Defining the modular Hamiltonian In Casini's proof of the Bekenstein bound a crucial step is the use of the modular Hamiltonian $K$ defined
$$\rho^0_V = \frac{e^{-K}}{\text{tr} e^{-K}}$$
where $\rho_V^0 = \text{tr}_{-V} |0\rangle\langle0|$ is the reduced density matrix of the vacuum state where we've traced over the complement of $V$, denoted $-V$. $\rho_V^0$ must be invertible for this to be valid. It is stated that $\rho_V^0$ has no zero eigenvalues, so this can always be defined-- unless $V$ is the whole space, in which case $\rho_V^0 = |0\rangle\langle0|$. 
How does one see that $\rho_V^0$ has no zero eigenvalues? This is certainly not true in finite-dimensional systems (either $\rho_V^0$ or $\rho_{-V}^0$ generically has zero eigenvalues)-- is this a property of QFT?
 A: The claim is that $\rho_V^0$ has no zero eigenvalues, which is a prerequisite for writing it as $\rho_V^0\propto e^{-K}$. 

is this a property of QFT?

Yes, it's a property of relativistic QFT, which is what Casini means by QFT. This property is closely related to the Reeh-Schlieder theorem, which is mentioned in Casini's paper and reviewed in 


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*Witten, "Notes On Some Entanglement Properties Of Quantum Field Theory," https://arxiv.org/abs/1803.04993
The RS theorem implies that in a conventional relativistic QFT in Minkowski spacetime, the vacuum state cannot be annihilated by any nonzero local operator. Here, "local" means localized within any finite region, not necessarily at a point.
Suppose that $\rho^0_V$ had a zero eigenvalue. This would imply the existence of some nonzero operator $A$, localized in $V$, such that $\text{trace}(XA\rho_V^0)=0$ for all operators $X$ localized in $V$. The operators $A$ and $X$ must be localized in $V$ for $\text{trace}(XA\rho_V^0)$ to be defined, because $\rho_V^0$ is the result of taking a partial trace over the complement of the region $V$. So if $\rho_V^0$ had a zero eigenvalue, then we would have a nonzero operator $A$ localized in $V$ that annihilates the vacuum state. If $V$ has a non-empty complement, then this would contradict the RS theorem. Thus $\rho_V^0$ cannot have any zero eigenvalues unless the complement of $V$ is empty.
The answer would be that easy, except that Casini is considering a cutoff theory. This is stated in the text above equation (6) in Casini's paper, which is the equation quoted in the OP. 
We could take the cutoff theory to be a lattice QFT, but a lattice QFT is never strictly relativistic, so the RS theorem does not apply here. In particular, lattice versions of the simplest non-relativistic QFTs do have local operators that annihilate the vacuum. 
Sometimes people impose something like the RS property as a necessary condition for a lattice QFTs to be considered "relativistic." If we did that here, then the question would be answered exactly as in the continuum case above. 
If we don't do that here, then some degree of hand-waving seems to be required. Here are a couple of hand-waving arguments:


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*A matrix generically has no zero eigenvalues: zero eigenvalues are exceptional. This suggests that among those lattice QFTs that are expected to flow to a particular relativistic QFT in the continuum limit, most of them probably do have the property $\rho_V^0\propto e^{-K}$ that Casini assumes, and we might as well use one that does.

*If removing the cutoff is a smooth process, then $\rho_V^0$ must smoothly approach the $\rho_V^0$ of the continuum theory, which has no non-zero eigenvalues thanks to the RS theorem, as explained above. Thus the eigenvalues of $\rho_V^0$ must all be non-zero even for some finite values of the cutoff.
