Eigenvector without calling eigenvalues

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AVM
AVM am 1 Feb. 2020
Kommentiert: Walter Roberson am 1 Feb. 2020
I would like to call a eigenvector of a matirx without calling its eigenvalues inside a function. Here I attach my code. Pl somebody help me.
function [out]=integration(hami1)
[V,L]=eig(hami1); %% Some error is here showing that L is unused in my code.
u=V(:,1)/sqrt(sum(V(:,1)));
w=diff(u,phi);
f=dot(u,w);
out=1/pi*1i*int(f,phi,0,2*pi);
end

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Walter Roberson
Walter Roberson am 1 Feb. 2020
Given your question as asked, you will need to write your own code to somehow determine eigenvectors without calculating the corresponding eigenvalues.
However what you are seeing is a warning not an error, and most people would deal with it by coding
[V,~]=eig(hami1);
which tells MATLAB to tell eig that two outputs are requested (so that it knows to return eigenvectors in the first output), but that the second output will be ignored by the code.
  3 Kommentare
AVM
AVM am 1 Feb. 2020
The Integration by using 'fucntion ' is faster than sciprt? Pl help me
Walter Roberson
Walter Roberson am 1 Feb. 2020
integral() cannot call scripts so the relative speeds of scripts and functions is not relevant to the situation.

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Weitere Antworten (1)

Vladimir Sovkov
Vladimir Sovkov am 1 Feb. 2020
  1. This is not an error but a warning that you do not use the eigenvalues, which influences nothing. If you want to avoid it, substitute the symbol "~" in place of "L".
  2. It looks that you use "phi" before defining it. This must be an error.
  3. Are you sure that your way of the eigenvector normalization is what you wanted? It looks quite unusual... The function "eig" is expected to produce the eigenvectors with unit algebraic norm already, at least for real symmetric matrices.
  7 Kommentare
AVM
AVM am 1 Feb. 2020
..." most probbably you would just divide by 1" . I didn't get your point. Pl help me to understand. But how I would be sure about the normalisation factor is unity?
Vladimir Sovkov
Vladimir Sovkov am 1 Feb. 2020
This is problem-dependent. Sometimes it is correct, sometimes not. You can just calculate the norm of your case and see if it equals 1 or not. Maybe, you are right and this re-normalization is really needed. Anyway, it would not spoil the results, and maybe safer to keep it in the program.

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