Diagonalization of a quadratic form in integers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:17:48Z http://mathoverflow.net/feeds/question/104339 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104339/diagonalization-of-a-quadratic-form-in-integers Diagonalization of a quadratic form in integers Anton 2012-08-09T10:30:24Z 2012-08-10T03:17:08Z <p>Hello,</p> <p>Recently I've been studying the problem of quadratic form diagonalization. Suppose that we have a form $F(x,y,z)$ with corresponding symmetric matrix $M$. This form is equivalent to another form $F'(x',y',z')$ which is in <a href="http://en.wikipedia.org/wiki/Smith_normal_form" rel="nofollow">Smith normal form</a>. The matrix that corresponds to $F'$ is $M'$ and it is diagonal. Note that both $F$ and $F'$ have integral coefficients. It is known that there exist two matrices $A$ and $B$ (which coefficients are integers as well) such that $M' = AMB$.</p> <p>Now, do I understand correctly that $F$ and $F'$ represent exactly the same set of integers? If so, am I right that there must exist a change of variables of the form</p> <p>$$x = a_{11}x' + a_{12}y' + a_{13} z'$$ $$y = a_{21}x' + a_{22}y' + a_{23} z'$$ $$z = a_{31}x' + a_{32}y' + a_{33} z'$$</p> <p>that transforms $F$ into $F'$? Finally, is it possible to deduce the matrix $(a_{ij})$ given $M'$, $M$, $A$ and $B$?</p> <p>P.S. Here's an example. I have the following matrix $M$ of determinant 1:</p> <p>$$5, 13, 1$$</p> <p>$$13, 34, 0$$</p> <p>$$1, 0, 35$$</p> <p>It corresponds to $M' = I$ where $I$ is an identity matrix. The Sage command M.smith_form() produced the following $A$:</p> <p>$$0, 0, 1$$ $$0, 1, 0$$</p> <p>$$1, 0, 0$$</p> <p>and $B$:</p> <p>$$-34, -455, 1190$$ $$13, 174, -455$$ $$1, 13, -34$$</p> <p>Now I need to find $(a_{ij})$.</p> http://mathoverflow.net/questions/104339/diagonalization-of-a-quadratic-form-in-integers/104382#104382 Answer by Will Jagy for Diagonalization of a quadratic form in integers Will Jagy 2012-08-10T02:38:41Z 2012-08-10T03:17:08Z <p>The actual correct manipulation is $$<br> \left( \begin{array}{rrr} 3 &amp; -1 &amp; 0 \\ -5 &amp; 2 &amp; 0 \\ -34 &amp; 13 &amp; 1<br> \end{array} \right) \cdot \left( \begin{array}{rrr} 5 &amp; 13 &amp; 1 \\ 13 &amp; 34 &amp; 0 \\ 1 &amp; 0 &amp; 35<br> \end{array} \right) \cdot \left( \begin{array}{rrr} 3 &amp; -5 &amp; -34 \\ -1 &amp; 2 &amp; 13 \\ 0 &amp; 0 &amp; 1<br> \end{array} \right) = \left( \begin{array}{rrr} 1 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1<br> \end{array} \right)$$</p> <p>Taking the sign for transpose as an apostrophe, we evaluate a quadratic form with symmetric matrix $A$ at a column vector $x$ as $$x' A x.$$ We change from one symmetric matrix to another by taking a matrix $P$ of determinant $1$ and finding $P' A P,$ as I do above. This is called an equivalence. Typically, for dimension $3$ or higher, most authors allow $\det P = \pm 1.$</p> <p>There is no guarantee that a for diagonalizes over the integers. There is also no guarantee of diagonalization over $\mathbb Q,$ as there may be the necessity for a few terms of type $xy$ or $x^2 + xy+y^2$ in $\mathbb Z_2,$ the $2$-adic integers. </p> <p>I'm not entirely sure what to recommend, but Rational Quadratic Forms by Cassels is an inexpensive Dover reprint.</p>