Eigenvectors not changing with constant parameter
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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
採用された回答
Bruno Luong
2022 年 4 月 23 日
編集済み: Bruno Luong
2022 年 4 月 23 日
"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 件のコメント
SHUBHAM PATEL
2022 年 4 月 23 日
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
2022 年 4 月 23 日
Bruno Luong
2022 年 4 月 23 日
編集済み: Bruno Luong
2022 年 4 月 24 日
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
2022 年 4 月 24 日
その他の回答 (1 件)
Walter Roberson
2022 年 4 月 23 日
0 投票
eigenvectors are geometrically directions. When you scale a matrix by a nonzero constant, the direction does not change.
1 件のコメント
SHUBHAM PATEL
2022 年 4 月 23 日
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