Convexity of the ellipsoid

Question

In Boyd & Vandenberghe, it is mentioned that the ellipsoid is defined by

\begin{equation} \mathcal{E} = \left\{ x \in \mathbb R^n \mid (x-x_c)^T P^{-1} (x-x_c) \leq 1 \right\} \end{equation}

where $P$ is positive definite. My questions are:

1. How to transform it to the normed (in)equation?

2. How to show that it is convex?

Thank you.

Answer 1

The equation 2 you mentioned is $$\{x_c + Au \mid \|u\|_2 \le 1 \}$$

Where $$A^TPA = I$$

Any and only $x$ satisfying the equation will satisfy the original $$(x_c + Au - x_c)^T P (x_c + Au - x_c) \\ = u^TA^T P Au \\ = u^T I u \\ = \|u\|_2$$

Answer 2

Regarding question 2: Any sublevel set of a convex function is convex. And the function $f(x) = (x - x_c)^T P^{-1} (x - x_c)$ is convex because its Hessian is $2P^{-1}$, which is positive definite.

Answer 3

Using the Schur complement, the inequality

$$1 - (\mathrm x - \mathrm x_c)^\top \mathrm P^{-1} (\mathrm x - \mathrm x_c) \geq 0$$

can be rewritten as follows

$$\begin{bmatrix} \mathrm P & \mathrm x - \mathrm x_c\\ (\mathrm x - \mathrm x_c)^\top & 1 \end{bmatrix} \succeq \mathrm O_{n+1}$$

Hence, the ellipsoid is a spectrahedron and, thus, convex.

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convex-analysis schur-complement linear-matrix-inequality spectrahedra ellipsoids