Is there an efficient method to find all the self-inverse matrices with integers in a given range?











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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 $[-10,10]$?








linear-algebra matrices inverse







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  • See also my related question, if the only eigenvalues of a self-inverse matrix are -1 and 1.
    – Peter
    Jun 13 '14 at 10:43












  • Without loss of generality, we can assume $a_{11}ge0$ and $a_{12}ge a_{21}$ to reduce the number of matrices.
    – Peter
    Jun 13 '14 at 10:53

















up vote
4
down vote

favorite
2












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 $[-10,10]$?








linear-algebra matrices inverse







share|cite|improve this question
























  • See also my related question, if the only eigenvalues of a self-inverse matrix are -1 and 1.
    – Peter
    Jun 13 '14 at 10:43












  • Without loss of generality, we can assume $a_{11}ge0$ and $a_{12}ge a_{21}$ to reduce the number of matrices.
    – Peter
    Jun 13 '14 at 10:53















up vote
4
down vote

favorite
2









up vote
4
down vote

favorite
2






2





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 $[-10,10]$?








linear-algebra matrices inverse







share|cite|improve this question















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 $[-10,10]$?








linear-algebra matrices inverse




linear-algebra matrices inverse






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edited Jun 9 '15 at 3:56









Ken

3,60151728




3,60151728










asked Jun 13 '14 at 10:39









Peter

46.3k1039125




46.3k1039125












  • See also my related question, if the only eigenvalues of a self-inverse matrix are -1 and 1.
    – Peter
    Jun 13 '14 at 10:43












  • Without loss of generality, we can assume $a_{11}ge0$ and $a_{12}ge a_{21}$ to reduce the number of matrices.
    – Peter
    Jun 13 '14 at 10:53




















  • See also my related question, if the only eigenvalues of a self-inverse matrix are -1 and 1.
    – Peter
    Jun 13 '14 at 10:43












  • Without loss of generality, we can assume $a_{11}ge0$ and $a_{12}ge a_{21}$ to reduce the number of matrices.
    – Peter
    Jun 13 '14 at 10:53


















See also my related question, if the only eigenvalues of a self-inverse matrix are -1 and 1.
– Peter
Jun 13 '14 at 10:43






See also my related question, if the only eigenvalues of a self-inverse matrix are -1 and 1.
– Peter
Jun 13 '14 at 10:43














Without loss of generality, we can assume $a_{11}ge0$ and $a_{12}ge a_{21}$ to reduce the number of matrices.
– Peter
Jun 13 '14 at 10:53






Without loss of generality, we can assume $a_{11}ge0$ and $a_{12}ge a_{21}$ to reduce the number of matrices.
– Peter
Jun 13 '14 at 10:53












2 Answers
2






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oldest

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up vote
2
down vote



accepted










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 $A$.






share|cite|improve this answer




























    up vote
    -3
    down vote













    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)






    share|cite|improve this answer





















    • the matrix you propose do not necessarily have integer coefficients (except for a finite number of $a$)
      – Surb
      Jan 6 '16 at 10:58






    • 1




      The number cannot be infinite because I have a finite range (with DISCRETE entries) and a finite matrix.
      – Peter
      Jan 6 '16 at 20:18













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

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    active

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    active

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    up vote
    2
    down vote



    accepted










    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 $A$.






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      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 $A$.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        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 $A$.






        share|cite|improve this answer












        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 $A$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 11 '15 at 0:04









        Christopher Carl Heckman

        3,917920




        3,917920






















            up vote
            -3
            down vote













            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)






            share|cite|improve this answer





















            • the matrix you propose do not necessarily have integer coefficients (except for a finite number of $a$)
              – Surb
              Jan 6 '16 at 10:58






            • 1




              The number cannot be infinite because I have a finite range (with DISCRETE entries) and a finite matrix.
              – Peter
              Jan 6 '16 at 20:18

















            up vote
            -3
            down vote













            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)






            share|cite|improve this answer





















            • the matrix you propose do not necessarily have integer coefficients (except for a finite number of $a$)
              – Surb
              Jan 6 '16 at 10:58






            • 1




              The number cannot be infinite because I have a finite range (with DISCRETE entries) and a finite matrix.
              – Peter
              Jan 6 '16 at 20:18















            up vote
            -3
            down vote










            up vote
            -3
            down vote









            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)






            share|cite|improve this answer












            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)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 '16 at 10:38









            Zeno

            1




            1












            • the matrix you propose do not necessarily have integer coefficients (except for a finite number of $a$)
              – Surb
              Jan 6 '16 at 10:58






            • 1




              The number cannot be infinite because I have a finite range (with DISCRETE entries) and a finite matrix.
              – Peter
              Jan 6 '16 at 20:18




















            • the matrix you propose do not necessarily have integer coefficients (except for a finite number of $a$)
              – Surb
              Jan 6 '16 at 10:58






            • 1




              The number cannot be infinite because I have a finite range (with DISCRETE entries) and a finite matrix.
              – Peter
              Jan 6 '16 at 20:18


















            the matrix you propose do not necessarily have integer coefficients (except for a finite number of $a$)
            – Surb
            Jan 6 '16 at 10:58




            the matrix you propose do not necessarily have integer coefficients (except for a finite number of $a$)
            – Surb
            Jan 6 '16 at 10:58




            1




            1




            The number cannot be infinite because I have a finite range (with DISCRETE entries) and a finite matrix.
            – Peter
            Jan 6 '16 at 20:18






            The number cannot be infinite because I have a finite range (with DISCRETE entries) and a finite matrix.
            – Peter
            Jan 6 '16 at 20:18




















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