AM 034 Applied Mathematics - II Brown University Spring 2018 Solutions to Homework, Set 3 Due February 28

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Rudolf Mosley
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1 AM 034 Applied Mathematics - II Brown University Spring 18 Solutions to Homework, Set 3 Due February ( points) Find all square roots of the following matrices: [ ] (a), (b) Solution Grading: PART A: +4 for resolvent, +2 for first correct sqrt, +2 for second correct sqrt PART B: +4 for resolvent, +2 for first correct sqrt, +2 for second correct sqrt, +2 for third correct sqrt, +2 for fourth correct sqrt (a) The resolvent of the given matrix is R λ (A) = 1 (λ 49)(λ 9) [ ] λ λ 31 Now square roots are sums of two residues: ( ) R 1 = Res +Res R λ (A) λ λ=9 λ=49 = 3 [ λ λ 31] 7 [ λ λ= λ 31] = 3 [ ] [ ] 11 9 = 1 [ ] Since Sylvester s auxiliary matrix at λ = a is Z (a) = Res λ=a R λ (A), we have Z (9) = Res R λ(a) = A 49I λ= = 1 [ ] 9 9, Z (49) = Res R λ(a) = A 9I λ= = 1 [ ] Another root is R 2 = 1 [ ] 5 9 = 3Z (9) +7Z (49) Obviously, other two roots are R 1 and R 2 Mathematica codes: A = {{31, 18}, {22, 27}} Eigenvalues[A] P[lambda_] = Simplify[Inverse[ lambda*identitymatrix[2] - A]*(lambda - 49)*(lambda - 9)] λ=49

2 (b) The matrix has three positive eigenvalues λ = 1,1,4 (so λ = 1 is a double eigenvalue), and its resolvent is R λ = (λi A) 1 = 1 λ λ 7 9, ψ(λ) 3 3 λ+5 where ψ(λ) = (λ 4)(λ 1) is the minimal polynomial for the given 3 3 matrix According to the resolvent method, the four roots of this matrix can be obtained by summing the residues: where Z (1) = Res λ=1 R λ (A) = (A 4I) = (1 4) R = Res +ResR λ(a) λ = Z (1) 1+Z(4) 4 λ=1 λ=4 = ±Z (1) ±2Z (4), , Z (4) = Res λ=4 R λ (A) = (A I) are Sylvester s auxiliary matrices For one choice of square roots, we have Another root: λ λ R 1 = 3 λ 7 9 (λ 4) 3 3 λ+5 λ=1 λ λ + 3 λ 7 9 (λ 1) 3 3 λ = = λ= R 2 = = Two other roots are negative of these ones Mathematica code to find the resolvent: A = {{-2, -3, 9}, {-3, -2, 9}, {-3, -3, 10}} Simplify[Inverse[lambda*IdentityMatrix[3] - A]] (4 1) = We obtain the same output using diagonalization procedure Using eigenvectors 3,0,1 T, 1,1,0 T and 1,1,1 T, corresponding to eigenvalues λ = 1 and λ = 4, respectively We build a matrix S of these eigenvectors, S = 0 1 1, using Mathematica:

3 SS = Eigenvectors[A] S = Transpose[SS] Then all eight square roots of matrix A are obtained by choosing certain combinations of signs in the formula: ±1 0 0 S 0 ±1 0 S ±2 Note that the diagonalization procedure provides two (actually four when count negative) more square roots that cannot be obtained using neither Sylvester s method nor the Resolvent method: R 3 = 3 4 9, R 4 = ( points) ( For each matrix from the previous exercise, find At ) sin ( ) Φ(t) = and Ψ(t) = cos At A Solution Grading: PART A: +5 for Φ, +5 for Ψ PART B: +5 for Φ, +5 for Ψ (a) Using Sylvester s formula, we define the matrix-functions: ( λt ) ( sin λt ) sin Φ(t) = λ Z (9) + λ Z (49) λ=9 λ=49 = sin3t Z (9) + sin7t Z (49) = sin3t [ ] sin7t [ ] and ( ) ( ) Ψ(t) = cos λt Z (9) + cos λt λ=9 λ=64 = cos3tz (9) +cos7tz (49) = cos3t Z (49) [ ] + cos7t [ ] (b) Similarly, independently which square root was chosen, we have that ( λt ) ( sin λt ) sin Φ(t) = λ Z (1) + λ Z (4) λ=1 λ=4 = sintz (1) + sin2t Z (4) = sint sin2t

