Equivalence of positivity Let us have complex matrices and their real decompositions as $H=H_1 + \imath H_2$ and $L = L_1 + i L_2$. Further, $H_1\ge 0$ and $H_2$ is skew symmetric. $L = I - P$ where $P$ is some positive matrix. $I$ is identity matrix. $L$ itself need not be positive. 
Lets further consider the dilation 
\begin{equation} 
\tilde{H} =\begin{bmatrix} H_1 & -H_2 \\ H_2 &H_1\end{bmatrix},
\end{equation}
and 
\begin{equation} 
\tilde{L} =\begin{bmatrix} L_1 & -L_2 \\ L_2 & L_1\end{bmatrix}.
\end{equation}
Let $J=\begin{bmatrix} 0 & I \\ -I & 0 \end{bmatrix}$. Please tell me whether the following statement is true:

$\tilde{H} + \imath J \tilde{L} \geq0$ if and only if  $\tilde{H}
> \pm\tilde{L} \geq 0$.

I was convinced that this must be wrong; but I could not construct a counterexample. Neither I could prove it. Advanced thanks for any help/suggestions.
 A: OK. So the whole thing was very simple. First note that both $\tilde{H}$ and $\tilde{L}$ commutes with $J$. Now $\tilde{H} + \imath J \tilde{L}$ \ge 0$ is equivalent to 
\begin{equation}
\begin{bmatrix}
\tilde{H} & -J \tilde{L} \\
J \tilde{L} & \tilde{H} \end{bmatrix} \ge 0.
\end{equation}
Which is possible if and only if $\tilde{H} \geq - J \tilde{L} \tilde{H}^{-1} J \tilde{L}$. Using commutativity of $J$ with both $\tilde{H}$ and $\tilde{L}$ we eventually get its equivalence with the form $\tilde{H} \geq  \tilde{L} \tilde{H}^{-1}  \tilde{L}$ which is equivalent to 
\begin{equation}
\begin{bmatrix}
\tilde{H} & \tilde{L} \\
\tilde{L} & \tilde{H} \end{bmatrix} \ge 0.
\end{equation}
Rest follows from taking $\tilde{H}^{-\frac{1}{2}}$ and writing the above matrix in the form 
\begin{equation}
\begin{bmatrix}
I & C \\
C & I \end{bmatrix} \ge 0,
\end{equation}
where $C$ is areal symmetric matrix. We ave assume $\tilde{H}$ to be invertible. If it is not, then we can take Moore- Penrose inverse and give the same argument. 
