Intersection of ellipsoid & hyperplane, inequality chain.

Question

If $S$ is an ellipsoid in $\mathbb{R}^n$ centered at the origin, then, in suitable orthonormal coordinates $x_1, \dots, x_n$ on $\mathbb{R}^n$, $S$ can be defined by the inequality$$\sum_{j = 1}^n a_j x_j^2 \le 1,$$where$$0 < a_1 \le a_2 \le \dots \le a_n$$are uniquely determined by $S$. Let $H$ denote a hyperplane in $\mathbb{R}^n$ passing through the origin; then $S \cap H$ is an ellipsoid in $H$, and $H$ is a Euclidean space of dimension $n-1$. Hence $S \cap H$ determines a sequence of coefficients$$0 < b_1 \le b_2 \le \dots \le b_{n-1}.$$What is the easiest way to see that$$a_1 \le b_1 \le a_2 \le b_2 \le \dots \le a_{n-1} \le b_{n-1} \le a_n?$$

Answer 1

Intuition

Think of intersection $\, S\cap H\,$ as orthogonal projection of ellipse $\,S\,$ on hyperplane $\,H.\,$

• Projection operator is injective, which means that diameter of projection is always less than or equal to diameter of original set. Therefore we conclude that for any coefficient in inequality for projection $\, S\cap H\, $ there is a larger coefficient in inequality for the original ellipsoid $\, S.\, $
• On the other hand, projecting onto hyperplane reduces dimension of geometrical figure. Using this fact we can show that for any coefficient in inequality for $\,S\cap H\,$ there will be a certain number of lesser coefficients in inequality for $\,S.\,$
• Above facts combined with pigeonhole principle result in $\,a_1 \le b_1 \le \dots \le a_{n-1} \le b_{n-1} \le a_n.\,$

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Proof

Let us rewrite ellipsoid inequalities in terms of semi-principle axis $\, \alpha_i:=a_i^{-1/2}\,$ and $\, \beta_i : = b_i^{-1/2}\,$ $$ \left. \begin{aligned} S &= \left\lbrace x = \left[ \begin{smallmatrix}x_1 \\ \vdots \\ x_n \end{smallmatrix} \right] \in \mathbb R^n \, \bigg\vert \; \ \sum_{j = 1}^n a_j x_j^2 \le 1 \right\rbrace && = \left\lbrace x \in \mathbb R^n \, \bigg\vert \; \ \sum_{j = 1}^n \left( \frac{x_j}{\alpha_j} \right)^2 \le 1 \right\rbrace \\ S\cap H &=\left\lbrace x = \left[ \begin{smallmatrix}x_1 \\ \vdots \\ x_{n-1} \end{smallmatrix} \right] \in H \subseteq \mathbb R^{n-1} \, \bigg\vert \;\ \sum_{j = 1}^{n-1} b_j x_j^2 \le 1 \right\rbrace && = \left\lbrace x \in H^{\phantom{n}} \, \bigg\vert \; \ \sum_{j = 1}^{n-1}\left( \frac{x_j}{\beta_j} \right)^2 \le 1 \right\rbrace \end{aligned} \right. $$ Coefficients $\,\alpha_i\,$ equal to semi-principle axis $\,\vec{\boldsymbol{v}}_i\,$ of ellipse $\,S\,$ and can be defined iteratively:

• $\, E_1 : = \mathbb R^n \,$ Euclidean subspace,
• $\, S_1 := S\,$ original ellipse in $\,\mathbb R^n\,$
• $\displaystyle\,\vec{\boldsymbol{v}}_1: = \left\lbrace \overrightarrow{UV} \; \ \Big\vert \; \ \ \max_{U,V \in S} \left\|V - U \right\|\right\rbrace\,$ – vector connecting two most distant points of ellipse $\,S.\,$
• $\, \alpha_1 := \left\| \vec{\boldsymbol{v}}_1 \right\|\,$ diameter of ellipse $\,S\,$ $%\,\vec{\boldsymbol{e}}_{1}:=\dfrac{\vec{\boldsymbol{v}}_1}{\left\| \vec{\boldsymbol{v}}_1 \right\|} = \dfrac{\vec{\boldsymbol{v}}_1}{\alpha_1}\,$

