Inverting variable size block matrix












1












$begingroup$


Given a block matrix $M$



$$bf M=left(begin{array}{cccc}
a mathbb{I}_{2} & boldsymbol{boldsymbol{A}}_{12} & boldsymbol{A}_{13} & boldsymbol{0_{(2,3)}}\
\
boldsymbol{A_{21}} & a mathbb{I}_{2} & boldsymbol{A_{23}} & boldsymbol{0_{(2,3)}}\
\
boldsymbol{A_{31}} & boldsymbol{A_{32}} & a mathbb{I}_{2} & boldsymbol{0_
{(2,3)}}\
boldsymbol{0_{(3,2)}} & boldsymbol{0_{(3,2)}} & boldsymbol{0_{(3,2)}} & b mathbb{I}_{3}
end{array}right)$$



where $bf 0_{(m,n)}$ is an $ mtimes n$ matrix with all entries equal to zero.



We need to find $bf M^{-1}$ and I wondered if there were any theorems that allow me to invert this matrix block by block?



Also, how does the problem change for $A_{kl}=A_{lk}, k,l=1,2,3$?






linear-algebra matrices symmetry







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Given a block matrix $M$



    $$bf M=left(begin{array}{cccc}
    a mathbb{I}_{2} & boldsymbol{boldsymbol{A}}_{12} & boldsymbol{A}_{13} & boldsymbol{0_{(2,3)}}\
    \
    boldsymbol{A_{21}} & a mathbb{I}_{2} & boldsymbol{A_{23}} & boldsymbol{0_{(2,3)}}\
    \
    boldsymbol{A_{31}} & boldsymbol{A_{32}} & a mathbb{I}_{2} & boldsymbol{0_
    {(2,3)}}\
    boldsymbol{0_{(3,2)}} & boldsymbol{0_{(3,2)}} & boldsymbol{0_{(3,2)}} & b mathbb{I}_{3}
    end{array}right)$$



    where $bf 0_{(m,n)}$ is an $ mtimes n$ matrix with all entries equal to zero.



    We need to find $bf M^{-1}$ and I wondered if there were any theorems that allow me to invert this matrix block by block?



    Also, how does the problem change for $A_{kl}=A_{lk}, k,l=1,2,3$?






    linear-algebra matrices symmetry







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Given a block matrix $M$



      $$bf M=left(begin{array}{cccc}
      a mathbb{I}_{2} & boldsymbol{boldsymbol{A}}_{12} & boldsymbol{A}_{13} & boldsymbol{0_{(2,3)}}\
      \
      boldsymbol{A_{21}} & a mathbb{I}_{2} & boldsymbol{A_{23}} & boldsymbol{0_{(2,3)}}\
      \
      boldsymbol{A_{31}} & boldsymbol{A_{32}} & a mathbb{I}_{2} & boldsymbol{0_
      {(2,3)}}\
      boldsymbol{0_{(3,2)}} & boldsymbol{0_{(3,2)}} & boldsymbol{0_{(3,2)}} & b mathbb{I}_{3}
      end{array}right)$$



      where $bf 0_{(m,n)}$ is an $ mtimes n$ matrix with all entries equal to zero.



      We need to find $bf M^{-1}$ and I wondered if there were any theorems that allow me to invert this matrix block by block?



      Also, how does the problem change for $A_{kl}=A_{lk}, k,l=1,2,3$?






      linear-algebra matrices symmetry







      share|cite|improve this question











      $endgroup$




      Given a block matrix $M$



      $$bf M=left(begin{array}{cccc}
      a mathbb{I}_{2} & boldsymbol{boldsymbol{A}}_{12} & boldsymbol{A}_{13} & boldsymbol{0_{(2,3)}}\
      \
      boldsymbol{A_{21}} & a mathbb{I}_{2} & boldsymbol{A_{23}} & boldsymbol{0_{(2,3)}}\
      \
      boldsymbol{A_{31}} & boldsymbol{A_{32}} & a mathbb{I}_{2} & boldsymbol{0_
      {(2,3)}}\
      boldsymbol{0_{(3,2)}} & boldsymbol{0_{(3,2)}} & boldsymbol{0_{(3,2)}} & b mathbb{I}_{3}
      end{array}right)$$



      where $bf 0_{(m,n)}$ is an $ mtimes n$ matrix with all entries equal to zero.



