Finding $max |A|$ with $a_ij=pm 1$

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$A$ is a matrix sized $ntimes n$, all elements in $A$ are $pm1$. Find $max |A|$.
My Attempt
Denote $f(n)=max |A_ntimes n|$.
$f(1)=1$.
$f(2)=2$ is also obviously.
If $nge2$, $|A|$ must be even.
For $n=3$, $f(3)ge4$ because $left|beginarray r1&1&1\1&-1&1\1&-1&-1
endarrayright|=4$. Also, $|A|=A_11 A_22 A_33+A_12 A_23 A_31+A_13 A_21 A_32-A_13 A_22 A_31-A_11 A_23 A_32-A_12 A_21 A_33$ can not be $6$ since $A_11 A_22 A_33A_12 A_23 A_31A_13 A_21 A_32A_13 A_22 A_31A_11 A_23 A_32A_12 A_21 A_33$ must be $1$.
$1ne 1cdot1cdot1cdot(-1)cdot(-1)cdot(-1)$.
Hence we have $f(3)ne 6$. $f(3)=4$.
For $nge4$, I have no idea where to start with $f(n)$.
A trivial bound is $0<f(n)le n!$.
EDIT
Related question:Maximum value of Determinant of $3 times 3$ Matrix with entries $pm 1$
It is not duplicated since I am discussing $ntimes n$ determinants, not $3times3$.
linear-algebra determinant
 |Â
show 5 more comments
up vote
13
down vote
favorite
$A$ is a matrix sized $ntimes n$, all elements in $A$ are $pm1$. Find $max |A|$.
My Attempt
Denote $f(n)=max |A_ntimes n|$.
$f(1)=1$.
$f(2)=2$ is also obviously.
If $nge2$, $|A|$ must be even.
For $n=3$, $f(3)ge4$ because $left|beginarray r1&1&1\1&-1&1\1&-1&-1
endarrayright|=4$. Also, $|A|=A_11 A_22 A_33+A_12 A_23 A_31+A_13 A_21 A_32-A_13 A_22 A_31-A_11 A_23 A_32-A_12 A_21 A_33$ can not be $6$ since $A_11 A_22 A_33A_12 A_23 A_31A_13 A_21 A_32A_13 A_22 A_31A_11 A_23 A_32A_12 A_21 A_33$ must be $1$.
$1ne 1cdot1cdot1cdot(-1)cdot(-1)cdot(-1)$.
Hence we have $f(3)ne 6$. $f(3)=4$.
For $nge4$, I have no idea where to start with $f(n)$.
A trivial bound is $0<f(n)le n!$.
EDIT
Related question:Maximum value of Determinant of $3 times 3$ Matrix with entries $pm 1$
It is not duplicated since I am discussing $ntimes n$ determinants, not $3times3$.
linear-algebra determinant
4
$f(n) leq n^2$ because any eigen value $lambda$ is bounded by $n$ and $|A|$ is the product of the eigen values.
– Kavi Rama Murthy
Sep 10 at 8:36
1
Related: oeis A003433
– Kemono Chen
Sep 10 at 9:01
5
This problem is called "Hadamard maximal determinant problem".
– Kemono Chen
Sep 10 at 9:17
3
Your question seems to be an open problem... Anyway, consider adding the first column to each of the others. This does not change the determinant, but since $-1equiv 1 pmod 2$, each column $c_i$ can now be written as $2c_i'$ where $c_i'$ has integer entries. Thus $|A|=2^n-1|A'|$, where $|A'|$ is an integer.
– Gabriel Romon
Sep 10 at 9:32
2
There's also a Hadamard bound for the determinant of a matrix. Look it up.
– Gerry Myerson
Sep 10 at 10:21
 |Â
show 5 more comments
up vote
13
down vote
favorite
up vote
13
down vote
favorite
$A$ is a matrix sized $ntimes n$, all elements in $A$ are $pm1$. Find $max |A|$.
