Does this form of matrix have a name?












0












$begingroup$


I'm looking for the name of this kind of $n$-by-$n$ matrix:



$$left(begin{array}{cccc}
-s_1 & b_{12} & b_{13} & b_{14} \
b_{21} & -s_2 & b_{23} & b_{24} \
b_{31} & b_{32} & -s_3 & b_{34} \
b_{41} & b_{42} & b_{43} & -s_4
end{array}right)$$



where $s_i = sum_{jneq i} b_{ij}$. This is basically so I can google it, so by name I really mean an appropriate "search string".










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$endgroup$












  • $begingroup$
    I think it's the Q-matrix of a continuous time Markov chain, but I'm not sure...
    $endgroup$
    – user71815
    Jun 12 '13 at 11:29










  • $begingroup$
    Sounds about right
    $endgroup$
    – Lucas
    Jun 12 '13 at 15:50










  • $begingroup$
    It reminds of a adjacency matrix (see wiki). Your matrix is minus the neighborhood matrix for the Besag model (slide 10).
    $endgroup$
    – Rafael
    Dec 14 '18 at 22:58
















0












$begingroup$


I'm looking for the name of this kind of $n$-by-$n$ matrix:



$$left(begin{array}{cccc}
-s_1 & b_{12} & b_{13} & b_{14} \
b_{21} & -s_2 & b_{23} & b_{24} \
b_{31} & b_{32} & -s_3 & b_{34} \
b_{41} & b_{42} & b_{43} & -s_4
end{array}right)$$



where $s_i = sum_{jneq i} b_{ij}$. This is basically so I can google it, so by name I really mean an appropriate "search string".










share|cite|improve this question









$endgroup$












  • $begingroup$
    I think it's the Q-matrix of a continuous time Markov chain, but I'm not sure...
    $endgroup$
    – user71815
    Jun 12 '13 at 11:29










  • $begingroup$
    Sounds about right
    $endgroup$
    – Lucas
    Jun 12 '13 at 15:50










  • $begingroup$
    It reminds of a adjacency matrix (see wiki). Your matrix is minus the neighborhood matrix for the Besag model (slide 10).
    $endgroup$
    – Rafael
    Dec 14 '18 at 22:58














0












0








0





$begingroup$


I'm looking for the name of this kind of $n$-by-$n$ matrix:



$$left(begin{array}{cccc}
-s_1 & b_{12} & b_{13} & b_{14} \
b_{21} & -s_2 & b_{23} & b_{24} \
b_{31} & b_{32} & -s_3 & b_{34} \
b_{41} & b_{42} & b_{43} & -s_4
end{array}right)$$



where $s_i = sum_{jneq i} b_{ij}$. This is basically so I can google it, so by name I really mean an appropriate "search string".










share|cite|improve this question









$endgroup$




I'm looking for the name of this kind of $n$-by-$n$ matrix:



$$left(begin{array}{cccc}
-s_1 & b_{12} & b_{13} & b_{14} \
b_{21} & -s_2 & b_{23} & b_{24} \
b_{31} & b_{32} & -s_3 & b_{34} \
b_{41} & b_{42} & b_{43} & -s_4
end{array}right)$$



where $s_i = sum_{jneq i} b_{ij}$. This is basically so I can google it, so by name I really mean an appropriate "search string".







linear-algebra matrices






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jun 12 '13 at 11:25









LucasLucas

1,032720




1,032720












  • $begingroup$
    I think it's the Q-matrix of a continuous time Markov chain, but I'm not sure...
    $endgroup$
    – user71815
    Jun 12 '13 at 11:29










  • $begingroup$
    Sounds about right
    $endgroup$
    – Lucas
    Jun 12 '13 at 15:50










  • $begingroup$
    It reminds of a adjacency matrix (see wiki). Your matrix is minus the neighborhood matrix for the Besag model (slide 10).
    $endgroup$
    – Rafael
    Dec 14 '18 at 22:58


















  • $begingroup$
    I think it's the Q-matrix of a continuous time Markov chain, but I'm not sure...
    $endgroup$
    – user71815
    Jun 12 '13 at 11:29










  • $begingroup$
    Sounds about right
    $endgroup$
    – Lucas
    Jun 12 '13 at 15:50










