Let $E = \{x\mid (x - c)^{T} P^{-1} (x-c) \le 1 \}$, where $P$ is symmetric positive definite. Show that $E$ is convex.
Here is what I did. It seems like $E$ is an ellipsoid. We want to show that if $x, y \in E$, then $(1-\theta)x+\theta y \in E$, for $0\le\theta\le1$.
Suppose $x, y \in E$. By definition we have
$$ (x-c)^{T}P^{-1}(x-c) \le 1 $$ $$ (y-c)^{T}P^{-1}(y-c) \le 1 $$
Expanding equations we get
$$ x^{T}P^{-1}x - 2c^{T}P^{-1}x +c^{T}P^{-1}c \le 1 $$
$$ y^{T}P^{-1}y - 2c^{T}P^{-1}y +c^{T}P^{-1}c \le 1 $$
Multiply both sides of first equation by $1-\theta$ and second by $\theta$, add them up, then we are done?
My question is why do we need the fact that $P$ is symmetric positive definite?
That won't do it. You need to consider $$ ((1-\theta)x+\theta y-c)^T P^{-1} ((1-\theta)x+\theta y-c). $$ This is equal to $$ \begin{align} & \phantom{={}} ((1-\theta)x+\theta y-[(1-\theta) c + \theta c])^T P^{-1} ((1-\theta)x+\theta y-[(1-\theta) c + \theta c]) \\[10pt] & = (1-\theta)^2 \underbrace{(x-c)^T P^{-1}(x-c)} + \theta^2 \underbrace{(y-c)^T P^{-1}(y-c)} + 2\theta(1-\theta)\underbrace{(x-c)^T P^{-1} (y-c)}. \end{align} $$ The third term is subject to a Cauchy–Schwarz inequality: $$ ((x-c)^T P^{-1} (y-c))^2 \le \Big((x-c)^T P^{-1} (x-c)\Big)\Big((y-c)^T P^{-1} (y-c) \Big). $$