# Positive semidefinite inequality

Let $$\mathbf{A}\in\mathbb{C}^{m\times m}$$, and $$\mathbf{B}\in\mathbb{H}^{m\times m}$$ be an $$m$$-dimension Hermitian matrix, solve $$\theta$$ that satisfies the condition $$\begin{equation*} e^{\jmath\theta}\mathbf{A}+e^{-\jmath\theta}\mathbf{A}^H+\mathbf{B}\succeq0, \end{equation*}$$ where $$\jmath=\sqrt{-1}$$.

Thank you so much.

• @user1551Sorry, I made a mistake in the dimension. I have fixed it. Thank you so much. – Alex Jul 3 at 7:53

The solution $$\theta$$ (which is presumably real) may not exist. This occurs, e.g., when $$A=0$$ and $$B=-I$$. I don't know any sufficient existential condition that isn't too demanding. (It is easy to cook up a sufficient condition, such as $$B\succeq2\|A\|_2I$$, but that would make every $$\theta$$ a solution.)
Here is a reformulation of the problem that may or may not be useful. Let $$c>0$$. Rewrite the inequality as $$\left(\frac1c e^{\jmath\theta}A + cI\right)\left(\frac1c e^{-\jmath\theta}A^H + cI\right) \succeq \frac{1}{c^2}AA^H + c^2I - B.\tag{1}$$ Choose a sufficiently large $$c$$ so that the RHS is positive definite. Let $$P$$ be the positive definite square root of the RHS. Then $$(1)$$ is equivalent to $$\sigma_\min\left(\frac1c e^{\jmath\theta}P^{-1}A + cP^{-1}\right)\ge1.\tag{2}$$ So, the problem boils down to maximising the minimum singular value of $$\frac1c e^{\jmath\theta}P^{-1}A + cP^{-1}$$ over $$\theta\in[0,2\pi]$$. (You don't need to find the global maximiser. Any $$\theta$$ that satisfies $$(2)$$ suffices as a solution to the original problem.)