Inside and outside of an ellipsoid

Question

Let $A$ be a positive definite matrix.

The equation $x^TAx=1$ defines an ellipsoid.

I would like to justify the fact that $x^T A x > 1$ implies that $x$ is outside of the ellispoid.

In other words, why is $\{x\ | \ x^TAx\leq1\}$ the "full" ellipsoid?

Answer 1

It may help to note that $x^\top A x = c^2$ defines an ellipsoid for any $c$. When $c = 0$, this is a single point $\{0\}$, and as $c^2$ grows, the ellipsoid grows and grows.

---

Alternatively, another way to see this is that if $x^\top A x = c^2 > 1$ then $x/|c|$ lies on the original ellipsoid since $(x/|c|)^\top A (x / |c|) = 1$. But $x/|c|$ is closer to zero than $x$.

Answer 2

The ellipsoid is the surface such that $x^TAx-1$ is zero, and by continuity this expression cannot change sign without $x$ crossing the surface. As the surface is closed, it has an inside and an outside. As it contains the origin, the inside is where $x^TAx-1$ is negative.