I would like to prove an ellipsoid $E_a$ center at $q_a$ with shape matrix $A$ and radius $r_a$ defined as: $(x-q_a)^T A^{-1} (x-qa) <= r_a$
is included into another ellipsoid $E_b$ -i.e. $E_a$ covers $E_b$- defined as $E_a$:
$(x-q_b)^T B^{-1} (x-qb) <= r_b$
with x defined on $R^n$
I was thinking about showing that all points defined by $E_a$ are also included into $E_b$ such that $E_a <= E_b$ but I don't know how to prove a condition satisfying this ellipsoid status. I don't really find paper or information about that ... link Because they target center at the origin ellipsoids.
Thank you a lot for your help and maybe if Jacob see this topic because he asked for the same question without any answers, Regards.
A possible approach to the problem is first to perform an affine transformation on $R^n$ that takes ellipsoid $E_b$ to the unit sphere. Writing $u$ for the transformed coordinates, we have $$u = \frac1{\sqrt{r_b}} B^{-1/2}(x - q_b).$$ Then $x = \sqrt{r_b} B^{1/2} u + q_b$ and $$x - q_a = \sqrt{r_b} B^{1/2} u + q_b - q_a = \sqrt{r_b} B^{1/2} \left( u - \frac1{\sqrt{r_b}} B^{-1/2}(q_a - q_b) \right).$$ Let $q_0 = \dfrac1{\sqrt{r_b}} B^{-1/2}(q_a - q_b),$ and we can write the equation above more concisely as $$x - q_a = \sqrt{r_b} B^{1/2} (u - q_0).$$
When we substitute the right-hand side of this equation for $x - q_a$ in the equation of ellipsoid $E_a,$ we get $$(u - q_0)^T \left(\sqrt{r_b} B^{1/2}\right)^T A^{-1} \sqrt{r_b} B^{1/2} (u - q_0) \leq r_a,$$ which simplifies to $$(u - q_0)^T B^{1/2} A^{-1} B^{1/2} (u - q_0) \leq \frac{r_a}{r_b}.$$ (Note that $B$ must be symmetric positive definite in order that $(x-q_b)^T B^{-1} (x-q_b) \leq r_b$ may define an ellipsoid, and therefore $(B^{1/2})^T = B^{1/2}.$)
Let $M = \dfrac{r_b}{r_a}B^{1/2} A^{-1} B^{1/2},$ and the problem reduces to a maximization problem in which we want to find $$s = \sup \{ u^T u : (u - q_0)^T M (u - q_0) \leq 1 \}.$$ Then $E_a$ is contained in $E_b$ if and only if $s \leq 1.$
You should be able to find helpful information on the solution of this maximization problem, for example on pages $13$-$14$ of this document.
As an aside, for an ellipsoid defined by $(x-q_b)^T B^{-1} (x-q_b) \leq r_b,$ I think it is misleading to call $r_b$ the "radius" of the ellipsoid. First of all, an ellipsoid does not have a single radius; furthermore, even if we do somehow single out one radius, for example the circumradius, you should find that as you vary $r_b$ in the equation above, the circumradius of the ellipsoid varies in proportion to $\sqrt{r_b},$ not $r_b.$
A more conventional way to write the equation of a general ellipsoid would be $(x-q_b)^T B^{-1} (x-q_b) \leq 1,$ which you can get by scalar multiplication of the matrix you would have used in your original formula. This would simplify the equations in this answer.