Convex set, quadratic form

Question

I'm trying to answer a question concerning convex sets

"Does the following constraint system define a convex set?

$x^T Qx ≤ 1$

$a^T x = 0$

Here, Q is a symmetric and positive definite matrix and a is a vector. Give a graphical motivation of your answer."

But I can't figure it out.

I'm guessing that $a^T x = 0$ is a line (at least in 2- or 3-dim). But have no idea of how the quadratic form would look like and how it is relevant that it is symmetrical. That it is positive definite must mean that $x^T Qx > 0$, but I don't see how that is relevant.

Any hints?

Answer 1

Yes. The first equation defines a convex set. The second constraint also defines a convex set. An intersection of convex sets is convex.

Answer 2

Considering matrix $Q$ is symmetric and positive-definite, each eigenvalue of $Q$ is positive. Then the geometric image of quadratic form $Q(x)\leq1$ is interior of an (hyper-)ellipsoid. Plus the condition $x^Ta=0$, the (hyper-space) consisting of all vectors $x$ (including the origin) perpendicular to vector $a$. Hence the intersection of (hyper-)ellipsoid and (hyper-)plane is certainly convex.