Square sub matrix with determinant $2.$ [on hold] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Determinant of a real skew-symmetric matrix is square of an integerDeterminant of block matrices with non square matricesInequality for subordinate norm of a submatrix : $Vert BVertleVert AVert$.Is a square matrix with positive determinant, positive diagonal entries and negative off-diagonal entries an M-matrix?How to describe a matrix with a mathematical notation or a terminology?Low-degree “determinant” for non-square matrices?Show that $r$ is the rank of the $n$x$n$ matrix $Aiff A$ has a nonsingular $r$x$r$ submatrixdeterminant inequality for Hermitian matrixDeterminant of submatrixDeterminant of the sum of a positive semi-definite matrix and a diagonal matrix

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Square sub matrix with determinant $2.$ [on hold]



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Determinant of a real skew-symmetric matrix is square of an integerDeterminant of block matrices with non square matricesInequality for subordinate norm of a submatrix : $Vert BVertleVert AVert$.Is a square matrix with positive determinant, positive diagonal entries and negative off-diagonal entries an M-matrix?How to describe a matrix with a mathematical notation or a terminology?Low-degree “determinant” for non-square matrices?Show that $r$ is the rank of the $n$x$n$ matrix $Aiff A$ has a nonsingular $r$x$r$ submatrixdeterminant inequality for Hermitian matrixDeterminant of submatrixDeterminant of the sum of a positive semi-definite matrix and a diagonal matrix










5












$begingroup$


If $A$ is a nonsingular matrix with entries from $0,1,-1$ and $vert det(A) vert$ is
not equal $1$, then show that $A$ has a square submatrix $B$ with $vertdet(B)vert=2$.



I have tried but could not proceed. What could be the possible way to solve it? Please share some idea.










share|cite|improve this question











$endgroup$



put on hold as off-topic by user21820, José Carlos Santos, mrtaurho, Cesareo, Saad 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, mrtaurho, Cesareo, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 1




    $begingroup$
    If the determinant is neither $pm 1$ nor $0$, then at least one of the columns and at least one of the rows must have at least two non-zero entries. Don't know exactly what to do with this information. But my gut tells me that there is a solution somewhere in that direction.
    $endgroup$
    – Arthur
    Apr 1 at 13:34











  • $begingroup$
    @J.Doe Question for clarification: this submatrix B must be some block matrix made from adjacent entries of A or it can be also made from entries not necessary adjacent, cut down from A by removing one or more of its rows and columns (like in calculating minors) ?
    $endgroup$
    – Widawensen
    Apr 2 at 16:46











  • $begingroup$
    nptel.ac.in/courses/122104018/node10.html. This definition can be used
    $endgroup$
    – J.Doe
    Apr 2 at 16:54















5












$begingroup$


If $A$ is a nonsingular matrix with entries from $0,1,-1$ and $vert det(A) vert$ is
not equal $1$, then show that $A$ has a square submatrix $B$ with $vertdet(B)vert=2$.



I have tried but could not proceed. What could be the possible way to solve it? Please share some idea.










share|cite|improve this question











$endgroup$



put on hold as off-topic by user21820, José Carlos Santos, mrtaurho, Cesareo, Saad 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, mrtaurho, Cesareo, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 1




    $begingroup$
    If the determinant is neither $pm 1$ nor $0$, then at least one of the columns and at least one of the rows must have at least two non-zero entries. Don't know exactly what to do with this information. But my gut tells me that there is a solution somewhere in that direction.
    $endgroup$
    – Arthur
    Apr 1 at 13:34











  • $begingroup$
    @J.Doe Question for clarification: this submatrix B must be some block matrix made from adjacent entries of A or it can be also made from entries not necessary adjacent, cut down from A by removing one or more of its rows and columns (like in calculating minors) ?
    $endgroup$
    – Widawensen
    Apr 2 at 16:46











  • $begingroup$
    nptel.ac.in/courses/122104018/node10.html. This definition can be used
    $endgroup$
    – J.Doe
    Apr 2 at 16:54













5












5








5


2



$begingroup$


If $A$ is a nonsingular matrix with entries from $0,1,-1$ and $vert det(A) vert$ is
not equal $1$, then show that $A$ has a square submatrix $B$ with $vertdet(B)vert=2$.



