Why do most definitions of the ellipsoid use the inverse of a positive definite matrix instead of the matrix itself?

Question

For instance, on Boyd and Vandenberghe's Convex Optimization, the ellipsoid is defined as follows

$$\left\{ x \mid(x-x_c)^T P^{-1} (x-x_c) \leq 1 \right\}$$

I've also seen that on other books, too. I don't know what difference it makes using $P^{-1}$ instead of just using $P$, since both are positive definite (PD) matrices. There must be some reason why most books insist on using the inverse.

Answer 1

I guess this is for consistency with the ball notation $B(x_c,r)=\{x: \|x-x_c\|\le r\}$. If we define $\mathcal{E}(x_c,P)=\{x:(x-x_c)^TP^{-1}(x-x_c)\le 1\}$, then $$ \mathcal{E}(x_c,r^2I)=B(x_c,r). $$ If we instead define $\widetilde{\mathcal{E}}(x_c,P)=\{x:(x-x_c)^TP(x-x_c)\le 1\}$, then we would have $$ \widetilde{\mathcal{E}}(x_c,r^2I)=B(x_c,1/r). $$

Answer 2

You are correct that $P > 0$ if and only if $P^{-1} > 0$.

The lengths of the semi-axes are $\sqrt{\lambda_k}$, where $(\lambda_k)_k$ are the eigenvalues of $P$, instead of $\sqrt{\frac{1}{\lambda_k}}$, which they would be if we chose $P$ in the definition of the ellipse.

This might seem more natural because then scaling up $P$ (that is, instead considering $a P$ for some $a > 1$) results in the eigenvalues being multiplied $a$ (i.e. increasing), so that we get a larger ellipsoid, not a smaller one. A special case of this is the answer by @Marlovo where $P = r^2 I$ and thus $\lambda_k = r^2 I$ for all $k$.