Ellipsoid contained in another ellipsoid: geometric proof of an inequality?

Question

Let $E_1$ and $E_2$ be two solid ellipsoids in $\mathbb R^n$, where

• both $E_1$ and $E_2$ are centered at the origin,
• $E_1\subset E_2$
• $E_1\ne E_2$

Let $0<a_1\le\cdots\le a_n$ be the semi-axes of $E_1$, and $0<b_1\le \cdots\le b_n$ be the semi-axes of $E_2$. Show that

$$ \frac1{b_1} + \cdots + \frac1{b_n} < \frac1{a_1} + \cdots + \frac1{a_n} \tag{$\star$} $$

---

The 2D case

The 2D case is rather simple: the minor axis of $E_2$ passes through two (antipodal) points of $E_1$, whose distance between must be greater than the minor axis of $E_1$, thus $a_1\le b_1$. Next, notice that the major axis of $E_1$ is contained within $E_2$. Thus, $a_2\le b_2$.

To prove strict inequality as in $(\star)$, notice that equality holds iff $a_1=b_1$ and $a_2=b_2$, i.e. $E_2$ is a rotation of $E_1$, and in such case $E_2$ contains $E_1$ iff $E_2=E_1$, which is assumed to be false.

In $n$ dimensions, we have $a_1\le b_1$ and $a_n\le b_n$ by the same argument.

---

Formulation in terms of matrices

The boundary of $E_1$ can be described by $x^T \mathbf{A}x=1$, where $\mathbf A$ is symmetric and has eigenvalues $\frac1{a_1^2},\cdots,\frac1{a_n^2}$, with corresponding eigenvectors pointing in the direction of the respective axes. Similarly, $\partial E_2=\{x:x^T \mathbf{B}x=1\}$.

Since $\mathbf{A}$ and $\mathbf{B}$ are positive definite and symmetric, let $\mathbf{A'}$ and $\mathbf{B'}$ be their square roots respectively. Then $$\partial E_1 = \{x:|\mathbf{A'} x|=1\}, \partial E_2 = \{x:|\mathbf{B'} x|=1\}$$

Then, $E_1\subset E_2$ translates to $$|\mathbf{B'} x|\le|\mathbf{A'} x|\qquad\forall x\in\mathbb R^n$$ and $E_1\ne E_2$ means strict inequality holds somewhere.

---

Proof of $(\star)$ using linear algebra

Let $\mathbf C=\mathbf{A'}-\mathbf{B'}$.

Then, \begin{align} |\mathbf{B'} x| &\le|\mathbf{A'} x| \\ |(\mathbf{A'}-\mathbf C) x|^2 &\le|\mathbf{A'} x|^2 \\ |\mathbf{A'}x|^2+|\mathbf C x|^2 - 2 x^T \mathbf {A'C} x &\le|\mathbf{A'} x|^2 \\ |\mathbf C x|^2 \le 2 x^T \mathbf {A'C} x \\ \end{align}

Note that $\mathbf C$ is symmetric. Take $x=e_\lambda$ to be the eigenvector of $\mathbf C$ corresponding to an eigenvalue $\lambda$. $$\lambda^2 \le 2 \lambda (x^T \mathbf {A'} x)$$

Since $\mathbf {A'}$ is positive definite, this gives $\lambda\ge 0$. In other words, $\mathbf C$ is non-negative definite, and $|\mathbf{B'} x|<|\mathbf{A'} x|$ somewhere implies $\mathbf A'\ne\mathbf B'$, thus $\mathbf C\ne \mathbf 0$. In particular, $\mathbf C$ has positive trace,

$$\text{trace}(\mathbf C)>0 \implies \text{trace}(\mathbf B')<\text{trace}(\mathbf A')$$

Notice that $\mathbf A'$, being the square root of $\mathbf A$, has eigenvalues $\frac1{a_1},\cdots,\frac1{a_n}$. Similarly for $\mathbf B'$. Hence, $$\frac1{b_1}+\cdots+\frac1{b_n}< \frac1{a_1}+\cdots+\frac1{a_n}\qquad $$ $\blacksquare$

---

Context

I encountered this problem while reading Luis Caffarelli's notes on obstacle problems (at PISA in 1998), where the original statement of the problem is:

Let $\mathbf A'$ and $\mathbf B'$ be non-negative definite matrices with $\text{tr}(\mathbf A')=\text{tr}(\mathbf B')=1$. If $|\mathbf{B'} x| \le|\mathbf{A'} x|+O(\epsilon)$, then $\|\mathbf A'-\mathbf B'\|\le O(\epsilon)$.

This appeared in connecting the ACF monotonicity result to the uniqueness of blowup at a singular point, and the proof above originates from p. 44 of his notes. While this proof is concise, it is too algebraic to reveal the underlying picture. In search of more intuition behind, I transformed the problem into an geometric one, and am looking for a geometric/any alternative proof.

Answer 1

A slight modification of the proof of Courant-Fischer variational characterisation will suffice.

