[Solved] Power method, eigenvalues.

Question

Kamil 2011-5-12

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/7327-solved-power-method-eigenvalues

回答： Akankshya 2024-2-13

function l = ww(A,E)
n = length(A);
y = [];
x = [];
for i = 1:n            % starting vector
    x(i) = A(i,1);
end;
l = 0;
blad = E;        % starting value of error
while blad>=E
      for i = 1:n    % A*x
          y(i) = 0;
          for j = 1:n
              y(i) = y(i) + A(i,j)*x(j);
          end;
      end;
      blad = l;
      l = 0;           % Rayleigh
      m = 0;
      for i = 1:n
          l = l + x(i)*y(i);
          m = m + x(i)*x(i);
      end;
      l = l/m;           % eigenvalue
      blad = abs(l - blad);         % error
      x = y;
  end;
end

That's how I've tried to compute eigenvalues. It works for some matrices, but for:

A =
           0   -0.3333   -0.3333
     -0.3333         0    0.3333
      0.6000    0.2000         0

it doesn't work. How can I fix that?

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Lorenzo Amabili 2017-3-17

编辑：Lorenzo Amabili 2017-3-17

@Kamil

@Teja Muppirala

how do you set E initially? I am sorry in case this question is silly but I am still new to this field. Thank you!

karim hamza 2017-4-29

i have an error [not enough input argument]

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Answer 1

Teja Muppirala 2011-5-13

3
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/7327-solved-power-method-eigenvalues#answer_10084

在 MATLAB Online 中打开

Simple power iteration only works when there is a single dominant eigenvalue. The matrix

A =[  0        -0.3333   -0.3333
     -0.3333         0    0.3333
      0.6000    0.2000         0];

has 3 eigenvalues,

-0.3333          
 0.1667 + 0.3249i
 0.1667 - 0.3249i

with absolute values:

3333
3651
3651

As you can see, the dominant eigenvalue is not unique. That is why your algorithm fails to converge.

One way to fix this is by using shifts (you can read all about it on Google).

But why not just use MATLAB's built in eigevnalue solver, EIG?

eig(A)

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Kamil 2011-5-13

Thank you, I forgot about that.

I didn't use EIG, because I'm studying Numerical Methods and I'm trying to exercise that way.

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Answer 2

Andrew Newell 2011-5-12

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/7327-solved-power-method-eigenvalues#answer_10072

在 MATLAB Online 中打开

While we wait for more information, here is a vectorized version of whatever your algorithm is doing:

function l = ww(A,E)
x = A(:,1);
l = 0;
blad = E;        % starting value of error
while blad>=E
    y = A*x;
    blad = l;
    l = x.*y;  % Rayleigh
    m = x.*x;
    l = l/m;           % eigenvalue
    blad = abs(l - blad);         % error
    x = y;
end;