4 and ( ) Ψ(t) = cos λt λ= = cost ( ) Z (1) + cos λt λ=4 +cos2t Z (4) 33 ( points) For each matrix from the previous exercise, show that the matrix-functions Φ(t) and Ψ(t) satisfy the matrix differential equation d 2 dt Φ(t)+AΦ(t) = 0 or d 2 Ψ(t)+AΨ(t) = 0 2 dt2 What initial conditions do these matrix-functions satisfy? Solution Grading: PART A: +5 for showing for Φ, +5 for showing for Ψ PART B: +5 for showing for Φ, +5 for showing for Ψ (a) First, we check the initial conditions: Φ(0) = sin(3 0) 3 Ψ(0) = cos(3 0) [ ] [ ] sin(7 0) 7 + cos(7 0) [ ] [ ] = = 0, [ ] 11 9 = [ For the first derivatives, we have d dt Φ(t) = Φ(t) = cos3t [ ] cos7t [ ] 11 9, d dt Ψ(t) = Ψ(t) = 3 sin3t [ ] sin7t [ ] ] = [ ] Setting t = 0, we get Φ(0) = 1 [ ] [ ] 11 9 = 11 9 [ ] Obviously, Ψ(0) = 0 Calculating second derivatives, we obtain d 2 dt 2 Φ(t) = 3 sin3tz (9) 7 sin7tz (64) d 2 dt 2 Ψ(t) = 9 cos3tz (9) 49 cos7tz (49) 4

5 To prove that matrix functions Φ(t) and Ψ(t) satisfy the second order matrix differential equation, we have to show that 9Z (9) = AZ (9) and 49Z (49) = AZ (49) Indeed, [ ] [ ] AZ (9) = = 1 [ ] [ ] [ ] AZ (49) = = 49 [ ] 11 9 = 49Z (64) = 9 [ ] 9 9 = 9Z (9), (b) Calculating derivatives, we get d Φ(t) = sint sin2t , dt d Ψ(t) = cost cos2t dt Next step is to evaluate products of matrices: AZ (1) = = = Z (1), AZ (4) = = = 4Z (4) Therefore, both matrix-functions, Φ(t) and Ψ(t), satisfy the second order differential equation The initial conditions follow from the following equations: Φ(0) = 0, Ψ(0) = Z (1) +Z (4) = = = I, Φ(0) = Z (1) +Z (4) = I, Ψ(0) = 0 5

6 ( points) For the matrix B = , construct Φ(t) = e Bt and Ψ(t) = cos(bt) Show that these matrix-functions are solutions of the following matrix differential equations: respectively d dt Φ(t) = BΦ(t) and d 2 dt 2 Ψ(t)+B2 Ψ(t) = 0, Solution Grading: +5 for constructing Φ +5 for constructing Ψ +5 for showing that Φ satisfies its differential equation +5 for showing that Ψ satisfies its differential equation Here is Matlab code to find these two functions: 1 I3 = [1 0 0; 0 1 0; 0 0 1] C = [ ; ; 7 3 9] 3 C Res = inv(x I3 C) C2 = x I3 C 5 C3 = cofactor (C2) C4 = transpose(c3) 7 C5 = factor (det(c4)) Z1 = (C 4 I3) (C 8 I3)/(1 4)/(1 8) 9 Z4 = (C I3) (C 8 I3)/(4 1)/(4 8) Z8 = (C I3) (C 4 I3)/(8 1)/(8 4) 11 IT = exp(t) Z1+exp(4 t) Z4+exp(8 t) Z8 WT = cos(t) Z1 + cos(4 t) Z4 + cos(8 t) Z8 13 isequal (exp(t) Z1+4 exp(4 t) Z5+8 exp(8 t) Z8, C IT) isequal( cos(t) Z1 16 cos(4 t) Z4 64 cos(8 t) Z8, (Cˆ2) WT) 15 function C = cofactor (A, i, j ) % cofactor matrix 17 % COFACTOR(A, i, j ) returns the cofactor of row i, column j % COFACTOR(A) returns the matrix C of cofactors 19 if nargin == 3 % Remove row i and column j to produce the minor 21 M = A; M(i,:) = []; 23 M(:, j ) = []; C = ( 1)ˆ( i+j ) det(m) ; 25 else [n,n] = size (A) ; 27 for i = 1:n for j = 1:n 29 C(i, j ) = cofactor (A, i, j ) ; end 31 end end 6