For $\, j=2, \dots, n\,$ compute

• $\, E_j := E_{j-1} \ominus \left\lbrace \vec{\boldsymbol{v}}_{j-1} \right\rbrace\,$ orthogonal complement of span of $\,\vec{\boldsymbol{v}}_{j-1}\,$ in $\, E_{j-1} \,$
• $\, S_j:= S_{j-1}\cap E_{j}\,$ projection of ellipse $\,S_{j-1}\,$ onto Euclidean space $\, E_{j-1} = \mathbb R^{n-j+1}.\,$
• $\, \vec{\boldsymbol{v}}_j\,$ vector connecting two most distant points of ellipse $\,S_j.\,$
• $\, \alpha_j:= \left\| \vec{\boldsymbol{v}}_j \right\|\,$ diameter of $\,S_{j}\,$

By replacing ellipses $\, S_j\,$ with their projections to hyperplane $\,h\,$ we can define coefficients $\,\beta_i\,$ in inequality for $\,S\cap H.\,$

Upper Bound

Note that coefficient $\,\alpha_j\,$ can be viewed as the size of ellipsoid along direction of $\,j^{\,\text{th}}$ basis unit vector $\,\vec{\boldsymbol{e}}_{j},\,$ i.e. as a size of orthogonal projection of $\, S\,$ on $\vec{\boldsymbol{e}}_{j}$.

The intersection $\,S\cap H\,$ of ellipsoid and hyperplane is also an orthogonal projection of $\,S\,$ onto $\,H,\,$ i.e. $\,S\cap H = P_H S\,$ where $\,P_H\,$ is the operator projecting its argument onto hyperplane $\,H.\,$ Since orthogonal projection is injective, the size of intersection $\,S\cap H\,$ along $\,\vec{\boldsymbol{e}}_{j}\,$ is always less than or equal to the diameter of the original ellipsoid $\,S.\,$ Therefore for any $ j=1, \dots, n-1,\,$ there exist $\, i = 1 , \dots, n\,$ such that $\,\beta_j \le \alpha_i.\,$

Lower Bound

Let us show now that for any $\, \beta_j, \ j = 1, \dots, n-1,\,$ there exist $\, n-1\,$ coefficients $\,\alpha_j\,$ which are less than or equal to $\,\beta_j.\,$ Assume $\,\vec{\boldsymbol{v}}_1\,$ is the smallest semi-principle axis of $\,S\,$ of the size $\,\alpha_1.\,$ Consider following possible cases:

1. $\,\vec{\boldsymbol{v}}_1 \cap H = \vec{\boldsymbol{v}}_1 \implies \vec{\boldsymbol{v}}_1 \in H\,$ – the smallest semi-principle axis of $\,S\,$ lies entirely in hyperplane.
   Projection does not have affect the smallest semi-principle axis of ellipse, so that $\,\alpha_1 = \beta_1.\,$
2. $\,\vec{\boldsymbol{v}}_1 \cap H \neq \vec{\boldsymbol{v}}_1 \implies \vec{\boldsymbol{v}}_1 \not \in S\cap H,\,$ i.e. $\,\vec{\boldsymbol{v}}_1\,$ is not the smallest semi-principle axis of the projection ellipse $\,S\cap H.\,$ But then the new smallest semi-principle axis of $\,S\cap H.\,$ will be larger or equal to $\,\vec{\boldsymbol{v}}_1,\,$ so that $\, \alpha_1 \le \beta_1.\,$
3. Combining results from two previous steps we conclude that $\, \bbox[3pt, border:1pt solid #001100]{\alpha_1 \le \beta_1.}\,$

We can repeat procedure above iteratively to show that for any coefficient $\,\beta_j, \ j=1,\dots, n-1,\,$ of projected ellipse $\,S\cap H\,$ there exist at least $\, j-1\,$ coefficients $\,\alpha_1, \dots, \alpha_{j-1}\,$ of original ellipse $\,S\,$ less than or equal to $\,\alpha_j.\,$ $$ \forall j=1,\dots, n-1, \ \exists \,\alpha_1, \dots, \alpha_{n-j} \, : \quad \bbox[5pt, border:1pt solid #000000]{\beta_j \ge \alpha_{j-1} \ge \cdots \ge \alpha_1} $$

Using Dirichlet's principle we get $$ \bbox[3pt, border:2pt solid #F90000]{\alpha_1 \le \beta_1 \le \alpha_2 \le \beta_2 \le \dots \le \alpha_{n-1} \le \beta_{n-1} \le \alpha_n} $$

Substituting back expressions for coefficient $\, a_i = 1/ \alpha_i^2, \ b_i = 1/ \beta_i^2\,$ and reverting enumeration we get $$ \bbox[3pt, border:2pt solid #F90000]{a_1 \le b_1 \le a_2 \le b_2 \le \cdots \le a_{n-1} \le b_{n-1} \le a_n} $$

Q.E.D.