Let $\mathbf{H}$ be an infinite dimensional, separable Hilbert space. Moreover, let $\{A_i\}_{i=1}^N$ be a set of linearly independent, invertible operators that act on $\mathbf{H}$. In other words, $A_i:\mathbf{H}\to \mathbf{H}$ for $i=1,\ldots, N$.
$\textbf{Question:}$ Are there any general conditions on $\{A_i\}_{i=1}^N$ such that all non-trivial linear combinations of $\{A_i\}_{i=1}^N$ results in an injective operator? Specifically, let $c = \{c_i\}_{i=1}^N\subseteq \mathbf{C}^N$ be any non-zero vector. Define $$ A_c = \sum_{i=1}^Nc_i A_i. $$ Are there any conditions on $A_i$ such that $A_c$ is injective for all $c\ne 0$?
I think there is no general condition and here is why:
The set of invertible operator between Banach spaces is open in the operator norm. Indeed, if we have $T_o$ to be invertible, then if $T$ is such that:
$$\lVert T-T_o \rVert<\frac{1}{\lVert T_o^{-1} \rVert}\Rightarrow \lVert T_o^{-1}T-I\rVert<1$$ Thus, by Neumann series, $T_o^{-1} T$ is invertible and because $T_o$ is invertible, so is $T$.
Thus, if we perturbe $A_1$ slightly it will remain invertible. Precisely, if we take:
$$|c_j|<\frac{1}{(N-1)\lVert A_j \rVert \lVert T_o^{-1}\rVert} $$
Then it is clear that for every $\lambda\not=0$, the following operator happens to be invertible:
$$ \lambda \left(A_1+\sum_{j=2}^N c_j A_j\right)$$
The same logic follows for any fixed $A_{i_0}$. However, if we do not have a given $c_{i_0}$ to dominate the others in absolute value, there is no guarantee it will be invertible.