  • $begingroup$
    It reminds of a adjacency matrix (see wiki). Your matrix is minus the neighborhood matrix for the Besag model (slide 10).
    $endgroup$
    – Rafael
    Dec 14 '18 at 22:58
















$begingroup$
I think it's the Q-matrix of a continuous time Markov chain, but I'm not sure...
$endgroup$
– user71815
Jun 12 '13 at 11:29




$begingroup$
I think it's the Q-matrix of a continuous time Markov chain, but I'm not sure...
$endgroup$
– user71815
Jun 12 '13 at 11:29












$begingroup$
Sounds about right
$endgroup$
– Lucas
Jun 12 '13 at 15:50




$begingroup$
Sounds about right
$endgroup$
– Lucas
Jun 12 '13 at 15:50












$begingroup$
It reminds of a adjacency matrix (see wiki). Your matrix is minus the neighborhood matrix for the Besag model (slide 10).
$endgroup$
– Rafael
Dec 14 '18 at 22:58




$begingroup$
It reminds of a adjacency matrix (see wiki). Your matrix is minus the neighborhood matrix for the Besag model (slide 10).
$endgroup$
– Rafael
Dec 14 '18 at 22:58










2 Answers
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$begingroup$

We have diagonally dominant matrices if absolute values of diagonal elements are larger than sum of absolute values of the others in that row.



Such matrices are invertible, ie. nonzero determinant.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    In your matrix the row sums are null.



    In a Stochastic Matrix they are equal to $1$, with the elements being non-negative.



    If your $b_{i,j}$'s are also non negative, then your matrix is essentially a diagonal matrix $times$ (a stochastic matrix minus $I$).

    And that in brackets is a matrix that , in a Markov chain, gives the difference between the configuration at step $n+1$ wrt that at step $n$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      We have diagonally dominant matrices if absolute values of diagonal elements are larger than sum of absolute values of the others in that row.



      Such matrices are invertible, ie. nonzero determinant.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        We have diagonally dominant matrices if absolute values of diagonal elements are larger than sum of absolute values of the others in that row.



        Such matrices are invertible, ie. nonzero determinant.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          We have diagonally dominant matrices if absolute values of diagonal elements are larger than sum of absolute values of the others in that row.



          Such matrices are invertible, ie. nonzero determinant.






          share|cite|improve this answer









          $endgroup$



          We have diagonally dominant matrices if absolute values of diagonal elements are larger than sum of absolute values of the others in that row.



          Such matrices are invertible, ie. nonzero determinant.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 4 '16 at 6:36









          Behnam EsmayliBehnam Esmayli

          1,956515




          1,956515























              0












              $begingroup$

              In your matrix the row sums are null.



              In a Stochastic Matrix they are equal to $1$, with the elements being non-negative.



              If your $b_{i,j}$'s are also non negative, then your matrix is essentially a diagonal matrix $times$ (a stochastic matrix minus $I$).

              And that in brackets is a matrix that , in a Markov chain, gives the difference between the configuration at step $n+1$ wrt that at step $n$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                In your matrix the row sums are null.



                In a Stochastic Matrix they are equal to $1$, with the elements being non-negative.



                If your $b_{i,j}$'s are also non negative, then your matrix is essentially a diagonal matrix $times$ (a stochastic matrix minus $I$).

                And that in brackets is a matrix that , in a Markov chain, gives the difference between the configuration at step $n+1$ wrt that at step $n$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  In your matrix the row sums are null.



                  In a Stochastic Matrix they are equal to $1$, with the elements being non-negative.



                  If your $b_{i,j}$'s are also non negative, then your matrix is essentially a diagonal matrix $times$ (a stochastic matrix minus $I$).

                  And that in brackets is a matrix that , in a Markov chain, gives the difference between the configuration at step $n+1$ wrt that at step $n$.






                  share|cite|improve this answer









                  $endgroup$



                  In your matrix the row sums are null.



                  In a Stochastic Matrix they are equal to $1$, with the elements being non-negative.



                  If your $b_{i,j}$'s are also non negative, then your matrix is essentially a diagonal matrix $times$ (a stochastic matrix minus $I$).

                  And that in brackets is a matrix that , in a Markov chain, gives the difference between the configuration at step $n+1$ wrt that at step $n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 14 '18 at 23:56









                  G CabG Cab

                  19.4k31238




                  19.4k31238






























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