Corrigendum to “Numerical methods for eigenvalues of singular polynomial eigenvalue problems” [Linear Algebra Appl. 719 (2025) 1–33]

In the appendix of [1], the following lemma was formulated. Lemma A.1 Let [Formula presented] be complex polynomials such that their greatest common divisor [Formula presented] has simple roots. Then there exists a generic set [Formula presented] such that for all [Formula presented] the polynomial [Formula presented] has only simple roots. In the proof given in [1], the argument only works for the particular case [Formula presented]. Below, we present a corrected proof. Case 1) mainly follows the lines from the original proof, but is adapted to the special case [Formula presented], while the proof of Case 2) is new. Proof First we note that the polynomial [Formula presented] has simple roots if and only if the determinant of the Sylvester matrix [Formula presented] of [Formula presented] and its derivative [Formula presented] is nonzero. Since this determinant is a polynomial in the variables [Formula presented] it remains to show that [Formula presented] is not the zero polynomial and to this end it is sufficient to find one example [Formula presented] such that the roots of the polynomial [Formula presented] are simple. Let [Formula presented] for [Formula presented], where [Formula presented], and let [Formula presented] be the roots of [Formula presented] that are simple by assumption. Case 1): [Formula presented]. If [Formula presented] and [Formula presented] are both constant polynomials, then [Formula presented] and [Formula presented] are both multiples of [Formula presented] and hence for any [Formula presented] such that [Formula presented] the roots of [Formula presented] are [Formula presented] and thus simple. Thus, in the following we may assume that one of the polynomials [Formula presented] and [Formula presented] is not constant, without loss of generality let this be the case for [Formula presented]. Suppose that ξ is a multiple zero of [Formula presented] for some [Formula presented]. Then we have [Formula presented] and [Formula presented]. It follows that s is in the kernel of the matrix [Formula presented], where [Formula presented] is defined as a [Formula presented] polynomial matrix and it does not depend on s. We know that [Formula presented] for each [Formula presented] because [Formula presented], so [Formula presented] for all [Formula presented]. Let us show that [Formula presented]. To this end, consider the polynomial [Formula presented] Suppose that [Formula presented] is identically zero. This implies that [Formula presented]. But then it follows from [Formula presented] that [Formula presented] divides its derivative [Formula presented], which is not possible, since the degree of [Formula presented] is at least one. Therefore [Formula presented] is not identically zero and this shows that [Formula presented]. Then there are only finitely many values [Formula presented] such that [Formula presented] for [Formula presented]. Now consider the [Formula presented] lines [Formula presented], [Formula presented], and [Formula presented], [Formula presented], in [Formula presented]. If s is not on any of these lines, the roots of [Formula presented] are simple and different from [Formula presented]. But then also the roots of [Formula presented] are simple. This proves the claim for the case [Formula presented]. Case 2): [Formula presented]. Let [Formula presented] be the roots of [Formula presented]. Consider the polynomial matrix [Formula presented] Since [Formula presented] it follows that [Formula presented] for [Formula presented]. Thus, if [Formula presented] is not from any of the m hyperplanes [Formula presented] for [Formula presented], it follows that [Formula presented] has no roots in common with [Formula presented], or, in other words, [Formula presented]. But then we can apply Case 1) to find a linear combination of [Formula presented] that has only simple roots that are distinct from the roots of [Formula presented]. Then we conclude that also the roots of [Formula presented] are simple.