The minimum rank of a matrix with a given pattern of zeros












4












$begingroup$


For real matrices $A=(a_{ij})$ and $B=(b_{ij})$ of the same size, I write $Aprec B$ if $a_{ij}=0$ whenever $b_{ij}=0$.



If
$$ B = begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 0 & 1 end{pmatrix}^{otimes10}, $$
then the matrix
$$ A = begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ -1 & 0 & 1 end{pmatrix}^{otimes10} $$
satisfies $mathrm{rk}(A)<mathrm{rk}(B)$ and $Iprec Aprec B$ (where $I$ is the identity matrix of order $3^{10}$) . Does there exist a square matrix $C$ of order $3^{10}$ such that $mathrm{rk}(C)<mathrm{rk}(A)$ and $Iprec Cprec A$?






linear-algebra matrices







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$endgroup$

















    4












    $begingroup$


    For real matrices $A=(a_{ij})$ and $B=(b_{ij})$ of the same size, I write $Aprec B$ if $a_{ij}=0$ whenever $b_{ij}=0$.



    If
    $$ B = begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 0 & 1 end{pmatrix}^{otimes10}, $$
    then the matrix
    $$ A = begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ -1 & 0 & 1 end{pmatrix}^{otimes10} $$
    satisfies $mathrm{rk}(A)<mathrm{rk}(B)$ and $Iprec Aprec B$ (where $I$ is the identity matrix of order $3^{10}$) . Does there exist a square matrix $C$ of order $3^{10}$ such that $mathrm{rk}(C)<mathrm{rk}(A)$ and $Iprec Cprec A$?






    linear-algebra matrices







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      1



      $begingroup$


      For real matrices $A=(a_{ij})$ and $B=(b_{ij})$ of the same size, I write $Aprec B$ if $a_{ij}=0$ whenever $b_{ij}=0$.



      If
      $$ B = begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 0 & 1 end{pmatrix}^{otimes10}, $$
      then the matrix
      $$ A = begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ -1 & 0 & 1 end{pmatrix}^{otimes10} $$
      satisfies $mathrm{rk}(A)<mathrm{rk}(B)$ and $Iprec Aprec B$ (where $I$ is the identity matrix of order $3^{10}$) . Does there exist a square matrix $C$ of order $3^{10}$ such that $mathrm{rk}(C)<mathrm{rk}(A)$ and $Iprec Cprec A$?






      linear-algebra matrices







      share|cite|improve this question









      $endgroup$




      For real matrices $A=(a_{ij})$ and $B=(b_{ij})$ of the same size, I write $Aprec B$ if $a_{ij}=0$ whenever $b_{ij}=0$.



      If
      $$ B = begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 0 & 1 end{pmatrix}^{otimes10}, $$
      then the matrix
      $$ A = begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ -1 & 0 & 1 end{pmatrix}^{otimes10} $$
      satisfies $mathrm{rk}(A)<mathrm{rk}(B)$ and $Iprec Aprec B$ (where $I$ is the identity matrix of order $3^{10}$) . Does there exist a square matrix $C$ of order $3^{10}$ such that $mathrm{rk}(C)<mathrm{rk}(A)$ and $Iprec Cprec A$?






      linear-algebra matrices




      linear-algebra matrices






      share|cite|improve this question













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      asked Dec 22 '18 at 11:01









      SevaSeva

      12.7k138103




      12.7k138103






















          1 Answer
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          4












          $begingroup$

          As Misha Muzychuck has observed, the answer is "no": since
          $$ begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ -1 & 0 & 1 end{pmatrix} $$ contains a non-degenerate upper-triangular submatrix of size $2$, the matrix $A$ contains a non-degenerate upper-triangular submatrix of size $2^{10}$, whence $mathrm{rk}(C)ge 2^{10}=mathrm{rk}(A)$ for any matrix $C$ with $Iprec Cprec A$.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            just curious: since ${rm rk}(A)=2^{10}$, it implies that $A$ does not contain a non-degenerate upper-triangular-after-permutation submatrix of size greater than $2^{10}$. Is it obvious a priori without linear algebra (on the size of directed graphs)?
            $endgroup$
            – Fedor Petrov
            Dec 22 '18 at 12:21












