Suppose that $$||M(\lambda)||_2 ||\Delta(\lambda)||_2 <1 \quad \forall \lambda \in \mathbb C^0 \cup \mathbb C^+ \cup \{\infty\},$$ where $M,\Delta$ are real rational matrices in $\lambda$ and $||\cdot||_2$ denotes the spectral norm.
I want to show that above implies $$\text{det}(I-M(\lambda)\Delta(\lambda)) \neq 0 \quad \forall \lambda \in \mathbb C^0 \cup \mathbb C^+ \cup \{\infty\}.$$
From inspection and just considering the spectral norms inside the determinant, it is trivial. But I also tried to derive it via the spectral norm properties, but am not sure about the conclusion.
Attempt of proof: For convenience I drop the argument $\lambda$. $$||M||_2 ||\Delta||_2 <1$$ $$\iff$$ $$ \sqrt{\text{eig}_{\text{max}}(M^HM) \text{eig}_{\text{max}}( \Delta^H\Delta)} < 1 $$ $$\iff$$ $$M^H M \Delta^H\Delta \prec I. $$ What I would need is instead $$M\Delta \prec I,$$ since then $$I-M(\lambda)\Delta(\lambda)\succ 0 \quad \forall \lambda \in \mathbb C^0 \cup \mathbb C^+ \cup \mathbb \{\infty\}$$ and thus non-singular.
The problem can be simplified as establishing whether $||M||_2<1\Rightarrow\det(I-M)\ne0$ is true for some square real/complex matrix $M$.
Assume that $||M||_2<1$, then we have that $\rho(M)\le||M||_2<1$ where $\rho(\cdot)$ denotes the spectral radius. As a consequence, $\det(I-M)\ne0$ since $\det(I-M)=0$ would require an eigenvalue of $M$ to be equal to 1, which is not possible by virtue of the spectral radius condition.