Let $C \subseteq \mathbb{R}^n$ be the solution set of a quadratic inequality,
$$ C = \{ x \in \mathbb{R}^n \mid x^TAx + b^Tx + c \le 0 \}$$
with $A \in \mathbb{R}^{n \times n}$ and $b \in \mathbb{R}^n$. Show that $C$ is convex if $A \succcurlyeq 0$
How does one prove it for the general case? I would think we would need to use the general definition of convexity somehow: if $\theta x_1 + (1-\theta)x_2 \in C$ where $\theta \in [0,1]$ and $x_1,x_2 \in C$ then $C$ is convex. But I don't know how to apply this.
If $\mathrm A \succeq \mathrm O_n$, then there exists an $n \times r$ matrix $\rm Q$ such that $\rm A = Q Q^\top$, where $r = \mbox{rank} (\rm A)$. Hence,
$$\mathrm x^\top \mathrm A \,\mathrm x + \mathrm b^\top \mathrm x + c = \mathrm x^\top \mathrm Q \mathrm Q^\top \mathrm x + \mathrm b^\top \mathrm x + c \leq 0$$
Using the Schur complement, the quadratic inequality above can be rewritten as the following linear matrix inequality (LMI)
$$\begin{bmatrix} \mathrm I_r & \mathrm Q^\top \mathrm x \\ \mathrm x^{\top} \mathrm Q & - c - \mathrm b^{\top} \mathrm x\end{bmatrix} \succeq \mathrm O_{r+1}$$
whose solution set is a spectrahedron and, thus, convex.
convex-analysis schur-complement linear-matrix-inequality spectrahedra