I have tried but could not proceed. What could be the possible way to solve it? Please share some idea.










share|cite|improve this question











$endgroup$




If $A$ is a nonsingular matrix with entries from $0,1,-1$ and $vert det(A) vert$ is
not equal $1$, then show that $A$ has a square submatrix $B$ with $vertdet(B)vert=2$.



I have tried but could not proceed. What could be the possible way to solve it? Please share some idea.







linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 1 at 13:33







J.Doe

















asked Apr 1 at 13:22









J.DoeJ.Doe

824




824




put on hold as off-topic by user21820, José Carlos Santos, mrtaurho, Cesareo, Saad 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, mrtaurho, Cesareo, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.







put on hold as off-topic by user21820, José Carlos Santos, mrtaurho, Cesareo, Saad 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, José Carlos Santos, mrtaurho, Cesareo, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    $begingroup$
    If the determinant is neither $pm 1$ nor $0$, then at least one of the columns and at least one of the rows must have at least two non-zero entries. Don't know exactly what to do with this information. But my gut tells me that there is a solution somewhere in that direction.
    $endgroup$
    – Arthur
    Apr 1 at 13:34











  • $begingroup$
    @J.Doe Question for clarification: this submatrix B must be some block matrix made from adjacent entries of A or it can be also made from entries not necessary adjacent, cut down from A by removing one or more of its rows and columns (like in calculating minors) ?
    $endgroup$
    – Widawensen
    Apr 2 at 16:46











  • $begingroup$
    nptel.ac.in/courses/122104018/node10.html. This definition can be used
    $endgroup$
    – J.Doe
    Apr 2 at 16:54












  • 1




    $begingroup$
    If the determinant is neither $pm 1$ nor $0$, then at least one of the columns and at least one of the rows must have at least two non-zero entries. Don't know exactly what to do with this information. But my gut tells me that there is a solution somewhere in that direction.
    $endgroup$
    – Arthur
    Apr 1 at 13:34











  • $begingroup$
    @J.Doe Question for clarification: this submatrix B must be some block matrix made from adjacent entries of A or it can be also made from entries not necessary adjacent, cut down from A by removing one or more of its rows and columns (like in calculating minors) ?
    $endgroup$
    – Widawensen
    Apr 2 at 16:46











  • $begingroup$
    nptel.ac.in/courses/122104018/node10.html. This definition can be used
    $endgroup$
    – J.Doe
    Apr 2 at 16:54







1




1




$begingroup$
If the determinant is neither $pm 1$ nor $0$, then at least one of the columns and at least one of the rows must have at least two non-zero entries. Don't know exactly what to do with this information. But my gut tells me that there is a solution somewhere in that direction.
$endgroup$
– Arthur
Apr 1 at 13:34





$begingroup$
If the determinant is neither $pm 1$ nor $0$, then at least one of the columns and at least one of the rows must have at least two non-zero entries. Don't know exactly what to do with this information. But my gut tells me that there is a solution somewhere in that direction.
$endgroup$
– Arthur
Apr 1 at 13:34













$begingroup$
@J.Doe Question for clarification: this submatrix B must be some block matrix made from adjacent entries of A or it can be also made from entries not necessary adjacent, cut down from A by removing one or more of its rows and columns (like in calculating minors) ?
$endgroup$
– Widawensen
Apr 2 at 16:46





$begingroup$
@J.Doe Question for clarification: this submatrix B must be some block matrix made from adjacent entries of A or it can be also made from entries not necessary adjacent, cut down from A by removing one or more of its rows and columns (like in calculating minors) ?
$endgroup$
– Widawensen
Apr 2 at 16:46













$begingroup$
nptel.ac.in/courses/122104018/node10.html. This definition can be used
$endgroup$
– J.Doe
Apr 2 at 16:54




$begingroup$
nptel.ac.in/courses/122104018/node10.html. This definition can be used
$endgroup$
– J.Doe
Apr 2 at 16:54










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