Let (Tin mathbb {C}^{nltimes nl}) and T be a block diagonal matrix where each block is a Jordan matrix corresponding to an eigenvalue of (L lambda )), also let (Xin mathbb {C}^{ntimes nl}) and column vectors of X are precisely the Jordan chains of (L lambda )) corresponding to the eigenvalues of (L lambda )).
Each matrix polynomial (H xi )) possesses a spectral pair or Jordan pair (X, T), where X is a matrix containing the generalized eigenvectors of (H xi )) and T is a block diagonal matrix where each block is a Jordan matrix corresponding to the eigenvalues of (H xi )).
Let λ̅ be a complex eigenvalue and (overline{y} t,lambda)= ( overline{y}_{1}(t),overline{y}_{2}(t) ) ^{mathbf{T}}) be an eigenfunction corresponding to the eigenvalue λ̅ of the problem (1.1 - 1.3 1.1 - 1.3
Since obviously and have the same eigenvectors corresponding to the eigenvalue 1, the index of 1, as an eigenvalue of, is also 1.
Let be an eigenvector of corresponding to the eigenvalue,, and let be an eigenvector of corresponding to the eigenvalue,, such that (4.8).
where the columns of ([boldsymbol {U}_{1}^{T},~boldsymbol {U}_{3}^{T}]^{T}) are the eigenvectors of (boldsymbol {R}_{widetilde u}) corresponding to the eigenvalue zero and the columns of ([boldsymbol {U}_{2}^{T},~boldsymbol {U}_{4}^{T}]^{T}) are the other eigenvectors.
(ii)The vector is the sum of the vectors and where is an eigenvector of corresponding to the eigenvalue and is either the zero vector or an eigenvector of corresponding to the eigenvalue.
Thus, is a temporal period of This is contrary to being the least among all periods and In conclusion, has eigenvalue and where is an eigenvector of corresponding to the eigenvalue and is either a zero vector or an eigenvector of corresponding to the eigenvalue Since are all distinct eigenvalues of there exists some such that.
where is either the zero vector or an eigenvector of corresponding to the eigenvalue and is either a zero vector or an eigenvector of corresponding to the eigenvalue 1. Suppose that is the zero vector, or, is not an eigenvalue of Then must be an eigenvalue of and must be an eigenvector corresponding to the eigenvalue 1; otherwise, and this is impossible.
For any, let be an eigenvector of corresponding to the eigenvalue,, such that (4.21).