My Attempt
Denote $f(n)=max |A_ntimes n|$.
$f(1)=1$.
$f(2)=2$ is also obviously.
If $nge2$, $|A|$ must be even.
For $n=3$, $f(3)ge4$ because $left|beginarray r1&1&1\1&-1&1\1&-1&-1
endarrayright|=4$. Also, $|A|=A_11 A_22 A_33+A_12 A_23 A_31+A_13 A_21 A_32-A_13 A_22 A_31-A_11 A_23 A_32-A_12 A_21 A_33$ can not be $6$ since $A_11 A_22 A_33A_12 A_23 A_31A_13 A_21 A_32A_13 A_22 A_31A_11 A_23 A_32A_12 A_21 A_33$ must be $1$.
$1ne 1cdot1cdot1cdot(-1)cdot(-1)cdot(-1)$.
Hence we have $f(3)ne 6$. $f(3)=4$.
For $nge4$, I have no idea where to start with $f(n)$.
A trivial bound is $0<f(n)le n!$.
EDIT
Related question:Maximum value of Determinant of $3 times 3$ Matrix with entries $pm 1$
It is not duplicated since I am discussing $ntimes n$ determinants, not $3times3$.
linear-algebra determinant
$A$ is a matrix sized $ntimes n$, all elements in $A$ are $pm1$. Find $max |A|$.
My Attempt
Denote $f(n)=max |A_ntimes n|$.
$f(1)=1$.
$f(2)=2$ is also obviously.
If $nge2$, $|A|$ must be even.
For $n=3$, $f(3)ge4$ because $left|beginarray r1&1&1\1&-1&1\1&-1&-1
endarrayright|=4$. Also, $|A|=A_11 A_22 A_33+A_12 A_23 A_31+A_13 A_21 A_32-A_13 A_22 A_31-A_11 A_23 A_32-A_12 A_21 A_33$ can not be $6$ since $A_11 A_22 A_33A_12 A_23 A_31A_13 A_21 A_32A_13 A_22 A_31A_11 A_23 A_32A_12 A_21 A_33$ must be $1$.
$1ne 1cdot1cdot1cdot(-1)cdot(-1)cdot(-1)$.
Hence we have $f(3)ne 6$. $f(3)=4$.
For $nge4$, I have no idea where to start with $f(n)$.
A trivial bound is $0<f(n)le n!$.
EDIT
Related question:Maximum value of Determinant of $3 times 3$ Matrix with entries $pm 1$
It is not duplicated since I am discussing $ntimes n$ determinants, not $3times3$.
linear-algebra determinant
linear-algebra determinant
edited Sep 10 at 8:43
asked Sep 10 at 8:31


Kemono Chen
68916
68916
4
$f(n) leq n^2$ because any eigen value $lambda$ is bounded by $n$ and $|A|$ is the product of the eigen values.
– Kavi Rama Murthy
Sep 10 at 8:36
1
Related: oeis A003433
– Kemono Chen
Sep 10 at 9:01
5
This problem is called "Hadamard maximal determinant problem".
– Kemono Chen
Sep 10 at 9:17
3
Your question seems to be an open problem... Anyway, consider adding the first column to each of the others. This does not change the determinant, but since $-1equiv 1 pmod 2$, each column $c_i$ can now be written as $2c_i'$ where $c_i'$ has integer entries. Thus $|A|=2^n-1|A'|$, where $|A'|$ is an integer.
– Gabriel Romon
Sep 10 at 9:32
2
There's also a Hadamard bound for the determinant of a matrix. Look it up.
– Gerry Myerson
Sep 10 at 10:21
 |Â
show 5 more comments
4
$f(n) leq n^2$ because any eigen value $lambda$ is bounded by $n$ and $|A|$ is the product of the eigen values.
– Kavi Rama Murthy
Sep 10 at 8:36
1
Related: oeis A003433
– Kemono Chen
Sep 10 at 9:01
5
This problem is called "Hadamard maximal determinant problem".
