Calculating the eigen vectors without using eig function but using a data set, the transpose of the data set and a unit vector

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Zachary David
Zachary David 2022-4-5
评论： Bruno Luong 2022-4-5
Hello,
I am trying to calculate the eigen vectors using a looped equation without using the eig function but using a dataset, the transpose, and a random unit vector with the magnitude of 1.
here is what I have so far:
clc
clear
A = importdata('mariana_depth.csv');
B=transpose (A);
C=B*A;
u=rand(1440,1); u=u/norm(u);
for i=1:10
u_n=u;
u=C*u
v(:,i)=u;
u_n(:,i) =u.*v(:,i)/norm(u.*v(:,i));
diffu = norm(u_n-u);
i
diffu
end
the problem I am having is that it is counting up but not down so I dont believe it it calculating the proper values of my eigen values.
  1 个评论
Zachary David
Zachary David 2022-4-5
编辑：Zachary David 2022-4-5
this is the equation i am trying to use
u= the unit vector with a magnitude of 1
u_n+1= transposeA * A* u / || transposeA * A * u||
repeating ||u_n+1 - u|| until it is very small

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回答（2 个）

Sam Chak
Sam Chak 2022-4-5
Hi @Zachary David
Can consider an alternative approach by finding the characteristic polynomial of a matrix using poly()
https://www.mathworks.com/help/matlab/ref/poly.html
and then use some root-finding algorithms or roots() to find them.
https://www.mathworks.com/help/matlab/ref/roots.html
For example:
A = [0 1; -2 -3]
p = poly(A)
r = roots(p)
A =
0 1
-2 -3
p =
1 3 2
r =
-2
-1
  7 个评论
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Sam Chak
Sam Chak 2022-4-5
Think your method is related to the Power method. But the script below can only calculate the real largest eigenvalue.
function Demo_Eigen
close all
clc
A = [1 3 4; 5 6 7; 2 9 8]
% A = [0 1; -1 -1]
% A = inv(A) % Inverse Iteration Method to find smallest eigenvalue
n = size(A, 1);
x1 = ones(n, 1);
epsilon = 1e-6;
kmax = 100;
x(:, 1) = x1./norm(x1, 2); % initial eigenvector
x(:, 2) = A*x(:, 1);
lambda(2) = x(:, 1).'*x(:, 2); % initial eigenvalue
x(:, 2) = x(:, 2)./norm(x(:, 2), 2);
for k = 2:kmax
x(:, k+1) = A*x(:, k);
lambda(k + 1) = x(:, k).'*x(:, k + 1);
x(:, k + 1) = x(:, k + 1)./norm(x(:, k + 1), 2);
if abs(lambda(k + 1) - lambda(k)) < epsilon
break
end
end
Eigen_val = lambda(end)
Eigen_vec = x(:,end)
end

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Bruno Luong
Bruno Luong 2022-4-5
Ran in:
Your method give you the largest singular vector not eigen vector:
A=rand(5);
C=A'*A;
u=randn(size(A,2),1);
u=u/norm(u);
for i=1:100
u_n=C*u;
u_n=u_n/norm(u_n);
diffu = norm(u_n-u);
if diffu < 1e-10
fprintf('converge afeter %d iterations\n', i)
break
end
u=u_n;
end
converge afeter 9 iterations
u
u = 5×1
-0.5840 -0.5132 -0.2270 -0.2861 -0.5121
[U,S,V]=svd(A);
V(:,1)
ans = 5×1
-0.5840 -0.5132 -0.2270 -0.2861 -0.5121
  1 个评论
Bruno Luong
Bruno Luong 2022-4-5
Ran in:
If you want a smallest singular vector you need to use backslash instead of time
A = rand(5)
A = 5×5
0.7672 0.0288 0.8776 0.3047 0.9679 0.8588 0.8712 0.5263 0.7778 0.4880 0.5808 0.5344 0.5108 0.4657 0.6664 0.8760 0.8231 0.0007 0.5201 0.3654 0.2871 0.6496 0.1216 0.0663 0.6528
C=A'*A;
u=randn(size(A,2),1);
u=u/norm(u);
for i=1:100
u_n=C\u;
u_n=u_n/norm(u_n);
diffu = norm(u_n-u);
if diffu < 1e-10
fprintf('converge afeter %d iterations\n', i)
break
end
u=u_n;
end
converge afeter 7 iterations
u
u = 5×1
0.2794 0.3037 0.4550 -0.6649 -0.4249
[U,S,V]=svd(A);
V(:,end)
ans = 5×1
0.2794 0.3037 0.4550 -0.6649 -0.4249

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