Do I really need to invert this matrix

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I need to calculate a matrix $A$ (at least some elements of it, see below) as defined by the following equation



$$ A=B(mathbb1-B)^-1 $$



where B is a square matrix of dimension $N$ and $mathbb1$ is $N times N$ identity matrix.



Inspired by this post:



https://www.johndcook.com/blog/2010/01/19/dont-invert-that-matrix/



I was wondering if I really need to invert $mathbb1-B$ in my case, of if there's some easier way. Keep in mind that:



  • $N$ in my case is quite large, its order of magnitude can be ten thousand.


  • I don't need to know the full matrix $A$, I just need a few elements in the upper left corner, let's say $A_00$, $A_01$, $A_11$, $A_02$, $A_12$, $A_22$ would be perfect.






matrix inverse-problem inverse







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    up vote
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    down vote

    favorite












    I need to calculate a matrix $A$ (at least some elements of it, see below) as defined by the following equation



    $$ A=B(mathbb1-B)^-1 $$



    where B is a square matrix of dimension $N$ and $mathbb1$ is $N times N$ identity matrix.



    Inspired by this post:



    https://www.johndcook.com/blog/2010/01/19/dont-invert-that-matrix/



    I was wondering if I really need to invert $mathbb1-B$ in my case, of if there's some easier way. Keep in mind that:



    • $N$ in my case is quite large, its order of magnitude can be ten thousand.


    • I don't need to know the full matrix $A$, I just need a few elements in the upper left corner, let's say $A_00$, $A_01$, $A_11$, $A_02$, $A_12$, $A_22$ would be perfect.






    matrix inverse-problem inverse







    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I need to calculate a matrix $A$ (at least some elements of it, see below) as defined by the following equation



      $$ A=B(mathbb1-B)^-1 $$



      where B is a square matrix of dimension $N$ and $mathbb1$ is $N times N$ identity matrix.



      Inspired by this post:



      https://www.johndcook.com/blog/2010/01/19/dont-invert-that-matrix/



      I was wondering if I really need to invert $mathbb1-B$ in my case, of if there's some easier way. Keep in mind that:



      • $N$ in my case is quite large, its order of magnitude can be ten thousand.


      • I don't need to know the full matrix $A$, I just need a few elements in the upper left corner, let's say $A_00$, $A_01$, $A_11$, $A_02$, $A_12$, $A_22$ would be perfect.






      matrix inverse-problem inverse







      share|cite|improve this question













      I need to calculate a matrix $A$ (at least some elements of it, see below) as defined by the following equation



      $$ A=B(mathbb1-B)^-1 $$



      where B is a square matrix of dimension $N$ and $mathbb1$ is $N times N$ identity matrix.



      Inspired by this post:



      https://www.johndcook.com/blog/2010/01/19/dont-invert-that-matrix/



      I was wondering if I really need to invert $mathbb1-B$ in my case, of if there's some easier way. Keep in mind that:



      • $N$ in my case is quite large, its order of magnitude can be ten thousand.


      • I don't need to know the full matrix $A$, I just need a few elements in the upper left corner, let's say $A_00$, $A_01$, $A_11$, $A_02$, $A_12$, $A_22$ would be perfect.






      matrix inverse-problem inverse




      matrix inverse-problem inverse






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      asked 4 hours ago









      zakk

      1623




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          1 Answer
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          Since
          $$
          A = B(I-B)^-1 = (I-B)^-1(I-B)B(I-B)^-1 = (I-B)^-1B(I-B)(I-B)^-1 =(I-B)^-1B
          $$
          So you want to solve
          $$
          (I-B)A=B
          $$
          You seem to need only the first three columns of $A$. Solve the matrix problems
          $$
          (I-B)a_i = b_i, qquad i=0,1,2
          $$
          where $b_0,b_1,b_2$ are first three columns of $B$. Then $a_0,a_1,a_2$ are the first three columns of $A$.






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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            Since
            $$
            A = B(I-B)^-1 = (I-B)^-1(I-B)B(I-B)^-1 = (I-B)^-1B(I-B)(I-B)^-1 =(I-B)^-1B
            $$
            So you want to solve
            $$
            (I-B)A=B
            $$
            You seem to need only the first three columns of $A$. Solve the matrix problems
            $$
            (I-B)a_i = b_i, qquad i=0,1,2
            $$
            where $b_0,b_1,b_2$ are first three columns of $B$. Then $a_0,a_1,a_2$ are the first three columns of $A$.






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              Since
              $$
              A = B(I-B)^-1 = (I-B)^-1(I-B)B(I-B)^-1 = (I-B)^-1B(I-B)(I-B)^-1 =(I-B)^-1B
              $$
              So you want to solve
              $$
              (I-B)A=B
              $$
              You seem to need only the first three columns of $A$. Solve the matrix problems
              $$
              (I-B)a_i = b_i, qquad i=0,1,2
              $$
              where $b_0,b_1,b_2$ are first three columns of $B$. Then $a_0,a_1,a_2$ are the first three columns of $A$.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                Since
                $$
                A = B(I-B)^-1 = (I-B)^-1(I-B)B(I-B)^-1 = (I-B)^-1B(I-B)(I-B)^-1 =(I-B)^-1B
                $$
                So you want to solve
                $$
                (I-B)A=B
                $$
                You seem to need only the first three columns of $A$. Solve the matrix problems
                $$
                (I-B)a_i = b_i, qquad i=0,1,2
                $$
                where $b_0,b_1,b_2$ are first three columns of $B$. Then $a_0,a_1,a_2$ are the first three columns of $A$.






                share|cite|improve this answer












                Since
                $$
                A = B(I-B)^-1 = (I-B)^-1(I-B)B(I-B)^-1 = (I-B)^-1B(I-B)(I-B)^-1 =(I-B)^-1B
                $$
                So you want to solve
                $$
                (I-B)A=B
                $$
                You seem to need only the first three columns of $A$. Solve the matrix problems
                $$
                (I-B)a_i = b_i, qquad i=0,1,2
                $$
                where $b_0,b_1,b_2$ are first three columns of $B$. Then $a_0,a_1,a_2$ are the first three columns of $A$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Praveen Chandrashekar

                894510




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