilbert – (53435) distinct eigenvalues) and hence for R2 . The row reduction  1 3 [ v1 v2 x0 ] = 1 0  1 0 ∼ 0 1 form a basis  −10 −1  −1 −3 shows that x0 = −v1 − 3v2 . Since v1 and v2 are eigenvectors (for the eigenvalues −1 and − 34 respectively): x1 = Ax0 = A(−v1 − 3v2 ) = = −Av1 − 3Av2 = −−1v1 − 3 · − 34 v2       13 12 1 . = + 1 0 1 Consequently, x1 = 019  13 1  . 10.0 points Let A be a 3 × 3 matrix with eigenvalues 1, −1, and −2 and corresponding eigenvectors 1 −3 −1 v1 = −1 , v2 = 2 , v3 = 2 . −3 6 8 If {xk } is the solution of the difference equation 2 xk+1 = Axk , x0 = −5 , −19 determine x1 . −17 1. x1 = −3 −2 2 2. x1 = −3 17 −2 3. x1 = −3 −17 12 2 4. x1 = 3 correct 17 −17 5. x1 = 3 −2 17 6. x1 = 3 2 Explanation: To find x1 we must compute Ax0 . First, we express express x0 in terms of v1 , v2 , and v3 : x0 = c1 v1 + c2 v2 + c3 v3 . This is certainly possible as the eigenvectors v1 , v2 , and v3 are linearly independent because the eigenvalues are distinct. Hence they form a basis for R3 . The row reduction 1 −3 −1 2 [ v1 v2 v3 x0 ] = −1 2 2 −5 −3 6 8 −19 1 0 0 3 ∼ 0 1 0 1 0 0 1 −2 shows that x0 = 3v1 + v2 − 2v3 . But v1 , v2 and v3 are eigenvectors for the respective eigenvalues 1, −1 and −2, so x1 = Ax0 = A(3v1 + v2 − 2v3 ) = 3Av1 + Av2 − 2Av3 = 3 · (1)v1 + (−1)v2 − 2 · (−2)v2 3 3 −4 2 = −3 + −2 + 8 = 3 . −9 −6 32 17 Consequently, 2 x1 = 3 . 17 ...
View Full Document

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture