SyDe312 (Winter 2005) Unit 1 - Solutions (continued)

Transcription

1 SyDe3 (Winter 5) Unit - Solutions (continued) March, 5 Chapter 6 - Linear Systems Problem b Iterative solution by the Jacobi and Gauss-Seidel iteration methods: Given: b = [ 77] T, x = [ ] T 9x + x + x 3 = x + x + 3x 3 = 7 3x + 4x + x 3 = 7 x = 9 ( x x 3 ) x = ( 7 x 3x 3 ) x 3 = (7 3x 4x ) Jacobi method: x (k+) = 9 x (k+) = x (k+) 3 = ( x(k) x (k) 3 ) ( 7 x(k) 3x (k) 3 ) (7 3x(k) 4x (k) ) Gauss-Seidel method: x (k+) = 9 ( x(k) x (k) x (k+) = x (k+) 3 = 3 ) ( 7 x(k+) 3x (k) 3 ) (7 3x(k+) 4x (k+) )

2 For example beginiing with the initial x = [,, ], in the Jacobi method, we get: x () = 9 ( ) = x () = x () 3 = ( 7 3 ) = 7 (7 3 4 ) = 7 and with the Gauss-Seidel method: ( ) = x () = 9 x () = x () 3 = ( 7 3 ) = 7 ( ) = 98 Iterating these procedure until x x k.5, we get: k x k x k x k 3 Error Ratio.E E E E E E E E E E E E E E E E E E E E-5.35 Table : Problem 6.6.b Jacobi Iteration k x k x k x k 3 Error Ratio.E E- 3.E- -.E E-.3E E E-3 8.8E E E E E E-5 5.7E- Table : Problem 6.6.b Gauss-Seidel Iteration

3 Problem In case a, for the matrix converge. [ 4 4 ], in which the M >, the iteration methods do not k x k x k Table 3: Problem 6.6.3a Jacobi Iteration ], which is diagonally dominant, both methods converge. k x k x k E E E E 7.33E+ -.93E Table 4: Problem 6.6.3a Gauss-Seidel Iteration So the Gauss-Seidel method diverges more rapidly. In case b, for the matrix [ 4 4 So Gauss-Seidel method converges faster. 3

4 Problem A = k x k x k Error E E E E-6 Table 5: Problem 6.6.3b Jacobi Iteration k x k x k Error E E E E E-6 Table 6: Problem 6.6.3b Gauss-Seidel Iteration b = x = [The following solutions use the notation of the textbook for the iteration matrices etc. You can easily translate to relate these to the matrices defined in the lecture notes.] Recall that in Jacobi: N = a... a ann P = N A and In Gauss-Seidel: 4

5 N = a... a a an an... ann P = N A and M = N P k x k x k x k 3 x k 4 Error Ratio E E E E E E E E E E E E E E E-5.5 Table 7: Problem Jacobi Iteration M = 4 M =.5. This is consistent with the column ratio in the above table. Problem 6.6-4b M = 4 /4 /4 /6 /6 /4 /6 /6 /4 /3 /3 /8 5 M =.5.

6 k x k x k x k 3 x k 4 Error Ratio E- -.44E E E- -.E E E- -.3E E E-3 -.E E E-3 -.E E E-4 -.E E E-4 -.E E E-5 -.E E E-6 -.E+..34E-5.5 Table 8: Problem Gauss-Seidel Iteration The actual convergence rate is better than.5. upper bound on the convergence or divergence rate. Keep in mind that M gives an Problem x x x 3 b b b a 4 3 = 3 since det(a), for any right-hand side vector b, the system has a unique solution x = [x, x, x 3 ] T b ()Jacobi method: N = 4 3 N = /4 / /3 P = 6

7 M = N P = /4 /4 / /3 M = /3 Since M <, the method converges. () Gauss-Seidel method: N = 4 3 N = /4 /8 / / /3 /3 P = M = N P = /4 /4 /8 /8 / / M = / Since M <, the method converges. Problem Consider A n = The iteration method Nx (k+) = b+p x (k) will converge, for all b vectors and all initial guesses x (), if and only if all eigenvalues λ of M = N P satisfy λ < 7

8 () Jacobi method n=5, A=[,-,,,;-,,-,,;,-,,-,;,,-,,-;,,,-,]; for i=:n for j=:n N_Jacobi(i,j)=(i==j)*A(i,j); end end P=N_Jacobi-A; M=inv(N_Jacobi)*P; lambda=eig(m) A 5 = N = P = M = N P = N = The eigenvalues of M are - using eig(m): 8

9 λ =.866 λ =.5 λ 3 = λ 4 =.5 λ 5 =.866 Likewise for n =, the eigenvalues of M are: λ =.9595 λ =.843 λ 3 =.6549 λ 4 =.454 λ 5 =.43 λ 6 =.43 λ 7 =.454 λ 8 =.6549 λ 9 =.843 λ =.9595 For n =, the eigenvalues of M are: λ =.9888 λ =.9556 λ 3 =.9 λ 4 =.86 λ 5 =.733 λ 6 =.635 λ 7 =.5 λ 8 =.3653 λ 9 =.5 λ =.747 λ =.747 λ =.5 λ 3 =.3653 λ 4 =.5 λ 5 =.635 λ 6 =.733 λ 7 =.86 λ 8 =.9 λ 9 =.9556 λ =.9888 We see that for n = 5,,, λ i <, so Jacobi method will converge. () Gauss-Seidel method Again for n = 5, we have: for i=:n for j=:n N_GS(i,j)=(i>=j)*A(i,j); end end P=N_GS-A; M=inv(Ngs)*P; lambda=eig(m) M = N P = The eigenvalues of M are - using eig(m): λ = λ =.75 λ 3 =.5 λ 4 = λ 5 = 9

