Question

4. Let A. Į2 2 01 carryonít 10 iterations of the power method with 4. Let A =-1 2-1 | . Carry out 10 iterations of the power method with normalization, starting with x( ) (1,1,1)T and using the linear functional p(x) = a1. What does the ratio Tk-(x(+)/x) approximate? Pxapproximate

Answer 1

%Matlab code for power method clear all close all %Matrix for which eigen value have to find A= [ 2-1 0;-1 2-1:0-1 21: fprintf A matrix is\n) disp(A) fprintf(x vector is\n') disp(x) %power method 10 iterations for i= 1:10 fprint f ("After %d iterations\n disp(x) fprint f ("\n \tphi(x)=\n") disp(x./x(1)) \tx-\n",i) x=x./x ( 1 ) ; fprintf( "\n \tEigen value is %f"\n\n.,xx) end 용욤욤응응욤욤응응욤욤各各 End of Code 욤各各음음各各음음各各음음各 A matrix is 2 x vector is After 1 iterations phi(x) Eigen value is 1.000000. After 2 iteration

phi(x)- Eigen value is 2.000000. After 3 iterations phi(x)- -1. 33333333333333 Eigen value is 3.000000. After 4 iterations 3.33333333333333 -4.66666666666667 3. 33333333333333 phi(x)- Eigen value is 3.333333. After 5 iterations 3.4 3.4 phi (x)= -1.41176470588235

Eigen value is 3.400000 After 6 iterations 3.41176470588235 -4. 82352941176471I 3.41176470588235 phi(x) 1.41379310344828 Eigen value is 3.411765. After 7 iteration 3.41379310344828 -4.82758620689655 3.41379310344828 phi(x)- -1.41414141414141 Eigen value is 3.413793. After 8 iterations 3. 41414141414141 -4. 82828282828283 3. 41414141414141 phi(x)- -1.41420118343195 Eigen value is 3.414141 After 9 iterations 3.41420118343195 -4. 82840236686391I 3.41420118343195

phi(x) -1.41421143847487 Eigen value is 3.414201. After 10 iterations 3.41421143847487 -4.82842287694974 3.41421143847487 phi(x). -1.41421319796954 Eigen value is 3.41421 Published with MATLAB® R2018a

%Matlab code for power method
clear all
close all
%Matrix for which eigen value have to find
A=[2 -1 0;-1 2 -1;0 -1 2];
fprintf('A matrix is\n')
disp(A)
x=[1;1;1];
fprintf('x vector is\n')
disp(x)

%power method 10 iterations
for i=1:10
    x=A*x;
    fprintf('After %d iterations\n \tx=\n',i)
    disp(x)
    fprintf('\n \tphi(x)=\n')
    disp(x./x(1));
    xx=x(1);
    x=x./x(1);
    fprintf('\n \tEigen value is %f.\n\n',xx)
end

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