How to show a sum of positive definite operators is still positive definite? Consider a Hamiltonian $H$ of the form
$$\begin{split}H&=\sum_{i,j}A_{ij}^\dagger A_{ij},\\ 
A_{ij}&=(1-\sigma_i^z\sigma_j^z)e^{-g (\sigma_i^x+\sigma_j^x)},\end{split}$$
where $g\in\mathbb{R}$ is a real parameter and $\sigma_i^x$ and $\sigma_i^z$ are the Pauli matrices $x$ and $z$ acting on the $i$th spin (qubit). Obviously, each term $A_{ij}^\dagger A_{ij}$ in the summation is positive definite. What I found in numerics is that the eigenvalues of $H$ are all non-negative. How should I prove (or disprove) that $H$ is positive definite?
I am aware that supersymmetric quantum mechanics Hamiltonians are positive definite. Can the above Hamiltonian be cast into a supersymmetric form explicitly?
 A: Three remarks. One, a sum of positive (semi)definite matrices is again positive (semi)definite. The proof is really easy. Let's work over the reals for simplicity. A matrix $H_I$ is positive definite iff for any vector $v$, we have
$$v^T \cdot H_I \cdot v > 0\,.$$
Now let $H = \sum_I H_I$ be a sum of a finite number of positive matrices $H_I$. Then for any vector $v$, we have
$$v^T \cdot H \cdot v = \sum_I v^T \cdot H_I \cdot v > 0$$
because a sum over positive numbers is again positive.
Two, the operator $A_{ij}$ is only semidefinite, because it annihilates certain states. Consider for instance the two-qubit state
$$ | \psi_{ij} \rangle = | + \rangle_i \otimes | + \rangle_j - | - \rangle_i \otimes | - \rangle_j\,.$$
Then I think that $A_{ij} | \psi_{ij} \rangle = 0$ (unless I made a mistake). To check this you first show that
$$\exp\left[-g( \sigma_i^x + \sigma_j^x)\right]$$
leaves $|\psi_{ij} \rangle$ invariant, and second that it's killed by $1 - \sigma_i^z \sigma_j^z$.
Third, you want to cast the Hamiltonian into a form $H = Q^\dagger Q$, right? In linear algebra this is known as a Cholesky decomposition. Since $H$ is positive semidefinite, it's certain that such a $Q$ exists (and that it's upper triangular), but it will not be unique in general. There are standard algorithms to compute (at least one realization of) $Q$ if $H$ is a concrete matrix.
