Why is this optimization problem nonconvex: ${\min}_{x \in \mathbb{C}^n} \ x^* A x \ $s.t.$|b^* x|^2 \geq 1$, $A \succeq 0$, and $b \in \mathbb{C}^n$?

Question

Given problem,

\begin{align} \text{minimize}_{x \in \mathbb{C}^n} \quad & x^* A x \\ \text{subject to }\quad & |b^* x|^2 \geq 1 \ ,\\ \end{align} where $A \in \mathbb{C}^{n \times n} \succeq 0$ and $b \in \mathbb{C}^n$ are known. Also, $(\cdot)^*$ means complex conjugate and transpose operation.

Question: Why is this problem nonconvex?

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My arguments:

1. The objective function is convex because it's a quadratic problem with positive definite matrix $A$ (or can be shown convex via computing Hessian which will be $A$ matrix)
2. I suspect that the constraint is nonconvex. My thinking is that if we take a scalar case and real-valued, say $b = 1$, then $|x|^2 \geq 1 \Rightarrow x \geq 1; x \leq -1$. It means the set formed by such constraint would be nonconvex. Is it correct? If my thinking is correct, then can we prove it rigorously?

Answer 1

Unless $b=0$ the feasible set is non-empty: large enough scalar multiples of $b$ are in it. If $x$ is in the feasible set, so is $-x$. But the $0$ vector is not, even though it is a convex combination of $x$ and $-x$. So the feasible set is not convex.