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Let $A$ be an $n \times n$ nondegenerate square matrix with real-valued entries. If we interpret the rows of $A$ as points in $\mathbb{R}^n$, then $A$ defines a simplex. We'll say $v \in \mathbb{R}^n$ is a normal vector to $A$ if $v$ is the normal vector to the hyperplane on which this simplex lies.

I am looking for a function $f_A: \mathbb{R}^n \to \mathbb{R}^{n \times n}$ that maps a normal vector to a rotation of $A$ that has that normal vector. It's easy to see that many rotations of $A$ might correspond to a single normal vector; thus, many implementations of $f_A$ might be possible. Any of them will do, as long as $f_A$ is continuous.

Thanks!

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    I have a feeling this is impossible in $\mathbb R^3$ due to the [hairy ball theorem](https://en.wikipedia.org/wiki/Hairy_ball_theorem), but I'm not sure yet.2012-11-16

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If I've understood the question correctly, this is impossible for odd $n$.

Any rotation of $A$ transforms its unit normal vector $v$ by the same rotation, so what you're really looking for is a continuous function that, for any vector $u$, gives a rotation matrix $R(u)$ which maps $v$ to $u$.

Let $w$ be a fixed unit vector orthogonal to $v$. Then $R(u)w$ is orthogonal to $R(u)v = u$. Now $R(u)w$, seen as a function over the unit sphere, defines a vector field which is everywhere tangential to the sphere and nowhere zero. By the hairy ball theorem, such a vector field cannot be continuous when $n$ is odd, so $R(u)$ cannot be either.

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    Yes, you've understood the question correctly. Thanks - this is very clever, if disappointing.2012-11-17