          • $begingroup$
            @FedorPetrov: In fact, I cannot even prove without using linear algebra / polynomials that $(A-A)cap{0,1}^nne{0}$ for any set $Asubsetmathbb F_3^n$ with $|A|>2^n$ (can you?)
            $endgroup$
            – Seva
            Dec 23 '18 at 19:35






          • 1




            $begingroup$
            I can not. But recently it was slightly improved arxiv.org/abs/1812.05989 (again with linear algebra)
            $endgroup$
            – Fedor Petrov
            Dec 23 '18 at 20:37










          • $begingroup$
            @FedorPetrov: Interestingly, there is a huge set $Asubsetmathbb F_3^n$ (of size $|A|ge 3^{n-1}$) and a very large subset $D_0subset{0,1}$ (of size $|D_0|gtrsim frac13cdot 2^n$) such that $A-A$ is disjoint from $D_0$.
            $endgroup$
            – Seva
            Dec 25 '18 at 18:06












          • $begingroup$
            probably even $frac 23 cdot 2^n$? If $A$ lies in some hyperplane.
            $endgroup$
            – Fedor Petrov
            Dec 26 '18 at 7:27











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes









          4












          $begingroup$

          As Misha Muzychuck has observed, the answer is "no": since
          $$ begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ -1 & 0 & 1 end{pmatrix} $$ contains a non-degenerate upper-triangular submatrix of size $2$, the matrix $A$ contains a non-degenerate upper-triangular submatrix of size $2^{10}$, whence $mathrm{rk}(C)ge 2^{10}=mathrm{rk}(A)$ for any matrix $C$ with $Iprec Cprec A$.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            just curious: since ${rm rk}(A)=2^{10}$, it implies that $A$ does not contain a non-degenerate upper-triangular-after-permutation submatrix of size greater than $2^{10}$. Is it obvious a priori without linear algebra (on the size of directed graphs)?
            $endgroup$
            – Fedor Petrov
            Dec 22 '18 at 12:21












          • $begingroup$
            @FedorPetrov: In fact, I cannot even prove without using linear algebra / polynomials that $(A-A)cap{0,1}^nne{0}$ for any set $Asubsetmathbb F_3^n$ with $|A|>2^n$ (can you?)
            $endgroup$
            – Seva
            Dec 23 '18 at 19:35






          • 1




            $begingroup$
            I can not. But recently it was slightly improved arxiv.org/abs/1812.05989 (again with linear algebra)
            $endgroup$
            – Fedor Petrov
            Dec 23 '18 at 20:37










          • $begingroup$
            @FedorPetrov: Interestingly, there is a huge set $Asubsetmathbb F_3^n$ (of size $|A|ge 3^{n-1}$) and a very large subset $D_0subset{0,1}$ (of size $|D_0|gtrsim frac13cdot 2^n$) such that $A-A$ is disjoint from $D_0$.
            $endgroup$
            – Seva
            Dec 25 '18 at 18:06












          • $begingroup$
            probably even $frac 23 cdot 2^n$? If $A$ lies in some hyperplane.
            $endgroup$
            – Fedor Petrov
            Dec 26 '18 at 7:27
















          4












          $begingroup$

          As Misha Muzychuck has observed, the answer is "no": since
          $$ begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ -1 & 0 & 1 end{pmatrix} $$ contains a non-degenerate upper-triangular submatrix of size $2$, the matrix $A$ contains a non-degenerate upper-triangular submatrix of size $2^{10}$, whence $mathrm{rk}(C)ge 2^{10}=mathrm{rk}(A)$ for any matrix $C$ with $Iprec Cprec A$.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            just curious: since ${rm rk}(A)=2^{10}$, it implies that $A$ does not contain a non-degenerate upper-triangular-after-permutation submatrix of size greater than $2^{10}$. Is it obvious a priori without linear algebra (on the size of directed graphs)?
            $endgroup$
            – Fedor Petrov
            Dec 22 '18 at 12:21