– Kemono Chen
Sep 10 at 9:17
3
Your question seems to be an open problem... Anyway, consider adding the first column to each of the others. This does not change the determinant, but since $-1equiv 1 pmod 2$, each column $c_i$ can now be written as $2c_i'$ where $c_i'$ has integer entries. Thus $|A|=2^n-1|A'|$, where $|A'|$ is an integer.
– Gabriel Romon
Sep 10 at 9:32
2
There's also a Hadamard bound for the determinant of a matrix. Look it up.
– Gerry Myerson
Sep 10 at 10:21
4
4
$f(n) leq n^2$ because any eigen value $lambda$ is bounded by $n$ and $|A|$ is the product of the eigen values.
– Kavi Rama Murthy
Sep 10 at 8:36
$f(n) leq n^2$ because any eigen value $lambda$ is bounded by $n$ and $|A|$ is the product of the eigen values.
– Kavi Rama Murthy
Sep 10 at 8:36
1
1
Related: oeis A003433
– Kemono Chen
Sep 10 at 9:01
Related: oeis A003433
– Kemono Chen
Sep 10 at 9:01
5
5
This problem is called "Hadamard maximal determinant problem".
– Kemono Chen
Sep 10 at 9:17
This problem is called "Hadamard maximal determinant problem".
– Kemono Chen
Sep 10 at 9:17
3
3
Your question seems to be an open problem... Anyway, consider adding the first column to each of the others. This does not change the determinant, but since $-1equiv 1 pmod 2$, each column $c_i$ can now be written as $2c_i'$ where $c_i'$ has integer entries. Thus $|A|=2^n-1|A'|$, where $|A'|$ is an integer.
– Gabriel Romon
Sep 10 at 9:32
Your question seems to be an open problem... Anyway, consider adding the first column to each of the others. This does not change the determinant, but since $-1equiv 1 pmod 2$, each column $c_i$ can now be written as $2c_i'$ where $c_i'$ has integer entries. Thus $|A|=2^n-1|A'|$, where $|A'|$ is an integer.
– Gabriel Romon
Sep 10 at 9:32
2
2
There's also a Hadamard bound for the determinant of a matrix. Look it up.
– Gerry Myerson
Sep 10 at 10:21
There's also a Hadamard bound for the determinant of a matrix. Look it up.
– Gerry Myerson
Sep 10 at 10:21
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
From comments:
This is a semi-well-known open problem known as Hadamard's maximum determinant problem.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
From comments:
This is a semi-well-known open problem known as Hadamard's maximum determinant problem.
add a comment |Â
up vote
5
down vote
accepted
From comments:
This is a semi-well-known open problem known as Hadamard's maximum determinant problem.
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
From comments:
This is a semi-well-known open problem known as Hadamard's maximum determinant problem.
From comments:
This is a semi-well-known open problem known as Hadamard's maximum determinant problem.
answered Sep 16 at 11:16
community wiki
Henning Makholm
add a comment |Â
add a comment |Â
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4
$f(n) leq n^2$ because any eigen value $lambda$ is bounded by $n$ and $|A|$ is the product of the eigen values.
– Kavi Rama Murthy
Sep 10 at 8:36
1
Related: oeis A003433
– Kemono Chen
Sep 10 at 9:01
5
This problem is called "Hadamard maximal determinant problem".
– Kemono Chen
Sep 10 at 9:17
3
Your question seems to be an open problem... Anyway, consider adding the first column to each of the others. This does not change the determinant, but since $-1equiv 1 pmod 2$, each column $c_i$ can now be written as $2c_i'$ where $c_i'$ has integer entries. Thus $|A|=2^n-1|A'|$, where $|A'|$ is an integer.
– Gabriel Romon
Sep 10 at 9:32
2
There's also a Hadamard bound for the determinant of a matrix. Look it up.
– Gerry Myerson
Sep 10 at 10:21