Algorithm to obtain the null matrix from a series of predefined operators Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is it possible to back solve this matrix?linear problem with $|.|_infty$ and $|.|_1$ norm constraintsConstructing a matrix from the null spacePossible ranks of a $n!times n$ matrix with permuted rowsGauss Transform with $LU$ factorizationNull space, column space and rank with projection matrixWhich properties do elementary transformations on a matrices not preserve?How many coefficients do you have to change to lower the rank of a matrix?Can't understand step in LU decomposition proofSlick and fast linear algebra treatment for finite field extensions?

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Algorithm to obtain the null matrix from a series of predefined operators



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is it possible to back solve this matrix?linear problem with $|.|_infty$ and $|.|_1$ norm constraintsConstructing a matrix from the null spacePossible ranks of a $n!times n$ matrix with permuted rowsGauss Transform with $LU$ factorizationNull space, column space and rank with projection matrixWhich properties do elementary transformations on a matrices not preserve?How many coefficients do you have to change to lower the rank of a matrix?Can't understand step in LU decomposition proofSlick and fast linear algebra treatment for finite field extensions?










1












$begingroup$


Consider a matrix $tilde A$ of size $(N+2)times (N+2)$ whose coefficients are either $0$ or $1$.



Denote $A$ the submatrix containing columns from 2 to $N$ of $tilde A$
and line from $2$ to $N$ of $tilde A$.
In short $A=tilde A(2...N,2...N)$.



Define the following elementary operator:
$$b_i^j(A) = A + E_i,j+ E_i,j+1+ E_i+1,j+ E_i-1,j+ E_i,j-1 quad textmod; [2]$$
for all $i,j$ of $mathbf1,ldots, N$ where $E_a,b$ is the elementary matrix (which has 0 everywhere except for the coefficient in position $(a,b)$).



Is it possible to transform any submatrix $Min M_N,N(0,1)$ of a matrix $tilde Min M_N+2,N+2(0,1)$ to the null matrix of $M_N,N(0,1)$ using these operations $b_i^j$ ?



Remark: Assuming the existence of such algorithm and applying it to the matrix $tilde M$, we obtain a matrix denoted as $tilde M_infty$. We do not care about the coefficients of the first and last columns/lines of $tilde M_infty$, we only care that the submatrix $M_infty=tilde M_infty(2...N,2...N)$ is null.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Do you really want to reduce anything modulo $2 pi$ ? That's a fairly foreign object in the algebraic setting in which the rest of your problem is happening. (Most likely, it changes nothing, since $pi$ is rational and so the fractional part of your entries will give away how often you have added $2 pi$...)
    $endgroup$
    – darij grinberg
    Apr 8 at 21:32







  • 1




    $begingroup$
    On the other hand, if your "$mod 2pi$" is meant to be a "$mod 2$", then you are studying "button madness" (aka "lights-out") on a square board, and this has seen some research (at least on a torus: win.tue.nl/~aeb/ca/madness/mad.html ).
    $endgroup$
    – darij grinberg
    Apr 8 at 21:33











  • $begingroup$
    indeed, it's mod 2 .... yeah I would like to solve the button madness? I didn't expect it was so complicated
    $endgroup$
    – Smilia
    Apr 9 at 0:14










  • $begingroup$
    The link doesn't study the same problem that you are studying (as the link works with a toroidal grid), so maybe yours is actually simpler.
    $endgroup$
    – darij grinberg
    Apr 9 at 1:52















1












$begingroup$


Consider a matrix $tilde A$ of size $(N+2)times (N+2)$ whose coefficients are either $0$ or $1$.



Denote $A$ the submatrix containing columns from 2 to $N$ of $tilde A$
and line from $2$ to $N$ of $tilde A$.
In short $A=tilde A(2...N,2...N)$.



Define the following elementary operator:
$$b_i^j(A) = A + E_i,j+ E_i,j+1+ E_i+1,j+ E_i-1,j+ E_i,j-1 quad textmod; [2]$$
for all $i,j$ of $mathbf1,ldots, N$ where $E_a,b$ is the elementary matrix (which has 0 everywhere except for the coefficient in position $(a,b)$).



Is it possible to transform any submatrix $Min M_N,N(0,1)$ of a matrix $tilde Min M_N+2,N+2(0,1)$ to the null matrix of $M_N,N(0,1)$ using these operations $b_i^j$ ?



