3

Prove that $\mathrm{O}_n(\mathbb{Q})$ is a dense subset of $\mathrm{O}_n(\mathbb{R})$.

Recall that $\mathrm{O}_n(\Bbbk)$ is the set of $n \times n$ matrices with coefficients in the field $\Bbbk$ and such that the columns form an orthonormal system in $\Bbbk^n$ endowed with the usual euclidian structure.

I managed to prove that $\mathrm{GL}_n(\mathbb{Q})$ is dense in $\mathrm{GL}_n(\mathbb{R})$, but I couldn't go further. Any ideas ?

Tlön Uqbar Orbis Tertius
  • 3,294
  • I'm not sure it works, but I would try the following. Let $v_1$ and $v_2$ be two vectors in $\Bbb{Q}^n$ that are close to each other, and have norms approximately, say $1$ (bounded away from zero). Let $H_1, H_2$ be the hyperplanes through the origin with respective normals $v_1, v_2$. If $s_1$ and $s_2$ are the orthogonal reflections w.r.t. these hyperplanes, then their matrices (w.r.t. the natural basis) are rational. Hence the same hold for the matrix of $s=s_1\circ s_2^{-1}$. – Jyrki Lahtonen Jun 08 '15 at 09:17
  • (cont'd) But as $||v_1-v_2||$ is small, $s$ is a small rotation in the plane that is the orthogonal complement of $V_{12}:=H_1\cap H_2$. Because we have plenty of freedom in getting the same $V_{12}$, we can get a dense set of rotations of $V_{12}^\perp$. The Givens rotations are known to generate $SO_n(\Bbb{R})$, and the above process allows us to approximate any Givens rotation arbitrarily well by a rotation in $SO_n(\Bbb{Q})$, so this feels promising. Density in $O_n$ follows, if we can handle $SO_n$. – Jyrki Lahtonen Jun 08 '15 at 09:20
  • Givens rotations = orthogonal symmetries with respect to a plane ? – Tlön Uqbar Orbis Tertius Jun 08 '15 at 09:21
  • Givens rotation. – Jyrki Lahtonen Jun 08 '15 at 09:22

0 Answers0