Why there is such a $\mu_0$ such that both $ A-\mu_0 B$ and $C+\mu_0 D$ are both nonsingular matrices?

Question

Let $A,B,C,D\in \mathbb{C}^{n\times n}$. If both $A-\lambda B$ and $C-\lambda D$ are both regular matrix pencils, there exists $\mu_0\in \mathbb{C}$ such that both $A-\mu_0 B$ and $C+\mu_0 D$ are both nonsingular matrices. I wonder how to prove the existence of such a $\mu_0$.

Answer 1

Based on Jin's comment, a regular matrix pencil is a line in $\mathbb {C} \times \mathbb {C} $, given by $\{A - \lambda B \, \mid \, \lambda \in \mathbb {C}\} $ such that $\det ( A -\lambda_0 B) \neq 0$ for some $\lambda_0 \in \mathbb {C}$.

Define $p(\lambda) = \det(A - \lambda B)$ and $q(\lambda) = \det(C - \lambda D)$. Then $p,q$ are polynomials. Moreover, since both pencils are regular, both $p $ and $q$ are non-zero. Consequently, the zeros of $p$ and $q$ form a finite subset of $\mathbb{C}$. On the other hand, $\mathbb{C}$ is uncountable. Therefore, $\{\lambda \in \mathbb{C} \, \mid \, p(\lambda) \neq 0\} \cap \{\lambda \in \mathbb{C} \, \mid \, q(\lambda) \neq 0\}$ is an uncountable set. Since a matrix is invertible if and only if its determinant is non-zero, the set $\{\mu_{0} \in \mathbb{C} \, \mid \, \det(A - \mu_{0} B) \neq 0\} \cap \{\mu_{0} \in \mathbb{C} \, \mid \, \det(C - \mu_{0} D) \neq 0\}$ is uncountable.

Explanation of original mistake: Before I edited my answer, I made the silly mistake of assuming that $p $ and $q $ were non-zero and wondered whether or not the regularity assumption is really necessary. In fact, as I indicate in italics above, $p$ is a non-zero polynomial if and only if the pencil $\{A - \lambda B \, \mid \, \lambda \in \mathbb{C}\}$ is regular. It is true that in the case of the characteristic polynomial $f _{A}(\lambda) = \det (A - \lambda)$ we always get a non-zero matrix, but only because the spectrum of a matrix is a bounded set. However, if the kernels of $A$ and $B $ have non-empty intersection, for example, then $\det (A - \lambda B) =0$ independently of $\lambda $. This immediately implies the pencil won't be regular when $A$ is singular and $B$ is a multiple of $A$. Actually, the most obvious example when a matrix pencil is not regular is $A = B = 0$.