Display only one eigenvalue of symbolic matrix

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Rebecca Müller
Rebecca Müller 2020 年 2 月 21 日
コメント済み: Steven Lord 2020 年 2 月 24 日
Hey :)
How to ask matlab to display only one (the first, biggest in magnitude) eigenvalue of a symbolic matrix (well the matrix only contains one strictly positive variable)?
The thing is that I need to insert the scalar(!) eigenvalue further into a function.
I tried eigs(A,1) but I get the error: "Error using eigs>checkInputs (line 214) First argument must be a double matrix or a function."
So i assumed it is because of the one variable in the matix...
Any help would be appreciated :)
Edit: Maybe here my code:
clear
a = 1.0; b = 0.0; m = 25.0; V = 13.0; Ecut = 50.0; w = 1.5*sqrt(m/(a+9/4*b)); Nx = 100; Ny = 100; Nz = 100; k = 1;
T = 1;
syms kx ky kz real
syms D positive
E = eye(4);
U = [0 0 0 1; 0 0 -1 0; 0 1 0 0; -1 0 0 0];
Jx = 1/2*[0 sqrt(3) 0 0; sqrt(3) 0 2 0; 0 2 0 sqrt(3); 0 0 sqrt(3) 0];
Jy = 1i/2*[0 -sqrt(3) 0 0; sqrt(3) 0 -2 0; 0 2 0 -sqrt(3); 0 0 sqrt(3) 0];
Jz = 1/2*[3 0 0 0; 0 1 0 0; 0 0 -1 0; 0 0 0 -3];
P = (D*(Jy*Jz+Jz*Jy)+1i*D*(Jx*Jz+Jz*Jx))*U/sqrt(3);
h = (a*(kx*kx+ky*ky+kz*kz)-m)*E+b*(kx*Jx+ky*Jy+kz*Jz)^2;
H = [h P; ctranspose(P) -transpose(h)];
dsum = 0;
counter = 0;
for i = 1:Nx
for j = 1:Ny
for l = 1:Nz
kx = w*i/Nx;
ky = w*j/Ny;
kz = w*l/Nz;
if abs(a*(kx*kx+ky*ky+kz*kz)^2-m)<Ecut
dsum = dsum + 8.0*D*D/V-8.0*k*T*ln(2.0*cosh(max(eig(H))/(2.0*k*T)));
counter = counter + 1;
end
end
end
end
F = @(D)dsum;
D = [0,10];
x = fminsearch(F,D)
  3 件のコメント
Rebecca Müller
Rebecca Müller 2020 年 2 月 24 日
Well but that is not the case as my matrix only depends on one symbolic variable which I defined to be positive(?)
Steven Lord
Steven Lord 2020 年 2 月 24 日
Okay, then what about:
syms x positive
A = [x 0; 0 x^2]
eig(A)
The eigenvalues of A are x and x^2.
If x is less than 1, x is greater than x^2.
If x is greater than 1, x^2 is greater than x.
Which eigenvalue would you want the function that returns "the largest" eigenvalue to return?

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回答 (1 件)

Christine Tobler
Christine Tobler 2020 年 2 月 21 日
The eigs function is not supported for symbolic values, as it is specifically based on getting a good approximation based on an iterative algorithm. For symbolic variables, only eig is provided, which computes all eigenvalues directly.
You can use max(eig(A)) to compute the largest eigenvalue for a symbolic matrix A. Note these computations can be very expensive for symbolic variables.
  1 件のコメント
Rebecca Müller
Rebecca Müller 2020 年 2 月 23 日
Thank you! Using max(eig(A)) unfortunately gives me the error : Error using sym/max (line 101)
Input arguments must be convertible to floating-point numbers. ..(?)

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