Let $\pm v_i$ be the two endpoints of the $i$-th shortest axis of $E_1$, so that $\|v_i\|=a_i$. For each $k$ between $1$ and $n$, the intersection of any $k$-dimensional subspace $S\subseteq\mathbb R^n$ and the $(n-k+1)$-dimensional subspace $\operatorname{span}\{v_k,\ldots,v_n\}$ must contain some line $L$, which meets the surface of $E_1$ at some $x_0\in\operatorname{span}\{v_k,\ldots,v_n\}\cap \partial E_1$. Thus $$ a_k \le\|x_0\| =\max_{x\in L\cap E_1}\|x\| \le\max_{x\in S\cap E_1}\|x\|. $$ This is true for all $k$-dimensional subspaces $S$ and equality holds when $S=\operatorname{span}\{v_1,\ldots,v_k\}$. It follows that $$ a_k=\min_{\dim S=k}\max_{x\in S\cap E_1}\|x\| $$ and the analogous holds for $b_k$. As $E_1\subseteq E_2$, we have $\max_{x\in E_1\cap S}\|x\|\le\max_{x\in E_2\cap S}\|x\|$. Therefore $$ a_k =\min_{\dim S=k}\max_{x\in S\cap E_1}\|x\| \le\min_{\dim S=k}\max_{x\in S\cap E_2}\|x\| =b_k $$ and the inequality $\sum_k\frac{1}{b_k}\le\sum_k\frac{1}{a_k}$ follows.

Answer 2

Let the $n \times n$ real matrices ${\bf A} \neq {\bf B}$ be symmetric and positive definite. Define the (non-degenerate) ellipsoids centered at the origin

$$ \begin{aligned} {\mathcal E}_{\mathcal A} &:= \left\{ {\bf x} \in {\Bbb R}^n : {\bf x}^\top {\bf A} \, {\bf x} \leq 1 \right\} \\ {\mathcal E}_{\mathcal B} &:= \left\{ {\bf x} \in {\Bbb R}^n : {\bf x}^\top {\bf B} \, {\bf x} \leq 1 \right\} \end{aligned} $$

It is known that

$$ {\mathcal E}_{\mathcal A} \subseteq {\mathcal E}_{\mathcal B} \iff {\bf A} \succeq {\bf B} \tag{1} $$

and that the square root preserves the Löwner (partial) order, i.e.,

$$ {\bf A} \succeq {\bf B} \implies {\bf A}^{\frac12} \succeq {\bf B}^{\frac12} \tag{2} $$

Hence, from $(1)$ and $(2)$,

$$ {\mathcal E}_{\mathcal A} \subseteq {\mathcal E}_{\mathcal B} \implies {\bf A}^{\frac12} \succeq {\bf B}^{\frac12} \tag{3} $$

Taking the trace,

$$ {\bf A}^{\frac12} \succeq {\bf B}^{\frac12} \implies \operatorname{tr} \left( {\bf A}^{\frac12} \right) \geq \operatorname{tr} \left( {\bf B}^{\frac12} \right) \tag{4} $$

Thus, from $(3)$ and $(4)$,

$$ {\mathcal E}_{\mathcal A} \subseteq {\mathcal E}_{\mathcal B} \implies \operatorname{tr} \left( {\bf A}^{\frac12} \right) \geq \operatorname{tr} \left( {\bf B}^{\frac12} \right) \tag{5} $$

Let $\alpha_i > 0$ and $\beta_i > 0$ be the lengths of the $i$-th semi-axes of ${\mathcal E}_{\mathcal A}$ and ${\mathcal E}_{\mathcal B}$, respectively.

$$ \alpha_i := \frac{1}{\sqrt{\lambda_i ({\bf A})}}, \qquad \beta_i := \frac{1}{\sqrt{\lambda_i ({\bf B})}} $$

Note that $\alpha_i, \beta_i > 0$ stem from the positive definiteness of matrices $\bf A$ and $\bf B$. Hence,

$$ \begin{align} \operatorname{tr} \left( {\bf A}^{\frac12} \right) &= \sum\limits_{i=1}^n \sqrt{\lambda_i ({\bf A})} = \sum\limits_{i=1}^n \frac{1}{\alpha_i} \tag{6} \\ \operatorname{tr} \left( {\bf B}^{\frac12} \right) &= \sum\limits_{i=1}^n \sqrt{\lambda_i ({\bf B})} = \sum\limits_{i=1}^n \frac{1}{\beta_i} \tag{7} \end{align} $$

Thus, from $(5)$, $(6)$ and $(7)$,

$$ \boxed{ {\mathcal E}_{\mathcal A} \subseteq {\mathcal E}_{\mathcal B} \implies \sum\limits_{i=1}^n \frac{1}{\alpha_i} \geq \sum\limits_{i=1}^n \frac{1}{\beta_i} } $$

---

Reference

• Stephen Boyd, Sanjay Lall, Ellipsoids [PDF], Stanford EE263.

---

Related

• Smallest axis-aligned bounding box of hyper-ellipsoid

• Does the square root preserve the Löwner order?

---

ellipsoids