Remarks on the perturbation of analytic matrix functions
Let \(A(\lambda)\) be a hermitian matrix function, analytic in a neighbourhood of \(\lambda_ 0=0\) and let \(B(\lambda,\epsilon) = A(\lambda) + \epsilon H(\lambda)\) be its perturbation. We say that \({\tilde\lambda}\) is an eigenvalue of \(B(\cdot,\epsilon)\) if det \(B({\tilde\lambda},\epsilon)=0.\) Assume that \(\lambda_ 0\) is an eigenvalue of \(B(\cdot,0)=A(\cdot)\) of partial multiplicities \(m_ 1\leq...\leq m_ r\) and let \(m=m_ 1+...+m_ r\). The main result states that there are exactly m eigenvalues \(\lambda_ j(\epsilon)\) of \(B(\lambda,\epsilon)\) for small \(\epsilon\) and that these m eigenvalues split into r groups of \(m_ j\) \((j=1,...,r)\) eigenvalues which have Puiseux expansions provided an algebraic condition on \(H(0)\) holds. No symmetry assumption on \(H(\lambda)\) is needed. The method of the proof is rather simple, particularly in comparison with an earlier result of I. Gohberg, P. Lancaster and L. Rodman where the same result is proved for hermitian \(H(\lambda)\). However in the paper is an error; the proof is correct only if \(m_ 1<...<m_ r\). If this is not the case, the result is still generically true, i.e. there is a dense open set of matrices \(H(0)\) for which the result holds as stated. Nevertheless it is possible to find examples where \(m_ 1=m_ 2\) and \(\lambda_ j(\epsilon)\) form one group of \(m_ 1+m_ 2\) eigenvalues.
scientific article; zbMATH DE number 3977735