Canonical form of symmetric integer matrix M - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T03:48:22Z http://mathoverflow.net/feeds/question/97448 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97448/canonical-form-of-symmetric-integer-matrix-m Canonical form of symmetric integer matrix M Xiao-Gang Wen 2012-05-20T01:09:45Z 2012-05-20T22:29:42Z <p>Let $M$, $N$ be a symmetric matrix over a ring $R$. $M$ and $N$ are said to be equivalent if there exist an invertible matrix $U$ (over the same ring $R$) such that $N=U M U^T$ ($U^T$ is the transpose of $U$). A question is that what is the simple canonical form of $M$ under such an equivalent relation.</p> <p>We know that when $R$ is the ring of real numbers, every real symmetric matrix is equivalent to an diagonal matrix with diagonal entries being 1, -1, or 0.</p> <p>When $R$ is the ring of integers, do we have a similar result?</p> <p>If there is no nice results, we may assume $M$ to satisfy additional conditions:</p> <p>(a) $|\det(M)|=1$</p> <p>(b) There exist a $J$ such that $J^2=1$ and $JMJ^T=-M$.</p> <p>Thanks!</p> <p>Edit: I am also interested in finding the simple canonical form of integer symmetric matrices $M$, that satisfy</p> <p>(a) $|\det(M)|=1$</p> <p>(b) There exist a $J$ such that $J^2=-1$ and $JMJ^T=-M$.</p> http://mathoverflow.net/questions/97448/canonical-form-of-symmetric-integer-matrix-m/97451#97451 Answer by Will Jagy for Canonical form of symmetric integer matrix M Will Jagy 2012-05-20T02:06:15Z 2012-05-20T02:06:15Z <p>I don't see how to diagonalize the quadratic form $2xy.$ As far as your conditions, we have $$M = \left( \begin{array}{cc} 0 &amp; 1 \\ 1 &amp; 0<br> \end{array} \right)$$ and $$J = \left( \begin{array}{cc} 1 &amp; 0 \\ 0 &amp; -1<br> \end{array} \right),$$ with $$\left( \begin{array}{cc} 1 &amp; 0 \\ 0 &amp; -1<br> \end{array} \right) \left( \begin{array}{cc} 0 &amp; 1 \\ 1 &amp; 0<br> \end{array} \right) \left( \begin{array}{cc} 1 &amp; 0 \\ 0 &amp; -1<br> \end{array} \right) = \left( \begin{array}{cc} 0 &amp; -1 \\ -1 &amp; 0<br> \end{array} \right)$$ However, the only diagonal matrices with determinant $-1$ are $\pm J,$ which does not work as $x^2 - y^2$ does not represent any numbers congruent to $2 \pmod 4.$ </p> <p>Let's see, Conway and Sloane refer to Watson for his 2-adic canonical form for their work on the Mass Formula, so I can recommend the book Integral Quadratic Forms by George Leo Watson. More recent, SPLAG, which is Sphere Packings, Lattices, and Groups by Conway and Sloane. </p> http://mathoverflow.net/questions/97448/canonical-form-of-symmetric-integer-matrix-m/97457#97457 Answer by Igor Rivin for Canonical form of symmetric integer matrix M Igor Rivin 2012-05-20T03:37:23Z 2012-05-20T03:37:23Z <p>For symplectic unimodular symmetric (or skew) matrices, such a result is <a href="http://dl.dropbox.com/u/5188175/zarrow.pdf" rel="nofollow">shown in</a> Zarrow, Robert A canonical form for symmetric and skew-symmetric extended symplectic modular matrices with applications to Riemann surface theory. Trans. Amer. Math. Soc. 204 (1975), 207–227. </p> <p>You might be able to extend it to the nonsymplectic case (though I am a bit skeptical).</p> http://mathoverflow.net/questions/97448/canonical-form-of-symmetric-integer-matrix-m/97458#97458 Answer by Will Jagy for Canonical form of symmetric integer matrix M Will Jagy 2012-05-20T05:00:04Z 2012-05-20T22:29:42Z <p>It's all in the correct reference. Cassels, <em>Rational Quadratic Forms</em>, chapter 9 "Integral Forms over the Rational Integers," pages 163-164, Examples 9-11. Example 11(i) says that, for "odd" matrices, we can cut down the dimension by 2 and write $y_1^2 - y_2^2 + g(z_1, \ldots , z_{n-2}).$ The determinant of $g$ is still $\pm 1,$ so the only problem is that $g$ may be "even." </p> <p>Next, if $f$ is "even" the quadratic form can, in fact, be written $2y_1 y_2 + g(z_1, \ldots , z_{n-2}).$</p> <p>So, all we really need is to show, as in Sylvester's Law of Inertia, that the resulting form $g$ continues to be indefinite. Presumably your condition with $J M J^T = -M$ can do this. </p> <p>Otherwise, without your $J$ condition, Example 11(vi) says that either $f$ or $-f,$ if "even," is equivalent to a sum of some $2x_j y_j$ terms along with a single $\mathbb E_8$ lattice. <a href="http://store.doverpublications.com/0486466701.html" rel="nofollow">CASSELS</a> </p> <p>I was uneasy about the possible need to mix 2 by 2 blocks of both types, despite Hahn's statement, but $$<br> \left( \begin{array}{cccc} 3 &amp; 4 &amp; 2 &amp; 2 \\ 2 &amp; 3 &amp; 1 &amp; 2 \\ 0 &amp; 1 &amp; 1 &amp; 1 \\ 2 &amp; 3 &amp; 2 &amp; 1<br> \end{array} \right) \left( \begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; -1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 0 &amp; 1 &amp; 0<br> \end{array} \right) \left( \begin{array}{cccc} 3 &amp; 2 &amp; 0 &amp; 2 \\ 4 &amp; 3 &amp; 1 &amp; 3 \\ 2 &amp; 1 &amp; 1 &amp; 2 \\ 2 &amp; 2 &amp; 1 &amp; 1<br> \end{array} \right) = \left( \begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; -1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; -1<br> \end{array} \right)$$</p>