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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

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

According to Boyd & Lall$^\color{magenta}{\star}$

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

A partial proof is presented below. I am looking for alternative proofs.

Proof

$$ \begin{align} {\mathcal E}_{\mathcal A} \subseteq {\mathcal E}_{\mathcal B} &\equiv \left( \forall {\bf x} \in {\Bbb R}^n \right) \left( {\bf x} \in {\mathcal E}_{\mathcal A} \longrightarrow {\bf x} \in {\mathcal E}_{\mathcal B} \right) \tag{from the definition of $\subseteq$} \\ &\equiv \left( \forall {\bf x} \in {\Bbb R}^n \right) \left( {\bf x} \notin {\mathcal E}_{\mathcal A} \lor {\bf x} \in {\mathcal E}_{\mathcal B} \right) \tag{from $\phi \rightarrow \psi \equiv \neg \phi \lor \psi$} \\ &\equiv \left( \forall {\bf x} \in {\Bbb R}^n \right) \neg \left( {\bf x} \in {\mathcal E}_{\mathcal A} \land {\bf x} \notin {\mathcal E}_{\mathcal B} \right) \tag{from a de Morgan's law} \\ &\equiv \left( \forall {\bf x} \in {\Bbb R}^n \right) \neg \left( {\bf x}^\top {\bf A} \, {\bf x} \leq 1 \land {\bf x}^\top {\bf B} \, {\bf x} > 1 \right) \\ &\equiv \left( \forall {\bf x} \in {\Bbb R}^n \right) \neg \underbrace{ \left( {\bf x}^\top {\bf A} \, {\bf x} \leq 1 \land 1 < {\bf x}^\top {\bf B} \, {\bf x} \right) }_{\implies {\bf x}^\top {\bf A} \, {\bf x} < {\bf x}^\top {\bf B} \, {\bf x} \, \equiv \, \neg ({\bf x}^\top {\bf A} \, {\bf x} \geq {\bf x}^\top {\bf B} \, {\bf x})} \end{align} $$

Hence,

$$ {\mathcal E}_{\mathcal A} \subseteq {\mathcal E}_{\mathcal B} \implies \left( \forall {\bf x} \in {\Bbb R}^n \right) \left( {\bf x}^\top {\bf A} \, {\bf x} \geq {\bf x}^\top {\bf B} \, {\bf x} \right) \equiv \color{blue}{{\bf A} \succeq {\bf B}} $$

However, at the moment, it is not obvious to me how to prove that ${\mathcal E}_{\mathcal A} \subseteq {\mathcal E}_{\mathcal B} \impliedby {\bf A} \succeq {\bf B}$.

Reference

$\color{magenta}{\star}$ Stephen Boyd, Sanjay Lall, Ellipsoids [PDF], Stanford EE263.

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Rodrigo de Azevedo
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2 Answers2

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This is for the backward direction. Let $A\succeq B$, with $A$ and $B$ symmetric and positive definite. For any positive-definite $A$ and Hermitian $B$, there exists a non-singular $C$ such that $A=C^{\ast }IC$ and $B=C^{\ast }DC$ with $D$ diagonal [1]. Since $A$ and $B$ are real, we can take $C$ to be real. Because $A-B=C^{T}(I-D)C$ is positive semidefinite and congruence preserves inertia, $I-D$ has nonnegative eigenvalues and $0<d_{i}\leq 1$, $i=1,\ldots ,n$, where $d_{i}$ are the diagonal entries of $D$.

Make a change in coordinates to $u=Cx$ where $x$ is given in the definition of the ellipsoids. Then the ellipsoid $\varepsilon _{A}$ is $x^{T}Ax=u^{T}Iu=\sum_{i}u_{i}^{2}\leq 1$; all semi-axes have length $1$. The ellipsoid $\varepsilon _{B}$ is $x^{T}Bx=u^{T}Du=\sum_{i}d_{i}u_{i}^{2}=\sum_{i}\left( u_{i}/d_{i}^{-1/2}\right) ^{2}\leq 1\,$, which has semi-axes of length $d_{i}^{-1/2}$. Since $0<d_{i}\leq 1$, $d_{i}^{-1/2}\geq 1$, and $\varepsilon _{A}\subseteq \varepsilon _{B}$.

[1] R.A. Horn, C.R. Johnson, Matrix Analysis, Cambridge University Press, 1985, Corollary 7.6.5, p. 466.

dharr
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Suppose $\mathcal E_A\subseteq\mathcal E_B$. For any $x\ne0$, we have $u=\frac{x}{\sqrt{x^TAx}}\in\mathcal E_A\subseteq\mathcal E_B$. Therefore $u^TBu\le1$, i.e., $x^TBx\le x^TAx$. Hence $A\succeq B$.

Conversely, suppose $A\succeq B$. For any $x\in\mathcal E_A$, we have $x^TBx\le x^TAx\le1$, i.e., $x\in\mathcal E_B$. Hence $\mathcal E_A\subseteq\mathcal E_B$.

user1551
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