Existence of Rational Orthogonal Matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:26:01Z http://mathoverflow.net/feeds/question/90070 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90070/existence-of-rational-orthogonal-matrices Existence of Rational Orthogonal Matrices Peter J.C. Dickinson 2012-03-02T20:24:36Z 2012-03-03T06:06:08Z <h2>Question:</h2> <p>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 $\|A-B\|&lt;\varepsilon$?</p> <h2>Definitions:</h2> <ul> <li>A matrix $A\in\mathbb{R}^{n \times n}$ is an <em>orthogonal matrix</em> if $A^T=A^{-1}$</li> <li>A matrix $A\in\mathbb{R}^{n \times n}$ is a <em>rational matrix</em> if every entry of it is rational.</li> </ul> http://mathoverflow.net/questions/90070/existence-of-rational-orthogonal-matrices/90071#90071 Answer by Qiaochu Yuan for Existence of Rational Orthogonal Matrices Qiaochu Yuan 2012-03-02T20:42:17Z 2012-03-02T20:42:17Z <p>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. </p> http://mathoverflow.net/questions/90070/existence-of-rational-orthogonal-matrices/90072#90072 Answer by Igor Rivin for Existence of Rational Orthogonal Matrices Igor Rivin 2012-03-02T20:43:55Z 2012-03-02T20:43:55Z <p>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 <a href="http://www.math.upenn.edu/~pemantle/Summer2007/Library/liebeck-osborne.pdf" rel="nofollow">the very nice paper by Liebeck and Osborne</a></p> http://mathoverflow.net/questions/90070/existence-of-rational-orthogonal-matrices/90073#90073 Answer by Denis Serre for Existence of Rational Orthogonal Matrices Denis Serre 2012-03-02T20:48:10Z 2012-03-02T20:48:10Z <p>I should say <strong>yes</strong>. 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 $n$.</p> <p>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.</p> <p>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 $$A_1\sim\begin{pmatrix} 1 &amp; 0^T \\ 0 &amp; R \end{pmatrix}.$$ 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 $$B\begin{pmatrix} 1 &amp; 0^T \\ 0 &amp; Q \end{pmatrix}$$ is arbitrarily close to $A$.</p>