Remark: Assuming the existence of such algorithm and applying it to the matrix $tilde M$, we obtain a matrix denoted as $tilde M_infty$. We do not care about the coefficients of the first and last columns/lines of $tilde M_infty$, we only care that the submatrix $M_infty=tilde M_infty(2...N,2...N)$ is null.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Do you really want to reduce anything modulo $2 pi$ ? That's a fairly foreign object in the algebraic setting in which the rest of your problem is happening. (Most likely, it changes nothing, since $pi$ is rational and so the fractional part of your entries will give away how often you have added $2 pi$...)
    $endgroup$
    – darij grinberg
    Apr 8 at 21:32







  • 1




    $begingroup$
    On the other hand, if your "$mod 2pi$" is meant to be a "$mod 2$", then you are studying "button madness" (aka "lights-out") on a square board, and this has seen some research (at least on a torus: win.tue.nl/~aeb/ca/madness/mad.html ).
    $endgroup$
    – darij grinberg
    Apr 8 at 21:33











  • $begingroup$
    indeed, it's mod 2 .... yeah I would like to solve the button madness? I didn't expect it was so complicated
    $endgroup$
    – Smilia
    Apr 9 at 0:14










  • $begingroup$
    The link doesn't study the same problem that you are studying (as the link works with a toroidal grid), so maybe yours is actually simpler.
    $endgroup$
    – darij grinberg
    Apr 9 at 1:52













1












1








1





$begingroup$


Consider a matrix $tilde A$ of size $(N+2)times (N+2)$ whose coefficients are either $0$ or $1$.



Denote $A$ the submatrix containing columns from 2 to $N$ of $tilde A$
and line from $2$ to $N$ of $tilde A$.
In short $A=tilde A(2...N,2...N)$.



Define the following elementary operator:
$$b_i^j(A) = A + E_i,j+ E_i,j+1+ E_i+1,j+ E_i-1,j+ E_i,j-1 quad textmod; [2]$$
for all $i,j$ of $mathbf1,ldots, N$ where $E_a,b$ is the elementary matrix (which has 0 everywhere except for the coefficient in position $(a,b)$).



Is it possible to transform any submatrix $Min M_N,N(0,1)$ of a matrix $tilde Min M_N+2,N+2(0,1)$ to the null matrix of $M_N,N(0,1)$ using these operations $b_i^j$ ?



Remark: Assuming the existence of such algorithm and applying it to the matrix $tilde M$, we obtain a matrix denoted as $tilde M_infty$. We do not care about the coefficients of the first and last columns/lines of $tilde M_infty$, we only care that the submatrix $M_infty=tilde M_infty(2...N,2...N)$ is null.










share|cite|improve this question











$endgroup$




Consider a matrix $tilde A$ of size $(N+2)times (N+2)$ whose coefficients are either $0$ or $1$.



Denote $A$ the submatrix containing columns from 2 to $N$ of $tilde A$
and line from $2$ to $N$ of $tilde A$.
In short $A=tilde A(2...N,2...N)$.



Define the following elementary operator:
$$b_i^j(A) = A + E_i,j+ E_i,j+1+ E_i+1,j+ E_i-1,j+ E_i,j-1 quad textmod; [2]$$
for all $i,j$ of $mathbf1,ldots, N$ where $E_a,b$ is the elementary matrix (which has 0 everywhere except for the coefficient in position $(a,b)$).



Is it possible to transform any submatrix $Min M_N,N(0,1)$ of a matrix $tilde Min M_N+2,N+2(0,1)$ to the null matrix of $M_N,N(0,1)$ using these operations $b_i^j$ ?



