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Is there an efficient method to find all the self-inverse matrices with integers in a given range?
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
edited Jun 9 '15 at 3:56
Ken
3,62151728
asked Jun 13 '14 at 10:39
PeterPeter
46.8k1039125
add a comment |
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
edited Jun 9 '15 at 3:56
Ken
3,62151728
asked Jun 13 '14 at 10:39
PeterPeter
46.8k1039125
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
add a comment |
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
edited Jun 9 '15 at 3:56
Ken
3,62151728
asked Jun 13 '14 at 10:39
PeterPeter
46.8k1039125
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
edited Jun 9 '15 at 3:56
Ken
3,62151728
asked Jun 13 '14 at 10:39
PeterPeter
46.8k1039125
edited Jun 9 '15 at 3:56
Ken
3,62151728
asked Jun 13 '14 at 10:39
PeterPeter
46.8k1039125
edited Jun 9 '15 at 3:56
Ken
3,62151728
edited Jun 9 '15 at 3:56
Ken
3,62151728
edited Jun 9 '15 at 3:56
Ken
3,62151728
3,62151728
asked Jun 13 '14 at 10:39
PeterPeter
46.8k1039125
asked Jun 13 '14 at 10:39
PeterPeter
46.8k1039125
asked Jun 13 '14 at 10:39
PeterPeter
46.8k1039125
46.8k1039125
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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:
- Swapping two rows
- Adding a multiple of one row to another row
- 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 $A$.
answered Aug 11 '15 at 0:04
Christopher Carl HeckmanChristopher Carl Heckman
3,907920
add a comment |
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)
answered Jan 6 '16 at 10:38
ZenoZeno
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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:
- Swapping two rows
- Adding a multiple of one row to another row
- 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 $A$.
answered Aug 11 '15 at 0:04
Christopher Carl HeckmanChristopher Carl Heckman
3,907920
add a comment |
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:
- Swapping two rows
- Adding a multiple of one row to another row
- 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 $A$.
answered Aug 11 '15 at 0:04
Christopher Carl HeckmanChristopher Carl Heckman
3,907920
add a comment |
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:
- Swapping two rows
- Adding a multiple of one row to another row
- 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 $A$.
answered Aug 11 '15 at 0:04
Christopher Carl HeckmanChristopher Carl Heckman
3,907920
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:
- Swapping two rows
- Adding a multiple of one row to another row
- 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 $A$.
answered Aug 11 '15 at 0:04
Christopher Carl HeckmanChristopher Carl Heckman
3,907920
answered Aug 11 '15 at 0:04
Christopher Carl HeckmanChristopher Carl Heckman
3,907920
answered Aug 11 '15 at 0:04
Christopher Carl HeckmanChristopher Carl Heckman
3,907920
answered Aug 11 '15 at 0:04
Christopher Carl HeckmanChristopher Carl Heckman
3,907920
3,907920
add a comment |
add a comment |
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)
answered Jan 6 '16 at 10:38
ZenoZeno
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
add a comment |
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)
answered Jan 6 '16 at 10:38
ZenoZeno
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
add a comment |
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)
answered Jan 6 '16 at 10:38
ZenoZeno
1
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)
answered Jan 6 '16 at 10:38
ZenoZeno
1
answered Jan 6 '16 at 10:38
ZenoZeno
1
answered Jan 6 '16 at 10:38
ZenoZeno
1
answered Jan 6 '16 at 10:38
ZenoZeno
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
add a comment |
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
add a comment |
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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