Integer matrices with integer inverses
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If all entries of an invertible matrix $A$ are rational, then all the entries of $A^{-1}$ are also rational. Now suppose that all entries of an invertible matrix $A$ are integers. Then it's not necessary that all the entries of $A^{-1}$ are integers. My question is:
What are all the invertible integer matrices such that their inverses are also integer?
linear-algebra matrices inverse unimodular-matrices
edited Jul 14 '17 at 7:11
Rodrigo de Azevedo
13k41959
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add a comment |
$begingroup$
If all entries of an invertible matrix $A$ are rational, then all the entries of $A^{-1}$ are also rational. Now suppose that all entries of an invertible matrix $A$ are integers. Then it's not necessary that all the entries of $A^{-1}$ are integers. My question is:
What are all the invertible integer matrices such that their inverses are also integer?
linear-algebra matrices inverse unimodular-matrices
edited Jul 14 '17 at 7:11
Rodrigo de Azevedo
13k41959
$endgroup$
2
$begingroup$
en.wikipedia.org/wiki/Unimodular_matrix
$endgroup$
– Martin Sleziak
Nov 23 '11 at 14:31
add a comment |
$begingroup$
If all entries of an invertible matrix $A$ are rational, then all the entries of $A^{-1}$ are also rational. Now suppose that all entries of an invertible matrix $A$ are integers. Then it's not necessary that all the entries of $A^{-1}$ are integers. My question is:
What are all the invertible integer matrices such that their inverses are also integer?
linear-algebra matrices inverse unimodular-matrices
edited Jul 14 '17 at 7:11
Rodrigo de Azevedo
13k41959
$endgroup$
If all entries of an invertible matrix $A$ are rational, then all the entries of $A^{-1}$ are also rational. Now suppose that all entries of an invertible matrix $A$ are integers. Then it's not necessary that all the entries of $A^{-1}$ are integers. My question is:
What are all the invertible integer matrices such that their inverses are also integer?
linear-algebra matrices inverse unimodular-matrices
linear-algebra matrices inverse unimodular-matrices
edited Jul 14 '17 at 7:11
Rodrigo de Azevedo
13k41959
edited Jul 14 '17 at 7:11
Rodrigo de Azevedo
13k41959
edited Jul 14 '17 at 7:11
Rodrigo de Azevedo
13k41959
edited Jul 14 '17 at 7:11
Rodrigo de Azevedo
13k41959
edited Jul 14 '17 at 7:11
Rodrigo de Azevedo
13k41959
13k41959
asked Jan 30 '11 at 5:22
anonymous
2
$begingroup$
en.wikipedia.org/wiki/Unimodular_matrix
$endgroup$
– Martin Sleziak
Nov 23 '11 at 14:31
add a comment |
2
$begingroup$
en.wikipedia.org/wiki/Unimodular_matrix
$endgroup$
– Martin Sleziak
Nov 23 '11 at 14:31
2
2
$begingroup$
en.wikipedia.org/wiki/Unimodular_matrix
$endgroup$
– Martin Sleziak
Nov 23 '11 at 14:31
$begingroup$
en.wikipedia.org/wiki/Unimodular_matrix
$endgroup$
– Martin Sleziak
Nov 23 '11 at 14:31
add a comment |
3 Answers
3
active
oldest
votes
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Exactly those whose determinant is $1$ or $-1$.
See the previous question about the $2times 2$ case. The determinant map gives necessity, the adjugate formula for the inverse gives sufficiency.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 30 '11 at 5:24
Arturo MagidinArturo Magidin
263k34587915
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$begingroup$
how? Please explain
$endgroup$
– anonymous
Jan 30 '11 at 5:27
3
$begingroup$
@Chandru1: See the link.
$endgroup$
– Arturo Magidin
Jan 30 '11 at 5:29
add a comment |
$begingroup$
The inverse of an integer matrix is again an integer matrix iff if the determinant of the matrix is $pm 1$. Integer matrices of determinant $pm 1$ form the General Linear Group $GL(n,mathbb{Z})$
$endgroup$
$begingroup$
Very interesting statement. Is it not possible to generate integer inverse for matrix with determinant for example $2$ ? Quite surprising but probably the statement is true.
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– Widawensen
Jul 14 '17 at 9:33
add a comment |
$begingroup$
Arturo and Sivaram have already given the general condition for integer matrices with integer inverses; here I only note this particular example due to Ericksen that the matrix $mathbf A$ with entries
$$a_{ij}=binom{n+j-1}{i-1}$$
where $n$ is an arbitrary nonnegative integer has an integer inverse.
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add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Exactly those whose determinant is $1$ or $-1$.
See the previous question about the $2times 2$ case. The determinant map gives necessity, the adjugate formula for the inverse gives sufficiency.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 30 '11 at 5:24
Arturo MagidinArturo Magidin
263k34587915
$endgroup$
$begingroup$
how? Please explain
$endgroup$
– anonymous
Jan 30 '11 at 5:27
3
$begingroup$
@Chandru1: See the link.
$endgroup$
– Arturo Magidin
Jan 30 '11 at 5:29
add a comment |
$begingroup$
Exactly those whose determinant is $1$ or $-1$.
See the previous question about the $2times 2$ case. The determinant map gives necessity, the adjugate formula for the inverse gives sufficiency.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 30 '11 at 5:24
Arturo MagidinArturo Magidin
263k34587915
$endgroup$
$begingroup$
how? Please explain
$endgroup$
– anonymous
Jan 30 '11 at 5:27
3
$begingroup$
@Chandru1: See the link.
$endgroup$
– Arturo Magidin
Jan 30 '11 at 5:29
add a comment |
$begingroup$
Exactly those whose determinant is $1$ or $-1$.