10 Likewise for n =, the eigenvalues of M are: λ = λ =.96 λ 3 =.777 λ 4 =.488 λ 5 =.76 λ 6 =.3 λ 7 = λ 8 = λ 9 = λ = For n =, the eigenvalues of M are: λ = λ =.9778 λ 3 =.93 λ 4 =.87 λ 5 =.687 λ 6 =.5374 λ 7 =.3887 λ 8 =.5 λ 9 =.335 λ =.495 λ =.68 λ = i λ 3 =.5.46i λ 4 =.8 +.7i λ 5 =.8.7i λ 6 = i λ 7 =.43.6i λ 8 = i λ 9 = i λ = So for n = 5,,, λ i <, so Gauss-Seidel method will converge. Problem Note: this is slightly different from the method described in the lecture notes, but equivalent. Careful with the sign of the correction vector. The residual correction method: A ɛ x = b where A ɛ = A + ɛb. A = B = Let N = A and P = N A ɛ = ɛb, so M = N P = A ( ɛb) M = A ( ɛb) = ɛ. A B) = ɛ /5.5 = ɛ. = ɛ In order to ensure the convergence of the residual correction method, α <.

11 k x k x k x k 3 Error Ratio.e+.e+ 9.e-.e- 9.99e-.e+.e+.e+.e-.e e-.e+.5e+ 5.e-4 5.e- 4.e e-.e+ 5.e-5.e- 5.e+.e+.e+.5e-6 5.e- Table 9: Problem ɛ =. In this case, we are not given the vector b, but the true answer x = [,, ]. Let x = [,, ] and repeat the procedure until x x (k) 5 k x k x k x k 3 Error Ratio.e+.e+ 8.e-.e-.6667e-.e+.4e+.e+ 4.e-.e e-.e+.4e+ 4.e-3.e- 4.e+ 9.99e-.e+ 8.e-4.e- 5.e+.e e- 8.e-5.e- 6.e+.e+.e+.6e-5.e- 7.e+.e+.e+.6e-6.e- Table : Problem ɛ =. The number of iterations increases as ɛ increases, consistent with expectations. Problem 7. - b [ ] 36 For, 36 3 f(λ) = λ 5λ 5 = (λ 5)(λ + 5), λ = 5, λ = 5 Using (A λi)v =, we get : An eigenvector corresponding to λ = 5: v = [, 5] T

12 k x k x k x k 3 Error Ratio.5e+.e+ 5.e- 5.e e-.e+.5e+.e+.5e- 5.e e-.e+.65e+ 6.5e-.5e- 4.e e-.e+ 3.5e- 5.e- 5.78e+.e+ 9.99e- 7.85e-3.5e- 6.e+.39e+.e e-3 5.e e-.e+.e e-4.5e- 8.e e-.e e-4 5.e- 9.e+.e e-.7e-4.5e-.e+.e+.e+ 6.35e-5 5.e e-.e+.e+.559e-5.5e-.e e-.e e-6 5.e- Table : Problem ɛ =.5 k x k x k x k 3 Error Ratio.8E+.E+.E- 8.E- 4.44E-.E+.64E+.E+ 6.4E- 8.E E-.E+.6E+.56E- 4.E- 4.E+ 7.95E-.E+.5E- 8.E- 5.8E+.E+ 9.8E- 8.9E- 4.E- 6.E+.7E+.E+ 6.55E- 8.E E-.E+.3E+.6E- 4.E- 8.E+ 9.79E-.E+.E- 8.E- 9.E+.E+ 9.9E- 8.39E-3 4.E-.E+.E+.E+ 6.7E-3 8.E- 9.97E-.E+.E+.68E-3 4.E-.E+ 9.98E-.E+.5E-3 8.E- 3.E+.E+ 9.99E- 8.59E-4 4.E- 4.E+.E+.E+ 6.87E-4 8.E- 5.E+.E+.E+.75E-4 4.E- 6.E+.E+.E+.E-4 8.E- 7.E+.E+.E+ 8.8E-5 4.E- 8.E+.E+.E+ 7.4E-5 8.E- 9.E+.E+.E+.8E-5 4.E-.E+.E+.E+.5E-5 8.E-.E+.E+.E+ 9.E-6 4.E- Table : Problem ɛ =.8

13 An eigenvector corresponding to λ = 5: v = [, ] T Problem 7. - For the symmetric matrix A = , eigenvalues and eigenvectors are: λ = { 36, 3, 9} v () = [ 3, 3, 3 ] T, v () = [,, ] T, v (3) = [ 6, 6, ] T,, U can be: U = [v () ; v () ; v (3) ] = Each column in U is one of the eigenvectors of A. The eigenvectors are normalized to get the columns of U which is supposed to be an orthogonal matrix. We get: U T AU = D = diag(λ, λ, λ 3 ) and U T U = UU T = I. A=[-7,3,-6; 3, -, 3; -6, 3, -7]; [U,lam]=eig(A) The result might be: λ = , U = Run the following command, in order to see: = [v () ; v (3) ; v () ] U *A*U

14 and also the commands U *U and U*U, in order to see U T U = UU T = I. Problem [ ] For, eigenvalues and eigenvectors are: f(λ) = λ λ + = (λ ) : λ =, v () = [, ] T λ =, v () = [, ] T For [ ], e-values and e-vectors are: f(λ) = λ λ + = (λ ) : λ =, v () = [, ] T In case b, this eigenvector and its multiples are the only eigenvectors of the matrix. It does NOT contradict Theorem 7..4, because the matrix is not symmetric. 4