Existence of Rational Orthogonal Matrices

Question

Question:

Let $A\in\mathbb{R}^{n \times n}$ be an orthogonal matrix and let $\varepsilon>0$. Then does there exist a rational orthogonal matrix $B\in\mathbb{R}^{n \times n}$ such that $\|A-B\|<\varepsilon$?

Definitions:

• A matrix $A\in\mathbb{R}^{n \times n}$ is an orthogonal matrix if $A^T=A^{-1}$
• A matrix $A\in\mathbb{R}^{n \times n}$ is a rational matrix if every entry of it is rational.

Answer 1

Yes. It is a theorem of Cayley that the mapping $S \rightarrow (S-I)^{-1}(S+1)$ gives a correspondence between the set of $n\times n$ skew-symmetric matrices over $\mathbb{Q}$ and the set of $n\times n$ orthogonal matrices which do not have one as an eigenvalue. Since the mapping is nice, and rational skew-symmetric matrices are dense in the set of all skew-symmetric matrices, you have your result. For more, see the very nice paper by Liebeck and Osborne

Answer 2

Sure. Consider matrices which fix $n-2$ of the standard basis vectors and describe a rotation in the plane spanned by the last two about an angle $\theta$ such that $\sin \theta, \cos \theta$ are both rational; these are dense in all such rotations, and all such rotations generate the orthogonal group, so the corresponding products (all of which are rational) are dense in the orthogonal group.

Answer 3

I should say yes. For this, I shall use the fact that in the unit sphere $\mathbb S^{d-1}$, the set of rational vectors is dense. I shall proceed by induction over $n$.

So let $A\in {\bf O}_n(\mathbb R)$ be given. Let $\vec v_1$ be its first column, an element of ${\mathbb S}^{n-1}$. We can choose a rational unit vector $\vec w_1$ arbitrarily close to $\vec v_1$. The first step is to construct a rational orthogonal matrix $B$ with first column $\vec w_1$. To this end we choose inductively rational unit vectors $\vec w_2,\ldots,\vec w_n$. This is possible because at each step, we may take a rational unit vector in the unit sphere of a "rational" subspace. Here, a subspace $F$ is rational if it admits a rational basis.

Now, let us form $A_1=B^{-1}A$. This is a orthogonal matrix, whose first column is arbitrarily close to $\vec e_1$. Hence its first line is close to $(1,0,\ldots,0)$ as well. Thus $$A_1\sim\begin{pmatrix} 1 & 0^T \\\\ 0 & R \end{pmatrix}.$$ The matrix $R$ is arbitrarily close to ${\bf O}_{n-1}({\mathbb R})$. By the induction hypothesis, there exists a rational orthogonal matrix $Q$ arbitraly close to $R$. Then $$B\begin{pmatrix} 1 & 0^T \\\\ 0 & Q \end{pmatrix}$$ is arbitrarily close to $A$.