Eigenvectors not changing with constant parameter

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SHUBHAM PATEL
SHUBHAM PATEL 2022-4-23
编辑： Bruno Luong 2022-4-24
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

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采纳的回答

Bruno Luong
Bruno Luong 2022-4-23
编辑：Bruno Luong 2022-4-23
"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.
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Bruno Luong
Bruno Luong 2022-4-23
编辑：Bruno Luong 2022-4-24
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
SHUBHAM PATEL 2022-4-24
Thank you @Bruno Luong for such a detailed explaination. I understood the gist.

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更多回答（1 个）

Walter Roberson
Walter Roberson 2022-4-23
eigenvectors are geometrically directions. When you scale a matrix by a nonzero constant, the direction does not change.
  1 个评论
SHUBHAM PATEL
SHUBHAM PATEL 2022-4-23
Thanks W. Roberson,
I understood your point that if we multiply the matrix by a constant it does not change the directions.
But I am seeing that if I include the diagonal term also and multiply the off-diagonal term only with a constant, even then the vectors are not changing, although the eigen values are changing.
Could you comment on that?
I am writing the code again including the diagonal term.
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 = (kx(i).^2+*ky(j).^2)*I + a.*(sx.*ky(i)-sy.*kx(j));
[V,D] = eig(M);
V
end
end

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