# VMath

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Satellite page for Distance evaluation between geometric objects

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

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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 $$X^{\top}{\mathbf A}_1X=t^2X_0^{\top}{\mathbf A}_1X_0=1 \mbox{ and } X^{\top}{\mathbf A}_2X=t^2X_0^{\top}{\mathbf A}_2X_0=t^2X_0^{\top}{\mathbf A}_1X_0=1\, .$$

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

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

I failed to find the original sourse for the result. It is contained as a problem in the problem book

. Proskuryakov, I.V. Problems in Linear Algebra. Mir. Moscow. 1978 