# How to find eigenvalues of problem that dont have the form of standard eigenvalue problem?

12 views (last 30 days)
Mehdi on 5 Apr 2017
Dear Math Professionals in Matlab.
My problem is exactly in the form of first above equation. Doing above method on my problem in Matlab I am trying to find eigenvalues of my equetion. But, I got several omegas for several LVs(the omegas continuesly increase when LV increases). While I expect my results converge to specific omegas when LV increases to high values.
How I can find the eigenvalues of my problem(first above equation) in Matlab?
How the value of LV must be choosen in above method? I am amazed.
clear
load M21;M21=mat; save M21; clear mat;
load M22;M22=mat; save M22; clear mat;
load K21;K21=mat; save K21; clear mat;
load K22;K22=mat; save K22; clear mat;
[M,N]=size(K22);
M22 =zeros(N);
Mt=inv(sqrtm(M11));n=1;
for LV=1:10:1000
A11=Mt*K11*Mt;
A12 = LV*Mt *K12;
A21 = LV*K21 * Mt;
A22 = K22;
AA =[A11, A12;A21, A22];
B=eig(AA);
C=sqrt(B);
D=C*9.598721848;
R=sort(real(D));
F=find(R>0);
result(1,n)=R(F(1));
n=n+1;
end
plot(n,result)

David Goodmanson on 7 Apr 2017
Edited: David Goodmanson on 7 Apr 2017
Hi Mehdi, I can't address whether there would be a successful solution, and there may well be better algorithms available now than this one, but I can comment on the method. The idea is to replace
[M11 0]
[ 0 0]
with
[M11 0]
[ 0 eps*I]
for some small value eps, small enough to perturb the problem only a small amount. Let N = M^(-1/2), multiply your equation on the left by
[N 0]
[ 0 c*I]
and insert the identity matrix in the form
[N 0 ] [N^-1 0]
[0 c*I] [ 0 (1/c)*I]
between the matrices and the eigenvector
[X ]
[lambda].
Multiply it all out blockwise and you get what they got, with
I22 = c^2*eps*I.
You want I22 to be the identity matrix, so
c = 1/sqrt(eps).
c is what they called LV. So how small do you have to make eps to not affect the problem too much? Having no direct experience with this I don't know, but I would hazard a guess of least 10 times smaller than the smallest eigenvalue of M11. In your case the smallest eigenvalue is 3e-11 so you can plan accordingly. It does lead to some pretty big LV numbers. (The largest eigenvalue is 4e-3, so scaling could be a problem). Good luck!
##### 2 CommentsShowHide 1 older comment
David Goodmanson on 7 Apr 2017
Hi Mehdi, I don't have anything new, except if you have not done so you could always let Matlab have a try at it directly with eig(K,M).

Mehdi on 6 Apr 2017
No comment yet???????