Eigenvectors not changing with constant parameter
Mostrar comentarios más antiguos
Dear All,
I am trying to calculate the eigenvectors(V) using eig() function. My 2x2 matrix(M) contains a constant parameter 'a' in it. But I see that changing a does not change my eigenvectors. Generally the eigevectors and eigenvalues change with the matrix elements. Here my eigenvalues are varying but not the eigenvectors. Could someone figure out the issue?
sx = [0 1; 1 0];
sy = [0 -1i; 1i 0];
a = 0.18851786;
kx = -0.5:0.1:0.5;
ky = kx;
for i = 1:length(kx)
for j = 1:length(ky)
M = a.*(sx.*ky(i)-sy.*kx(j));
[V,D] = eig(M);
V
end
end
Respuesta aceptada
Bruno Luong
el 23 de Abr. de 2022
Editada: Bruno Luong
el 23 de Abr. de 2022
"Could someone figure out the issue?"
But there is no issue beside thet fact that you expect something that not going to happen.
if V and diagonal D the eigen decomposition of A1
A1*V = V*D
then for any constant a
(a*A1)*V = V*(a*D)
Meaning V and a*D (still diagonal) are eigen decomposition of Aa := a*A1.
So A1 and Aa respective eigen decomposition can have the same V (eigen vectors, MATALB always normalized them to have norm(V(:,k),2)=1 for all k) but eigen values are proportional to a.
5 comentarios
SHUBHAM PATEL
el 23 de Abr. de 2022
Bruno Luong
el 23 de Abr. de 2022
Editada: Bruno Luong
el 23 de Abr. de 2022
Look like they change, so what is your point?
sx = [0 1; 1 0];
sy = [0 -1i; 1i 0];
I = eye(2);
a = 0.18851786;
kx = -0.5:0.1:0.5;
ky = kx;
for i = 1:length(kx)
for j = 1:length(ky)
M = (kx(i).^2+ky(j).^2)*I + a.*(sx.*ky(i)-sy.*kx(j));
[V,D] = eig(M);
V
end
end
SHUBHAM PATEL
el 23 de Abr. de 2022
Bruno Luong
el 23 de Abr. de 2022
Editada: Bruno Luong
el 24 de Abr. de 2022
This is just a coincidence in YOUR parametrization.
Your matrix M has the same eigen vectors as this matrix
A := M / (kx(i).^2+ky(j).^2) % see your original question
A = I + K(a);
with
K = [0 conj(z);
z 0]
z(a) := a*(kx(i) + 1i*ky(j)) / (kx(i).^2+ky(j)).^2 % but it doesn't matter, any complex still works
So
A = [1 conj(z);
z 1];
This (Hermitian) matrix A has two eigen values and (unormalized) eigen vectors
d1 = 1 - abs(z); V1 = [conj(z); -abs(z)];
d2 = 1 + abs(z); V2 = [conj(z); +abs(z)];
So when you scale z up and down by a (a non zero real value) the vector V1 and V2 remains the same after normalization (by MATLAB.
V1 = [conj(z); -abs(z)] / sqrt(2*abs(z)^2) = [conj(z)/abs(z), -1] / sqrt(2)
V2 = [conj(z); abs(z)] / sqrt(2*abs(z)^2) = [conj(z)/abs(z), +1] / sqrt(2)
Quod erat demonstrandum
SHUBHAM PATEL
el 24 de Abr. de 2022
Más respuestas (1)
Walter Roberson
el 23 de Abr. de 2022
0 votos
eigenvectors are geometrically directions. When you scale a matrix by a nonzero constant, the direction does not change.
1 comentario
SHUBHAM PATEL
el 23 de Abr. de 2022
Categorías
Más información sobre Linear Algebra en Centro de ayuda y File Exchange.
el 23 de Abr. de 2022
el 24 de Abr. de 2022
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
