Non-determinantal positive-definiteness conditions for $4 \times 4$ Hermitian matrices with certain null entries I have classes of $4 \times 4$ Hermitian matrices $D$, the two $2 \times 2$ diagonal blocks of which are themselves diagonal. That is, the (1,2),(2,1),(3,4) and (4,3) entries of $D$ are zero. (The other off-diagonal entries of $D$ can be real, complex or quaternionic.) I want necessary and sufficient (maybe minimal) conditions that $D$ be positive definite, that do not involve the full determinant of $D$. (In the quaternionic case, this would be the "Moore determinant", but the real and complex cases are the ones of immediate interest to me.) 
As some background, this pertains to a quantum-information-theoretic problem, in which I am also interested in the "partial transpose" $D^{PT}$ of $D$, obtained, in general, by transposing in place the four $2 \times 2$ blocks of $D$. The determinants of both $D$ and $D^{PT}$ are quite cumbersome, while the difference of the two determinants simplifies greatly, which is the basic motivation for my question. (I want to enforce the positive-definite nature of $D^{PT}$, subject to $D$ itself being positive-definite. Hopefully, the $2 \times 2$ and $3 \times 3$ minors of these matrices are more computationally amenable than their determinants. This pertains to the problem of determining the probability that two quantum bits ["qubits"] are disentangled/separable.)
 A: Let's go and derive first an equivalent condition on a related $2\times 2$ matrix which guarantees positive-definiteness of $\,D$, valid in either the complex or the real case.
(1) Fixing notation:
$$D\:=\;\begin{pmatrix} D_1& M\\ M^*& D_2\end{pmatrix}
 \:=\;\begin{pmatrix} \begin{pmatrix}d_1& 0\\ 0& d_2\end{pmatrix}& \begin{pmatrix}m_{13}& m_{14}\\ m_{23}& m_{24}\end{pmatrix}\\[1ex]
M^*& \begin{pmatrix}d_3& 0\\ 0& d_4\end{pmatrix}\end{pmatrix}$$
There are no a priori restrictions on the entries of $M$, and their indices stick to the indexation as of $D$.
It is henceforth assumed that $\,d_1,d_2,d_3,d_4>0$.
This feature is a necessary condition to $\,D\,$ being positive-definite, because
$\,d_i =\langle e_i| De_i\rangle>0$, with $e_i$ being a standard unit vector.
Using it right away one has the equivalence
$$0<D\;\iff\;
0\,<\:\begin{pmatrix} D_1^{-1/2}& \mathbb{0}\\ \mathbb{0}& D_2^{-1/2}\end{pmatrix}
 D\begin{pmatrix} D_1^{-1/2}& \mathbb{0}\\ \mathbb{0}& D_2^{-1/2}\end{pmatrix}
\;=\; \begin{pmatrix}\mathbb{1}& D_1^{-1/2}MD_2^{-1/2}\\
 D_2^{-1/2}M^*D_1^{-1/2}& \mathbb{1}\end{pmatrix}$$
since positive-definiteness is preserved under congruences
$A\mapsto T^*AT$ with invertible $T$. Now use that
$$0 < \begin{pmatrix}\mathbb{1}& C\\ C^*& \mathbb{1}\end{pmatrix}
\;\iff\;\|C\|<1\tag{1}$$
where $\|\cdot\|$ denotes the operator norm. It equals the largest singular value.
The little sister of $(1)$ within $M_2(\mathbb C)$ is the equivalence
$\;0<\left(\begin{smallmatrix}1& z\\ \overline z& 1\end{smallmatrix}\right)$
iff $|z|<1$ (and she usually shows up in the proof(*) of the general case).

Thus the equivalence criterion for positive-definiteness of $\,D\,$ reads
  $\,\left\|D_1^{-1/2}MD_2^{-1/2}\right\|\,\stackrel{!}{<}\, 1$
  put another way: $\,D_1^{-1/2}MD_2^{-1/2} = \left(\frac{m_{ij}}{\sqrt{d_id_j}}\right)_{i=1,2\;j=3,4}$ is a strict contraction.

(2) An alternative, more hands-on approach would be based on the 
Cholesky decomposition. It is the "$\Rightarrow$" in
$$A>0\:\iff\,\exists! \text{ lower-triangular matrix }L\text{ with }\,LL^*=A\,,$$
and many (to most) implementations of the Cholesky algorithm accept as input a matrix without definiteness properties, and return $L$, or some educated error message if the input was not positive (enough). Cholesky still works for positive-semidefinite input and yields a lower-triangular $L$ but its uniqueness is lost.
As $n=4$ this "positivity criterion" is checked within a blink of the eye $\:\ddot\smile$
* e.g. Proposition I.3.5 in R. Bhatia's book "Matrix Analysis"
(Springer, GTM 169, 1997)
