Convexity of Quadratic equation Inequality?

Question

Solving an inequality of the form $x^TAx\geq0$ or $x^TAx\leq0$ is straightforward. I mean we have to check if A is positive semidefinite or negative semidefinite. But what would be the solution to the inequality $x^TAx+b^Tx+c\leq0$ and $x^TAx+b^Tx+c\geq0$ ? Specifically I need to know when either of the inequality would be convex. If someone can share a good resource that talks about quadratic equation(not quadratic form) with matrices as coefficients besides Wikipedia, it would be great. Thank you.

Answer 1

Given an arbitrary line $L = \{ x_0 + tv \in R^n \ | \ t \in R \}$, we know that the set $C = \{ x \in R^n \ | \ x^TAx + b^Tx + c \le 0 \}$ is convex if and only if the intersection with this line is convex. For convexity, we want to prove that $L$ intersects $C$ as a continuous line segment where $t$ is continuous in a bounded range (e.g, stretching $v$ in a bounded range).

Substituting $L$ into $C$ we get the intersection $f(t) = (v^TAv)t^2 + (2x_0^TAv + b^Tv)t + (x_0^TAx_0 + b^Tx_0 + c) \le 0$, which is a concave upward function of $t$ if $A \ge 0 => v^TAv \ge 0$ (i.e., $ \ f''= v^TAv \ge 0$ ). $f(t)$ is bounded above by zero and bounded between a fixed range of $t$ (i.e., the two roots of f(t)=0). Therefore, $t$ is continuous in a bounded range implies that the intersection is a convex line segment.

In conclusion, $C$ is convex when $A \ge 0$.

Answer 2

If we substitute $x = y + v$ in the inequality $x^T A x + b^T x + c \le 0$ we get $y^T A y + (b^T + 2 v^T A) y + v^T A v + b^T v + c \le 0$. Assuming $A$ is invertible, we can choose $v$ so that $b^T + 2 v^T A = 0$, so the inequality becomes $y^T A y \le k$. If, for example, $A$ is positive definite and $k > 0$, this defines an ellipsoid.