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 $3times3$.










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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














up vote
13
down vote

favorite
1













$A$ 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 $3times3$.










share|cite|improve this question



















  • 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












up vote
13
down vote

favorite
1









up vote
13
down vote

favorite
1






1






$A$ 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 $3times3$.










share|cite|improve this question
















$A$ 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






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edited Sep 10 at 8:43

























asked Sep 10 at 8:31









Kemono Chen

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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












  • 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










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This is a semi-well-known open problem known as Hadamard's maximum determinant problem.






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    active

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    up vote
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    accepted










    From comments:



    This is a semi-well-known open problem known as Hadamard's maximum determinant problem.






    share|cite|improve this answer


























      up vote
      5
      down vote



      accepted










      From comments:



      This is a semi-well-known open problem known as Hadamard's maximum determinant problem.






      share|cite|improve this answer
























        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.






        share|cite|improve this answer














        From comments:



        This is a semi-well-known open problem known as Hadamard's maximum determinant problem.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        answered Sep 16 at 11:16


























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