iterative estimation of eigenvalues and eigenvectors by the inverse power method. Your MATLAB/Octave code shouldn’t be hardcoded. ● Calculate the eigenvectors and eigenvalues
function [lambda, v] = inverse_power_method(A, tol, max_iter)
% This function uses the inverse power method to estimate the smallest
% eigenvalue and corresponding eigenvector of matrix A.
% Inputs:
% A – The input square matrix
% tol – The tolerance for convergence (e.g., 1e-6)
% max_iter – The maximum number of iterations
% Outputs:
% lambda – The estimated smallest eigenvalue
% v – The estimated corresponding eigenvector
n = size(A, 1); % Get the size of the matrix
v = rand(n, 1); % Initialize a random vector
v = v / norm(v); % Normalize the vector
% Perform the inverse power iteration
for k = 1:max_iter
w = (A v); % Solve A * w = v
v_new = w / norm(w); % Normalize the vector
% Check for convergence
if norm(v_new – v) < tol
break;
end
v = v_new; % Update the vector
end
% Calculate the corresponding eigenvalue
lambda = v’ * A * v;
% Normalize the eigenvector to have unit length
v = v / norm(v);
end
% Define the matrix A
A = [1, -3, 3;
3, -5, 3;
6, -6, 4];
% Set the tolerance and maximum number of iterations
tol = 1e-6;
max_iter = 1000;
% Call the inverse power method function
[lambda, v] = inverse_power_method(A, tol, max_iter);
% Display the results
fprintf(‘Estimated smallest eigenvalue: %.6fn’, lambda);
fprintf(‘Estimated corresponding eigenvector:n’);
disp(v);function [lambda, v] = inverse_power_method(A, tol, max_iter)
% This function uses the inverse power method to estimate the smallest
% eigenvalue and corresponding eigenvector of matrix A.
% Inputs:
% A – The input square matrix
% tol – The tolerance for convergence (e.g., 1e-6)
% max_iter – The maximum number of iterations
% Outputs:
% lambda – The estimated smallest eigenvalue
% v – The estimated corresponding eigenvector
n = size(A, 1); % Get the size of the matrix
v = rand(n, 1); % Initialize a random vector
v = v / norm(v); % Normalize the vector
% Perform the inverse power iteration
for k = 1:max_iter
w = (A v); % Solve A * w = v
v_new = w / norm(w); % Normalize the vector
% Check for convergence
if norm(v_new – v) < tol
break;
end
v = v_new; % Update the vector
end
% Calculate the corresponding eigenvalue
lambda = v’ * A * v;
% Normalize the eigenvector to have unit length
v = v / norm(v);
end
% Define the matrix A
A = [1, -3, 3;
3, -5, 3;
6, -6, 4];
% Set the tolerance and maximum number of iterations
tol = 1e-6;
max_iter = 1000;
% Call the inverse power method function
[lambda, v] = inverse_power_method(A, tol, max_iter);
% Display the results
fprintf(‘Estimated smallest eigenvalue: %.6fn’, lambda);
fprintf(‘Estimated corresponding eigenvector:n’);
disp(v); function [lambda, v] = inverse_power_method(A, tol, max_iter)
% This function uses the inverse power method to estimate the smallest
% eigenvalue and corresponding eigenvector of matrix A.
% Inputs:
% A – The input square matrix
% tol – The tolerance for convergence (e.g., 1e-6)
% max_iter – The maximum number of iterations
% Outputs:
% lambda – The estimated smallest eigenvalue
% v – The estimated corresponding eigenvector
n = size(A, 1); % Get the size of the matrix
v = rand(n, 1); % Initialize a random vector
v = v / norm(v); % Normalize the vector
% Perform the inverse power iteration
for k = 1:max_iter
w = (A v); % Solve A * w = v
v_new = w / norm(w); % Normalize the vector
% Check for convergence
if norm(v_new – v) < tol
break;
end
v = v_new; % Update the vector
end
% Calculate the corresponding eigenvalue
lambda = v’ * A * v;
% Normalize the eigenvector to have unit length
v = v / norm(v);
end
% Define the matrix A
A = [1, -3, 3;
3, -5, 3;
6, -6, 4];
% Set the tolerance and maximum number of iterations
tol = 1e-6;
max_iter = 1000;
% Call the inverse power method function
[lambda, v] = inverse_power_method(A, tol, max_iter);
% Display the results
fprintf(‘Estimated smallest eigenvalue: %.6fn’, lambda);
fprintf(‘Estimated corresponding eigenvector:n’);
disp(v); can you please see this accurate this to get answe MATLAB Answers — New Questions