See the previous question about the $2times 2$ case. The determinant map gives necessity, the adjugate formula for the inverse gives sufficiency.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 30 '11 at 5:24
Arturo MagidinArturo Magidin
263k34587915
$endgroup$
Exactly those whose determinant is $1$ or $-1$.
See the previous question about the $2times 2$ case. The determinant map gives necessity, the adjugate formula for the inverse gives sufficiency.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 30 '11 at 5:24
Arturo MagidinArturo Magidin
263k34587915
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Jan 30 '11 at 5:24
Arturo MagidinArturo Magidin
263k34587915
answered Jan 30 '11 at 5:24
Arturo MagidinArturo Magidin
263k34587915
answered Jan 30 '11 at 5:24
Arturo MagidinArturo Magidin
263k34587915
263k34587915
$begingroup$
how? Please explain
$endgroup$
– anonymous
Jan 30 '11 at 5:27
3
$begingroup$
@Chandru1: See the link.
$endgroup$
– Arturo Magidin
Jan 30 '11 at 5:29
add a comment |
$begingroup$
how? Please explain
$endgroup$
– anonymous
Jan 30 '11 at 5:27
3
$begingroup$
@Chandru1: See the link.
$endgroup$
– Arturo Magidin
Jan 30 '11 at 5:29
$begingroup$
how? Please explain
$endgroup$
– anonymous
Jan 30 '11 at 5:27
$begingroup$
how? Please explain
$endgroup$
– anonymous
Jan 30 '11 at 5:27
3
3
$begingroup$
@Chandru1: See the link.
$endgroup$
– Arturo Magidin
Jan 30 '11 at 5:29
$begingroup$
@Chandru1: See the link.
$endgroup$
– Arturo Magidin
Jan 30 '11 at 5:29
add a comment |
$begingroup$
The inverse of an integer matrix is again an integer matrix iff if the determinant of the matrix is $pm 1$. Integer matrices of determinant $pm 1$ form the General Linear Group $GL(n,mathbb{Z})$
$endgroup$
$begingroup$
Very interesting statement. Is it not possible to generate integer inverse for matrix with determinant for example $2$ ? Quite surprising but probably the statement is true.
$endgroup$
– Widawensen
Jul 14 '17 at 9:33
add a comment |
$begingroup$
The inverse of an integer matrix is again an integer matrix iff if the determinant of the matrix is $pm 1$. Integer matrices of determinant $pm 1$ form the General Linear Group $GL(n,mathbb{Z})$
$endgroup$
$begingroup$
Very interesting statement. Is it not possible to generate integer inverse for matrix with determinant for example $2$ ? Quite surprising but probably the statement is true.
$endgroup$
– Widawensen
Jul 14 '17 at 9:33
add a comment |
$begingroup$
The inverse of an integer matrix is again an integer matrix iff if the determinant of the matrix is $pm 1$. Integer matrices of determinant $pm 1$ form the General Linear Group $GL(n,mathbb{Z})$
$endgroup$
The inverse of an integer matrix is again an integer matrix iff if the determinant of the matrix is $pm 1$. Integer matrices of determinant $pm 1$ form the General Linear Group $GL(n,mathbb{Z})$
answered Jan 30 '11 at 5:27
user17762
$begingroup$
Very interesting statement. Is it not possible to generate integer inverse for matrix with determinant for example $2$ ? Quite surprising but probably the statement is true.
$endgroup$
– Widawensen
Jul 14 '17 at 9:33
add a comment |
$begingroup$
Very interesting statement. Is it not possible to generate integer inverse for matrix with determinant for example $2$ ? Quite surprising but probably the statement is true.
$endgroup$
– Widawensen
Jul 14 '17 at 9:33
$begingroup$
Very interesting statement. Is it not possible to generate integer inverse for matrix with determinant for example $2$ ? Quite surprising but probably the statement is true.
$endgroup$
– Widawensen
Jul 14 '17 at 9:33
$begingroup$
Very interesting statement. Is it not possible to generate integer inverse for matrix with determinant for example $2$ ? Quite surprising but probably the statement is true.
$endgroup$
– Widawensen
Jul 14 '17 at 9:33
add a comment |
$begingroup$
Arturo and Sivaram have already given the general condition for integer matrices with integer inverses; here I only note this particular example due to Ericksen that the matrix $mathbf A$ with entries
$$a_{ij}=binom{n+j-1}{i-1}$$
where $n$ is an arbitrary nonnegative integer has an integer inverse.
$endgroup$
add a comment |
$begingroup$
Arturo and Sivaram have already given the general condition for integer matrices with integer inverses; here I only note this particular example due to Ericksen that the matrix $mathbf A$ with entries
$$a_{ij}=binom{n+j-1}{i-1}$$
where $n$ is an arbitrary nonnegative integer has an integer inverse.
$endgroup$
add a comment |
$begingroup$
Arturo and Sivaram have already given the general condition for integer matrices with integer inverses; here I only note this particular example due to Ericksen that the matrix $mathbf A$ with entries
$$a_{ij}=binom{n+j-1}{i-1}$$
where $n$ is an arbitrary nonnegative integer has an integer inverse.
$endgroup$
Arturo and Sivaram have already given the general condition for integer matrices with integer inverses; here I only note this particular example due to Ericksen that the matrix $mathbf A$ with entries
$$a_{ij}=binom{n+j-1}{i-1}$$
where $n$ is an arbitrary nonnegative integer has an integer inverse.
answered Jan 31 '11 at 5:29
user6429
add a comment |
add a comment |
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2
$begingroup$
en.wikipedia.org/wiki/Unimodular_matrix
$endgroup$
– Martin Sleziak
Nov 23 '11 at 14:31