I have to compute a matrix $P$ that defines an Linear Matrix Inequality region in the following way $$L_P = \{ s\in \mathbb{C}|\begin{pmatrix}I \\sI \end{pmatrix}^* P\begin{pmatrix}I \\sI \end{pmatrix}\prec 0\} $$ for the constraint $|\Re (s)|> |\Im (s)|$.
I have used the conic region matrix, knowing that for $\Re (s) \tan(\theta)> |\Im (s)|$ we get the matrix $$\begin{pmatrix}0 & 0&\sin(\theta) &\cos(\theta) \\0 & 0&-\cos(\theta)&\sin(\theta) \\\sin(\theta) & -\cos(\theta)&0 &0\\\cos(\theta) & \sin(\theta)&0 &0 \end{pmatrix}$$ and that the inequality $|\Re (s)|> |\Im (s)|$ can be split into $\Re (s)\tan(\pi/4)> |\Im (s)|$, $\Re (s)\tan(-\pi/4)> |\Im (s)|$.
So I get two different matrices $P_1$,$P_2$, but the problem requires to get only one. How can I put them together? Is it correct to build a matrix $$\begin{pmatrix}P_1 &0 \\0&P_2 \end{pmatrix}$$ and use it as solution?