Linear combination of matrix functions is invertible

Question

We are given two smooth $m \times m$ matrix functions $F_1, F_2: \mathbb{R}^n \to \mathbb{C}^{m \times m}$. We are also given bounded open sets $U_1$ and $U_2$, such that $F_i$ is non-singular (invertible) in a slightly larger open set containing $\overline{U_i}$ for $i = 1, 2$. Is there a linear combination $C = a_1F_1 + a_2F_2$ for some $a_1, a_2\in \mathbb{C}$ such that $C$ is invertible on $U_1 \cup U_2$?

Answer 1

If $M$ and $A$ are any matrix then $det(M-tA)$ is a polynomial in $t$. If in addition $M$ is invertible, this polynomial is non zero hence has only finitely many zeroes. So $M-tA$ is invertible for all but finitely $t$. Moreover the $t$ making this determinant vanish locally depends continously on $M$ and $A$ (local inversion theorem).

Therefore for all $x \in U_1$ we can find an open neighborhood$N_x$ of $x$ and a real $M_x$ such that for $t>M_x$, $F_1+tF_2$ is invertible on $N_x$.

Now the closure of $U_1$ is closed and bounded so we can cover it by a finite number of neighborhoods of the form $N_{x_i}$. Taking $t$ larger than the maximum of the corresponding $M_{x_i}$ we ensure that $F_1+tF_2$ is invertible on $U_1$.

Next we do an analogous trick on $U_2$ and we get that for $t$ sufficiently large, $F_1+tF_2$ is invertible.