          • $begingroup$
            @FedorPetrov: In fact, I cannot even prove without using linear algebra / polynomials that $(A-A)cap{0,1}^nne{0}$ for any set $Asubsetmathbb F_3^n$ with $|A|>2^n$ (can you?)
            $endgroup$
            – Seva
            Dec 23 '18 at 19:35






          • 1




            $begingroup$
            I can not. But recently it was slightly improved arxiv.org/abs/1812.05989 (again with linear algebra)
            $endgroup$
            – Fedor Petrov
            Dec 23 '18 at 20:37










          • $begingroup$
            @FedorPetrov: Interestingly, there is a huge set $Asubsetmathbb F_3^n$ (of size $|A|ge 3^{n-1}$) and a very large subset $D_0subset{0,1}$ (of size $|D_0|gtrsim frac13cdot 2^n$) such that $A-A$ is disjoint from $D_0$.
            $endgroup$
            – Seva
            Dec 25 '18 at 18:06












          • $begingroup$
            probably even $frac 23 cdot 2^n$? If $A$ lies in some hyperplane.
            $endgroup$
            – Fedor Petrov
            Dec 26 '18 at 7:27














          4












          4








          4





          $begingroup$

          As Misha Muzychuck has observed, the answer is "no": since
          $$ begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ -1 & 0 & 1 end{pmatrix} $$ contains a non-degenerate upper-triangular submatrix of size $2$, the matrix $A$ contains a non-degenerate upper-triangular submatrix of size $2^{10}$, whence $mathrm{rk}(C)ge 2^{10}=mathrm{rk}(A)$ for any matrix $C$ with $Iprec Cprec A$.






          share|cite|improve this answer









          $endgroup$



          As Misha Muzychuck has observed, the answer is "no": since
          $$ begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ -1 & 0 & 1 end{pmatrix} $$ contains a non-degenerate upper-triangular submatrix of size $2$, the matrix $A$ contains a non-degenerate upper-triangular submatrix of size $2^{10}$, whence $mathrm{rk}(C)ge 2^{10}=mathrm{rk}(A)$ for any matrix $C$ with $Iprec Cprec A$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 22 '18 at 12:02









          SevaSeva

          12.7k138103




          12.7k138103








          • 2




            $begingroup$
            just curious: since ${rm rk}(A)=2^{10}$, it implies that $A$ does not contain a non-degenerate upper-triangular-after-permutation submatrix of size greater than $2^{10}$. Is it obvious a priori without linear algebra (on the size of directed graphs)?
            $endgroup$
            – Fedor Petrov
            Dec 22 '18 at 12:21












          • $begingroup$
            @FedorPetrov: In fact, I cannot even prove without using linear algebra / polynomials that $(A-A)cap{0,1}^nne{0}$ for any set $Asubsetmathbb F_3^n$ with $|A|>2^n$ (can you?)
            $endgroup$
            – Seva
            Dec 23 '18 at 19:35






          • 1




            $begingroup$
            I can not. But recently it was slightly improved arxiv.org/abs/1812.05989 (again with linear algebra)
            $endgroup$
            – Fedor Petrov
            Dec 23 '18 at 20:37










          • $begingroup$
            @FedorPetrov: Interestingly, there is a huge set $Asubsetmathbb F_3^n$ (of size $|A|ge 3^{n-1}$) and a very large subset $D_0subset{0,1}$ (of size $|D_0|gtrsim frac13cdot 2^n$) such that $A-A$ is disjoint from $D_0$.
            $endgroup$
            – Seva
            Dec 25 '18 at 18:06












          • $begingroup$
            probably even $frac 23 cdot 2^n$? If $A$ lies in some hyperplane.
            $endgroup$
            – Fedor Petrov
            Dec 26 '18 at 7:27














          • 2




            $begingroup$
            just curious: since ${rm rk}(A)=2^{10}$, it implies that $A$ does not contain a non-degenerate upper-triangular-after-permutation submatrix of size greater than $2^{10}$. Is it obvious a priori without linear algebra (on the size of directed graphs)?
            $endgroup$
            – Fedor Petrov
            Dec 22 '18 at 12:21












