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Given two symmetric integer unimodular matrices $A$ and $B$ with $\det A = \det B = \pm 1$. How do we find any integer unimodular matrices $T$ such that $$ T^\top A T = B? $$

Here $T^\top$, denotes the transpose of $T$. As an example, here are the data from the Mathematica Table:

A = ({{2, -1, 0, 0, 0, 0, 0, 0, 0, 0},
      {-1, 2, -1, 0, 0, 0, 0, 0, 0, 0},
      {0, -1, 2, -1, 0, 0, 0, -1, 0, 0},
      {0, 0, -1, 2, -1, 0, 0, 0, 0, 0},
      {0, 0, 0, -1, 2, -1, 0, 0, 0, 0},
      {0, 0, 0, 0, -1, 2, -1, 0, 0, 0},
      {0, 0, 0, 0, 0, -1, 2, 0, 0, 0},
      {0, 0, -1, 0, 0, 0, 0, 2, 0, 0},
      {0, 0, 0, 0, 0, 0, 0, 0, 1, 0},
      {0, 0, 0, 0, 0, 0, 0, 0, 0, -1}    });

B = ({{1, 0, 0, 0, 0, 0, 0, 0, 0, 0},
      {0, 1, 0, 0, 0, 0, 0, 0, 0, 0},
      {0, 0, 1, 0, 0, 0, 0, 0, 0, 0},
      {0, 0, 0, 1, 0, 0, 0, 0, 0, 0},
      {0, 0, 0, 0, 1, 0, 0, 0, 0, 0},
      {0, 0, 0, 0, 0, 1, 0, 0, 0, 0},
      {0, 0, 0, 0, 0, 0, 1, 0, 0, 0},
      {0, 0, 0, 0, 0, 0, 0, 1, 0, 0},
      {0, 0, 0, 0, 0, 0, 0, 0, 1, 0},
      {0, 0, 0, 0, 0, 0, 0, 0, 0, -1}    });

In matrix form,

$$A= \left( \begin{array}{cccccccccc} 2 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 2 & -1 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 2 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 2 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 2 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 2 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ \end{array} \right) $$ $$ B=\left( \begin{array}{cccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ \end{array} \right) $$ such that $\det A=\det B=-1$ here. Can you solve integer matrix $T$ with $T^\top A T = B$?

Rodrigo de Azevedo
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zeta
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1 Answers1

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For positive definite integer matrices, one can compute an integer transformation matrix using Magma's IsIsometric(L, LL) command, where L and LL are two lattices. Relevant pieces of documentation are here and (about the Automorphism Group and Isometry Testing) here.

Matrices in your example are not positive definite. I will briefly explain why the determination of the transformation matrix for them is a very difficult problem. But for the sake of example consider first the isometrism of the positive definite Gram matrices $A^2$ and $B^2=I$. $$ \left( \begin{array}{cccccccccc} 5 & -4 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -4 & 6 & -4 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & -4 & 7 & -4 & 1 & 0 & 0 & -4 & 0 & 0 \\ 0 & 1 & -4 & 6 & -4 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -4 & 6 & -4 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & -4 & 6 & -4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -4 & 5 & 0 & 0 & 0 \\ 0 & 1 & -4 & 1 & 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) $$

It can be found in the online Magma calculator as follows:

L := LatticeWithGram(Matrix(Integers(), 10, 10, [5, -4, 1, 0, 0, 0, 0, 0, 0, 0, -4, 6, -4, 1, 0, 0, 0, 1, 0, 0, 1, -4, 7, -4, 1, 0, 0, -4, 0, 0, 0, 1, -4, 6, -4, 1, 0, 1, 0, 0, 0, 0, 1, -4, 6, -4, 1, 0, 0, 0, 0, 0, 0, 1, -4, 6, -4, 0, 0, 0, 0, 0, 0, 0, 1, -4, 5, 0, 0, 0, 0, 1, -4, 1, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]));
LL := LatticeWithGram(Matrix(Integers(), 10, 10, [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]));
IsIsometric(L, LL);

$$\left( \begin{array}{cccccccccc} 2 & 4 & 6 & 5 & 4 & 3 & 2 & 3 & 0 & 0 \\ 4 & 7 & 10 & 8 & 6 & 4 & 2 & 5 & 0 & 0 \\ 4 & 8 & 12 & 10 & 8 & 6 & 3 & 6 & 0 & 0 \\ 5 & 10 & 15 & 12 & 9 & 6 & 3 & 8 & 0 & 0 \\ 6 & 12 & 18 & 15 & 12 & 8 & 4 & 9 & 0 & 0 \\ 7 & 14 & 20 & 16 & 12 & 8 & 4 & 10 & 0 & 0 \\ 8 & 16 & 24 & 20 & 15 & 10 & 5 & 12 & 0 & 0 \\ 10 & 20 & 30 & 24 & 18 & 12 & 6 & 15 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right)$$ We can verify that

$$ T A^2 T^T=I.$$

Why this does not work if matrices are not positive definite? Let us come back to your example. For the sake of making a stronger point, invert the original equation:

$$T^T BT=A^{-1}.$$

Here I use that $B^{-1}=B$. The inverse of $A$ is

$$ \left( \begin{array}{cccccccccc} 4 & 7 & 10 & 8 & 6 & 4 & 2 & 5 & 0 & 0 \\ 7 & 14 & 20 & 16 & 12 & 8 & 4 & 10 & 0 & 0 \\ 10 & 20 & 30 & 24 & 18 & 12 & 6 & 15 & 0 & 0 \\ 8 & 16 & 24 & 20 & 15 & 10 & 5 & 12 & 0 & 0 \\ 6 & 12 & 18 & 15 & 12 & 8 & 4 & 9 & 0 & 0 \\ 4 & 8 & 12 & 10 & 8 & 6 & 3 & 6 & 0 & 0 \\ 2 & 4 & 6 & 5 & 4 & 3 & 2 & 3 & 0 & 0 \\ 5 & 10 & 15 & 12 & 9 & 6 & 3 & 8 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ \end{array} \right)$$

Consider the diagonal elements, i.e., $d=\{4,14,30,20,12,6,2,8,1,-1\}$. They should be obtained as values of the quadratic form

$$ x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2+x_7^2+x_8^2+x_9^2-x_{10}^2=d_k. $$

Last minus sign is the problem--it prevents an efficient searching procedure. For the all positive case (positive definiteness), $x_i\le\sqrt{d_k}$ and therefore exhaustive search can be performed by brute force, or by more efficient methods such as implemented in Magma or described in the paper Congruence of rational matrices defined by an integer matrix (doi). If matrices are not positive semidefinite, the search space is unlimited: $x_i$ can be a very big number.

yarchik
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