An issue with eigenvectors...
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When i execute this it only shows the last vector, not the first or second one.
clc
clear variables;
prompt = ("Ingrese su matriz de coeficientes ");
w=input(prompt);
fprintf("La matriz ingresada es\n");
disp(w)
sz=size(w);
tm=sz(1,1);
p=round(poly(w));
e = eig(w);
fprintf('**Los valores propios del sistema son: \n');
disp(e);
v1=e(1,1);
v2=e(2,1);
v3=e(3,1);
tf=isreal(e);
if(tf==1)
fprintf('**Los vectores propios del sistema son: \n');
L1=null(w-(v1*eye(tm)), 'r');
disp(L1);
L2=null(w-(v2*eye(tm)), 'r');
disp(L2);
L3=null(w-(v3*eye(tm)), 'r');
disp(L3);
回答（1 个）
Remove the rational option "r" (not reliable) you'll be fine
w = magic(3)
disp(w)
sz=size(w);
tm=sz(1,1);
p=round(poly(w));
e = eig(w);
fprintf('**Los valores propios del sistema son: \n');
disp(e);
v1=e(1,1);
v2=e(2,1);
v3=e(3,1);
tf=isreal(e);
if(tf==1)
fprintf('**Los vectores propios del sistema son: \n');
L1=null(w-(v1*eye(tm)));
disp(L1);
L2=null(w-(v2*eye(tm)));
disp(L2);
L3=null(w-(v3*eye(tm)));
disp(L3);
end
3 个评论
Santiago Piñon
2022-11-27
编辑：Santiago Piñon
2022-11-27
Bruno Luong
2022-11-27
编辑：Bruno Luong
2022-11-27
No "r" is NOT correct in general, as stated by the offiial doc page
Z = null(A,"rational") returns a rational basis for the null space of A that is typically not orthonormal. If A is a small matrix with small integer elements, then the elements of Z are ratios of small integers. This method is numerically less accurate than null(A).
It is just a toy option forr students who works with academic examples. That's why it fails to return the eigen vector randomly.
Santiago Piñon
2022-11-27
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2022-11-27
2022-11-27
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