7 33 end Here is MuPad code to find these two functions: 1 B := matrix(3,3,[[15,3, 17],[7,7, 13],[7,3, 9]]) V3 := linalg :: eigenvectors(b) 3 S3 := matrix (3,3,[[2,1,1],[1,2,1],[1,1,1]]) linalg :: eigenvalues(b) 5 Dia3 := matrix (3,3,[[8,0,0],[0,4,0],[0,0,1]]) B = S3 Dia3 inverse (S3) 7 I3 := matrix (3,3,[[1,0,0],[0,1,0],[0,0,1]]) / or / I3 := matrix :: identity (3) 9 factor ( linalg :: det(s I3 B)) / or / factor ( linalg :: charpoly(b, s)) 11 z8 := (B 4 I3) (B I3)/((8 4) (8 1)) z4 := (B 8 I3) (B I3)/((4 8) (4 1)) 13 z1 := (B 4 I3) (B 8 I3)/((1 4) (1 8)) Phi 3 := simplify (exp(8 t) z8 + exp(4 t) z4 + exp(t) z1) 15 Psi 3 := simplify (cos(t) z1 + cos(4 t) z4 + cos(8 t) z8) simplify ( diff (Phi 3, t) = B Phi 3) 17 diff ( diff (Psi 3, t),t) + B B Psi 3 = 0 Maxima codes: The matrix B is diagonalizable because it has three distinct eigenvalues, so the minimal polynomial is ψ(λ) = (λ 8)(λ 4)(λ 1) Sylvester s auxiliary matrices become Z (8) = (B 4I)(B I) = 1 0 1, (8 4)(8 1) Z (4) = (B 8I)(B I) = 0 2 2, (4 8)(4 1) Z (1) = (B 4I)(B 8I) (1 4)(1 8) =, where I is the identity matrix 7

8 Now we construct the exponential and cosine (3 3) matrices: Φ(t) = e Bt = e t Z (1) +e 4t Z (4) +e 8t Z (8) = e t +e 4t e 8t ; Ψ(t) = cos(bt) = costz (1) +cos(4t)z (4) +cos(8t)z (8) = cost +cos4t cos8t Since B 2 = , and we have d 2 Ψ(t) = cost dt cos4t cos7t 1 0 1, B 2 Ψ(t) = ( ) costz (1) +cos(4t)z (4) +cos(8t)z (8) Calculations show that B 2 Z (1) = = = Z (1), B 2 Z (4) = = = 16Z (4), B 2 Z (8) = = = 64Z (8) Therefore, the function Ψ(t) = cos(b t) satisfies the second order matrix differential equation Obviously, we have the initial conditions: Ψ(0) = I and Ψ(0) = 0 8

9 To show that the matrix-function Φ(t) = e Bt is a solution of the differential equation ẏ = By, we first differentiate it d dt Φ(t) = d dt ebt = e t +e 4t +e 8t, and then calculate the products: BZ (1) = =, BZ (4) = = , BZ (8) = = (7 3 = 21 points) Answer the same questions as in the previous exercise for the matrix C = Solution Grading: +5 for constructing Φ +5 for constructing Ψ +5 for showing that Φ satisfies its differential equation +5 for showing that Ψ satisfies its differential equation Since the resolvent of the matrix C is λ (λi C) 1 1 = 3 λ+52 9, (λ 2)(λ 1) λ 53 the given matrix is diagonalizable So we can calculate Sylvester s auxiliary matrices: Z 1 = C 2I = , Z 2 = C I 2 1 =

10 Now we construct the required matrices: Φ(t) = e Ct = e t Z 1 +e 2t Z 2 = e t Ψ(t) = cos(ct) = costz 1 +cos2tz = cost cos2t e 2t Obviously, Ψ(0) = I and Ψ(0) = 0 because Z 1 +Z 2 = I Now we calculate Φ(0) = Z 1 +Z 2 = I, Φ(0) = = = C ; 10

1. Diagonalize the matrix A if possible, that is, find an invertible matrix P and a diagonal

1. Diagonalize the matrix A if possible, that is, find an invertible matrix P and a diagonal . Diagonalize the matrix A if possible, that is, find an invertible matrix P and a diagonal 3 9 matrix D such that A = P DP, for A =. 3 4 3 (a) P = 4, D =. 3 (b) P = 4, D =. (c) P = 4 8 4, D =. 3 (d) P