Remark: Assuming the existence of such algorithm and applying it to the matrix $tilde M$, we obtain a matrix denoted as $tilde M_infty$. We do not care about the coefficients of the first and last columns/lines of $tilde M_infty$, we only care that the submatrix $M_infty=tilde M_infty(2...N,2...N)$ is null.







linear-algebra abstract-algebra matrices algorithms modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 9 at 0:13







Smilia

















asked Apr 8 at 20:20









SmiliaSmilia

733617




733617











  • $begingroup$
    Do you really want to reduce anything modulo $2 pi$ ? That's a fairly foreign object in the algebraic setting in which the rest of your problem is happening. (Most likely, it changes nothing, since $pi$ is rational and so the fractional part of your entries will give away how often you have added $2 pi$...)
    $endgroup$
    – darij grinberg
    Apr 8 at 21:32







  • 1




    $begingroup$
    On the other hand, if your "$mod 2pi$" is meant to be a "$mod 2$", then you are studying "button madness" (aka "lights-out") on a square board, and this has seen some research (at least on a torus: win.tue.nl/~aeb/ca/madness/mad.html ).
    $endgroup$
    – darij grinberg
    Apr 8 at 21:33











  • $begingroup$
    indeed, it's mod 2 .... yeah I would like to solve the button madness? I didn't expect it was so complicated
    $endgroup$
    – Smilia
    Apr 9 at 0:14










  • $begingroup$
    The link doesn't study the same problem that you are studying (as the link works with a toroidal grid), so maybe yours is actually simpler.
    $endgroup$
    – darij grinberg
    Apr 9 at 1:52
















  • $begingroup$
    Do you really want to reduce anything modulo $2 pi$ ? That's a fairly foreign object in the algebraic setting in which the rest of your problem is happening. (Most likely, it changes nothing, since $pi$ is rational and so the fractional part of your entries will give away how often you have added $2 pi$...)
    $endgroup$
    – darij grinberg
    Apr 8 at 21:32







  • 1




    $begingroup$
    On the other hand, if your "$mod 2pi$" is meant to be a "$mod 2$", then you are studying "button madness" (aka "lights-out") on a square board, and this has seen some research (at least on a torus: win.tue.nl/~aeb/ca/madness/mad.html ).
    $endgroup$
    – darij grinberg
    Apr 8 at 21:33











  • $begingroup$
    indeed, it's mod 2 .... yeah I would like to solve the button madness? I didn't expect it was so complicated
    $endgroup$
    – Smilia
    Apr 9 at 0:14










  • $begingroup$
    The link doesn't study the same problem that you are studying (as the link works with a toroidal grid), so maybe yours is actually simpler.
    $endgroup$
    – darij grinberg
    Apr 9 at 1:52















$begingroup$
Do you really want to reduce anything modulo $2 pi$ ? That's a fairly foreign object in the algebraic setting in which the rest of your problem is happening. (Most likely, it changes nothing, since $pi$ is rational and so the fractional part of your entries will give away how often you have added $2 pi$...)
$endgroup$
– darij grinberg
Apr 8 at 21:32





$begingroup$
Do you really want to reduce anything modulo $2 pi$ ? That's a fairly foreign object in the algebraic setting in which the rest of your problem is happening. (Most likely, it changes nothing, since $pi$ is rational and so the fractional part of your entries will give away how often you have added $2 pi$...)
$endgroup$
– darij grinberg
Apr 8 at 21:32





1




1




$begingroup$
On the other hand, if your "$mod 2pi$" is meant to be a "$mod 2$", then you are studying "button madness" (aka "lights-out") on a square board, and this has seen some research (at least on a torus: win.tue.nl/~aeb/ca/madness/mad.html ).
$endgroup$
– darij grinberg
Apr 8 at 21:33





$begingroup$
On the other hand, if your "$mod 2pi$" is meant to be a "$mod 2$", then you are studying "button madness" (aka "lights-out") on a square board, and this has seen some research (at least on a torus: win.tue.nl/~aeb/ca/madness/mad.html ).
$endgroup$
– darij grinberg
Apr 8 at 21:33













$begingroup$
indeed, it's mod 2 .... yeah I would like to solve the button madness? I didn't expect it was so complicated
$endgroup$
– Smilia
Apr 9 at 0:14




$begingroup$
indeed, it's mod 2 .... yeah I would like to solve the button madness? I didn't expect it was so complicated
$endgroup$
– Smilia
Apr 9 at 0:14












$begingroup$
The link doesn't study the same problem that you are studying (as the link works with a toroidal grid), so maybe yours is actually simpler.
$endgroup$
– darij grinberg
Apr 9 at 1:52




$begingroup$
The link doesn't study the same problem that you are studying (as the link works with a toroidal grid), so maybe yours is actually simpler.
$endgroup$
– darij grinberg
Apr 9 at 1:52










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