          • $begingroup$
            @FedorPetrov: In fact, I cannot even prove without using linear algebra / polynomials that $(A-A)cap{0,1}^nne{0}$ for any set $Asubsetmathbb F_3^n$ with $|A|>2^n$ (can you?)
            $endgroup$
            – Seva
            Dec 23 '18 at 19:35






          • 1




            $begingroup$
            I can not. But recently it was slightly improved arxiv.org/abs/1812.05989 (again with linear algebra)
            $endgroup$
            – Fedor Petrov
            Dec 23 '18 at 20:37










          • $begingroup$
            @FedorPetrov: Interestingly, there is a huge set $Asubsetmathbb F_3^n$ (of size $|A|ge 3^{n-1}$) and a very large subset $D_0subset{0,1}$ (of size $|D_0|gtrsim frac13cdot 2^n$) such that $A-A$ is disjoint from $D_0$.
            $endgroup$
            – Seva
            Dec 25 '18 at 18:06












          • $begingroup$
            probably even $frac 23 cdot 2^n$? If $A$ lies in some hyperplane.
            $endgroup$
            – Fedor Petrov
            Dec 26 '18 at 7:27








          2




          2




          $begingroup$
          just curious: since ${rm rk}(A)=2^{10}$, it implies that $A$ does not contain a non-degenerate upper-triangular-after-permutation submatrix of size greater than $2^{10}$. Is it obvious a priori without linear algebra (on the size of directed graphs)?
          $endgroup$
          – Fedor Petrov
          Dec 22 '18 at 12:21






          $begingroup$
          just curious: since ${rm rk}(A)=2^{10}$, it implies that $A$ does not contain a non-degenerate upper-triangular-after-permutation submatrix of size greater than $2^{10}$. Is it obvious a priori without linear algebra (on the size of directed graphs)?
          $endgroup$
          – Fedor Petrov
          Dec 22 '18 at 12:21














          $begingroup$
          @FedorPetrov: In fact, I cannot even prove without using linear algebra / polynomials that $(A-A)cap{0,1}^nne{0}$ for any set $Asubsetmathbb F_3^n$ with $|A|>2^n$ (can you?)
          $endgroup$
          – Seva
          Dec 23 '18 at 19:35




          $begingroup$
          @FedorPetrov: In fact, I cannot even prove without using linear algebra / polynomials that $(A-A)cap{0,1}^nne{0}$ for any set $Asubsetmathbb F_3^n$ with $|A|>2^n$ (can you?)
          $endgroup$
          – Seva
          Dec 23 '18 at 19:35




          1




          1




          $begingroup$
          I can not. But recently it was slightly improved arxiv.org/abs/1812.05989 (again with linear algebra)
          $endgroup$
          – Fedor Petrov
          Dec 23 '18 at 20:37




          $begingroup$
          I can not. But recently it was slightly improved arxiv.org/abs/1812.05989 (again with linear algebra)
          $endgroup$
          – Fedor Petrov
          Dec 23 '18 at 20:37












          $begingroup$
          @FedorPetrov: Interestingly, there is a huge set $Asubsetmathbb F_3^n$ (of size $|A|ge 3^{n-1}$) and a very large subset $D_0subset{0,1}$ (of size $|D_0|gtrsim frac13cdot 2^n$) such that $A-A$ is disjoint from $D_0$.
          $endgroup$
          – Seva
          Dec 25 '18 at 18:06






          $begingroup$
          @FedorPetrov: Interestingly, there is a huge set $Asubsetmathbb F_3^n$ (of size $|A|ge 3^{n-1}$) and a very large subset $D_0subset{0,1}$ (of size $|D_0|gtrsim frac13cdot 2^n$) such that $A-A$ is disjoint from $D_0$.
          $endgroup$
          – Seva
          Dec 25 '18 at 18:06














          $begingroup$
          probably even $frac 23 cdot 2^n$? If $A$ lies in some hyperplane.
          $endgroup$
          – Fedor Petrov
          Dec 26 '18 at 7:27




          $begingroup$
          probably even $frac 23 cdot 2^n$? If $A$ lies in some hyperplane.
          $endgroup$
          – Fedor Petrov
          Dec 26 '18 at 7:27


















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