Vrftsjtryk

Given $n$ and a range, for example $[-10,10]$, is there an efficient method to find
all $n times n$-matrices $A$ with integers in the given range, which are self-inverse (that means the equation $A=A^{-1}$ holds)?

Some necessary conditions for $A$:

• $det(A)=-1$ or $det(A)=1$


• $A$ has no eigenvalues other than $-1$ and $1$


• 
  

  The minimal polynomial of $A$ divides $x^2-1$

  
  
  

  With $A$, the matrices $-A$ , $A^T$ and $B^{-1}AB$ for any invertible matrix B
  are also self-inverse.

  
  
  

  So, is there a method to find the matrices systematically without checking
  all possible matrices, which would be infeasible for, lets say $n = 4$ and
  range $[-10,10]$?

  
  

This doesn't seem to have been studied much in the literature. The latest paper at MathSciNet was by Robert Hanson, titled "Self-Inverse Integer Matrices" (College Mathematics Journal, Vol 16, No 3 (Jun 1985), pp. 198-198). He proves that you can generate all self-inverse integer matrices by starting with a matrix of the form $left[begin{array}{c}I&A\ 0&-Iend{array}right]$ ($A$ is a rectangular matrix), and calculating $BAB^{-1}$, where $B$ ranges over all matrices you get from the identity matrix by doing the following row operations:

1. Swapping two rows


2. Adding a multiple of one row to another row


3. Multiplying all the entries in a row by -1

I'm not sure this helps much. Maybe you can limit the row operations once you know the matrix $A$.

I think the number is infinite...
because all possible rotation angles (a) are possible (analogy with spin in Quantum mechanics) for n=2 we can write in general the matrix [cos(a), sin(a)](line 1) [sin(a), -cos(a)] (line 2)

Zeno Toffano (France)