Author — Русская версия

Satellite page for ☞ Distance evaluation between geometric objects

We consider here the manifolds in $ \mathbb R^{n} $.

Theorem. Let the matrix $ {\mathbf A}_1 $ be positive definite. The quadrics $$ X^{\top}{\mathbf A}_1X=1 \quad and \quad X^{\top}{\mathbf A}_2X=1 $$ do not intersect iff the matrix $ {\mathbf A}_1 - {\mathbf A}_2 $ is sign-definite.

Proof. Let there exist a point $ X=X_0 \in \mathbb{R}^n $ such that $ X_0^{\top} {\mathbf A}_1X_0=1 $ and $ X_0^{\top} {\mathbf A}_2X_0=1 $. Thus $ X_0^{\top} ( {\mathbf A}_1 -{\mathbf A}_2) X_0=0 $ for $ X_0 \ne \mathbb{O} $. Therefore, $ {\mathbf A}_1 - {\mathbf A}_2 $ is not sign-definite.

Conversely, if $ {\mathbf A}_1 - {\mathbf A}_2 $ is not sign-definite then there exists $ X_0 \ne \mathbb{O} $ such that $ X_0^{\top} ( {\mathbf A}_1 -{\mathbf A}_2) X_0=0 $. Alternatively, $ X_0^{\top}{\mathbf A}_1X_0=X_0^{\top}{\mathbf A}_2X_0 $. Multiply the latter by a scalar $ t^2 $ with $ t \in \mathbb{R} $: $$ t^2X_0^{\top}{\mathbf A}_1X_0=t^2X_0^{\top}{\mathbf A}_2X_0 .$$ Set $$ t=1/\sqrt{X_0^{\top}{\mathbf A}_1X_0} $$ (the radicand is positive due to the positive definiteness of $ {\mathbf A}_1 $). The point $ X=tX_0 $ is an intersection point for both manifolds since $$X^{\top}{\mathbf A}_1X=t^2X_0^{\top}{\mathbf A}_1X_0=1 \mbox{ and } X^{\top}{\mathbf A}_2X=t^2X_0^{\top}{\mathbf A}_2X_0=t^2X_0^{\top}{\mathbf A}_1X_0=1\, .$$

If $ n_{} $ is even and $ \det (A_2-A_1)<0 $ then ellipsoid $ X^{\top} A_{1} X =1 $ intersects the quadric $ X^{\top} A_{2} X =1 $.

If the ellipsoid $ X^{\top} A_{1} X =1 $ touches the quadric $ X^{\top} A_{2} X =1 $ at least in one point then $ \det (A_2-A_1) = 0 $.

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