2.5 Bifurcation of Double Eigenvalue with Single Eigenvector

Let us consider the one-parameter matrix family

(2.62)

Its eigenvalues are

(2.63)

which are real for positive p, complex conjugate for negative p, and double zero at p=0. It is easy to see that at p=0 the double eigenvalue ?=0 is nonderogatory, i.e., it has a single eigenvector. Expression (2.63) shows that the eigenvalues are not differentiate functions of the parameter at p=0, where the double eigenvalue appears; and derivatives of the eigenvalues tend to infinity as p approaches zero. Therefore, perturbation of a nonderogatory double eigenvalue is singular and needs special analysis.

Let us consider an arbitrary family of matrices A(p). Let p ₀ be a point in the parameter space, where the matrix A ₀ =A(p ₀ ) has a double nonderogatory eigenvalue ? ₀. Let u ₀, u ₁ and v ₀, v ₁ be, respectively, the right and left Jordan chains of length 2 corresponding to ? ₀ and satisfying the equations

(2.64)

and the normalization conditions

(2.65)

Recall that these Jordan chains have the properties

(2.66)

The right Jordan chain is not unique. The vectors c ₀ u ₀ and c ₀ u ₁+ c ₁ u ₀ with arbitrary coefficients c ₀ ?0 and c ₁ form a right Jordan chain, which can be easily verified by the substitution into equations (2.64). If the vectors