Distance Between Two Ellipsoids

In general, the minimum and the maximum distances (with respect to the standard norm $\|\bullet\|$ on $\mathbb{R}^n$) between two $n$-dimensional ellipsoids given by $$A:=\Biggl\{\left(x_1,x_2,\ldots,x_n\right)\in\mathbb{R}^n\,\Big|\,\sum_{i=1}^n\,\frac{x_i^2}{a_i^2}=1\Biggr\}$$ and $$B:=\Biggl\{\left(x_1,x_2,\ldots,x_n\right)\in\mathbb{R}^n\,\Big|\,\sum_{i=1}^n\,\frac{x_i^2}{b_i^2}=1\Biggr\}\,,$$ where $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$ are positive real numbers such that $a_i\geq b_i$ for every $i=1,2,\ldots,n$, are $$\min\big\{a_1-b_1,a_2-b_2,\ldots,a_n-b_n\big\}$$ and $$\max\big\{a_1+b_1,a_2+b_2,\ldots,a_n+b_n\big\}\,,$$ respectively. (Of course, the minimum distance is $0$ if there exist $j,k\in\{1,2,\ldots,n\}$ such that $a_j\geq b_j$ and $a_k<b_k$, whilst the maximum distance is still given by the same formula above.) To show this, one only needs to observe that, if $\textbf{u}\in A$ and $\textbf{v}\in B$ are such that $\|\textbf{u}-\textbf{v}\|$ is optimal, then $\textbf{u}-\textbf{v}$ is parallel to the normal vector of $A$ at $\textbf{u}$ as well as the normal vector of $B$ at $\textbf{v}$.

Let $\frac{x_1^2}{a_1^2}+\dots+\frac{x_n^2}{a_n^2} = 1$ and $\frac{y_1^2}{a_1^2}+\dots+\frac{y_n^2}{a_n^2} = 2$. Then by Cauchy-Schwarz $$ 1 = \sum\limits_{k=1}^{n}{\frac{y_k^2-x_k^2}{a_k^2}} = \sum\limits_{k=1}^{n}{(y_k-x_k)\frac{y_k+x_k}{a_k^2}}\le\left(\sum\limits_{k=1}^{n}{(y_k-x_k)^2}\right)^{\frac{1}{2}}\left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^4}}\right)^{\frac{1}{2}}. $$ Since $a_k^4\ge(\min{|a_i|})^2a_k^2$ for all $k$, it follows that $\left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^4}}\right)^{\frac{1}{2}}\le \frac{1}{\min{|a_i|}}\left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^2}}\right)^{\frac{1}{2}}$, and applying the triangle inequality to the vectors $\vec{x} = \left(\frac{x_1}{a_1},\dots,\frac{x_k}{a_k}\right)$ and $\vec{y} = \left(\frac{y_1}{a_1},\dots,\frac{y_k}{a_k}\right)$ yields $$ \left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^2}}\right)^{\frac{1}{2}} = \|\vec{x}+\vec{y}\|\le\|\vec{x}\|+\|\vec{y}\| = 1+\sqrt{2}. $$ Hence, $1\le\frac{1+\sqrt{2}}{\min{|a_i|}}\left(\sum\limits_{k=1}^{n}{(y_k-x_k)^2}\right)^{\frac{1}{2}}$, i.e. $\left(\sum\limits_{k=1}^{n}{(y_k-x_k)^2}\right)^{\frac{1}{2}}\ge(\sqrt{2}-1)\min{|a_i|}$.

If we assume w.l.o.g. that $\min{|a_i|} = |a_1|$, then $(x_1,x_2,\dots,x_n) = (a_1,0,\dots,0)$ and $(y_1,y_2,\dots,y_n) = (\sqrt{2}a_1,0,\dots,0)$ attains equality in the above inequality.

Here is a solution using Lagrange multipliers: The key is to note that if $(x,y)$ solves the problem (with $x,y$ being on the two ellipses), then $x$ is a closest point to the '$y$'-ellipse, and $y$ is a closest point to the '$x$'-ellipse.

Let $f(x) = \sum_k ({x_k \over a_k } ) ^2$. Let $E_k = f^{-1} (\{ k \})$, $k=1,2$ The $E_k$ are compact, so we know there are minimisers. Suppose $(x,y)$ is a minimiser with $x \in E_1, y \in E_2$.

Since $y$ minimises the distance to $E_1$, we have $x-y\parallel \nabla f(y)$. Similarly, since $x$ minimises the distance to $E_2$, we have $y-x \parallel \nabla f(x)$.

Hence $\nabla f(y) = \lambda \nabla f(x)$ for some $\lambda$, and so $y = \lambda x$. Since $y \in E_2, x \in E_1$, we find that $\lambda = \sqrt{2}$.

Hence any minimiser $(x',y')$ satisfies $y' = \sqrt{2} x'$. If $x' \in E_1$, we see that $\sqrt{2}x' \in E_2$, hence we can restrict our search to minimising the distance between $x'$ and $\sqrt{2} x'$, with $x' \in E_1$.

That is, we need to solve $\min \{ (\sqrt{2}-1) \|x'\| | x' \in E_1 \}$, and it is clear that the minimising point is a closest point on $E_1$ to the origin. It is clear that if $i$ is such that $|a_i| = \min_k |a_k|$, then $x'=a_i e_i$ is a minimiser, from which we conclude that the minimum distance is $(\sqrt{2}-1) \min_k |a_k|$.