- #1
member 428835
Hi PF!
I'm trying to solve the polynomial eigenvalue problem ##M \lambda^2 + \Phi \lambda + K## such that
I verify the output of the first eigenvalue via
and the output has zero real component but non-zero imaginary: ##[44i,88i]##. Any ideas what's happening? Condition number is 5, which is evidently sufficiently low to output accurate answers?
I'm trying to solve the polynomial eigenvalue problem ##M \lambda^2 + \Phi \lambda + K## such that
Code:
K = [5.92 -.9837;-0.3381 109.94];
I*[14.3 24.04;24.04 40.4];
M = [1 0;0 1];
[f lambda cond] = polyeig(M,Phi,K)
Code:
(M*lambda(1)^2 + Phi*lambda(1) + K)*f(:,1)