      We need to find $bf M^{-1}$ and I wondered if there were any theorems that allow me to invert this matrix block by block?



      Also, how does the problem change for $A_{kl}=A_{lk}, k,l=1,2,3$?






      linear-algebra matrices symmetry




      linear-algebra matrices symmetry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 18 at 15:22







      usumdelphini

















      asked Jan 16 at 16:18









      usumdelphiniusumdelphini

      323111




      323111






















          1 Answer
          1






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          2












          $begingroup$

          I do not think that there is any practical useful formula for this general case.



          For block $2times 2$ matrices one generally use the Schure complements (which is nothing more than the block version of Gaussian elimination). Theoretically you can recursively use this for a $ntimes n$ ($n>2$) block matrix. However this is not a good idea because you will get overly complex formula without any practical value (with nested inverses etc...).



          Beside this comment, observe that:
          $$
          M^{-1}=left(
          begin{array}{cc}
          A^{-1}& 0 \
          0 & b^{-1}I
          end{array}
          right)
          $$
          with
          $$
          A=left(
          begin{array}{ccc}
          aI & A_{12} & A_{13} \
          A_{21} & aI & A_{23} \
          A_{31} & A_{32} & aI
          end{array}
          right)
          $$
          which is certainly the first simplification to considere.






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            I do not think that there is any practical useful formula for this general case.



            For block $2times 2$ matrices one generally use the Schure complements (which is nothing more than the block version of Gaussian elimination). Theoretically you can recursively use this for a $ntimes n$ ($n>2$) block matrix. However this is not a good idea because you will get overly complex formula without any practical value (with nested inverses etc...).



            Beside this comment, observe that:
            $$
            M^{-1}=left(
            begin{array}{cc}
            A^{-1}& 0 \
            0 & b^{-1}I
            end{array}
            right)
            $$
            with
            $$
            A=left(
            begin{array}{ccc}
            aI & A_{12} & A_{13} \
            A_{21} & aI & A_{23} \
            A_{31} & A_{32} & aI
            end{array}
            right)
            $$
            which is certainly the first simplification to considere.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              I do not think that there is any practical useful formula for this general case.



              For block $2times 2$ matrices one generally use the Schure complements (which is nothing more than the block version of Gaussian elimination). Theoretically you can recursively use this for a $ntimes n$ ($n>2$) block matrix. However this is not a good idea because you will get overly complex formula without any practical value (with nested inverses etc...).



              Beside this comment, observe that:
              $$
              M^{-1}=left(
              begin{array}{cc}
              A^{-1}& 0 \
              0 & b^{-1}I
              end{array}
              right)
              $$
              with
              $$
              A=left(
              begin{array}{ccc}
              aI & A_{12} & A_{13} \
              A_{21} & aI & A_{23} \
              A_{31} & A_{32} & aI
              end{array}
              right)
              $$
              which is certainly the first simplification to considere.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                I do not think that there is any practical useful formula for this general case.



                For block $2times 2$ matrices one generally use the Schure complements (which is nothing more than the block version of Gaussian elimination). Theoretically you can recursively use this for a $ntimes n$ ($n>2$) block matrix. However this is not a good idea because you will get overly complex formula without any practical value (with nested inverses etc...).



                Beside this comment, observe that:
                $$
                M^{-1}=left(
                begin{array}{cc}
                A^{-1}& 0 \
                0 & b^{-1}I
                end{array}
                right)
                $$
                with
                $$
                A=left(
                begin{array}{ccc}
                aI & A_{12} & A_{13} \
                A_{21} & aI & A_{23} \
                A_{31} & A_{32} & aI
                end{array}
                right)
                $$
                which is certainly the first simplification to considere.






                share|cite|improve this answer









                $endgroup$



                I do not think that there is any practical useful formula for this general case.



                For block $2times 2$ matrices one generally use the Schure complements (which is nothing more than the block version of Gaussian elimination). Theoretically you can recursively use this for a $ntimes n$ ($n>2$) block matrix. However this is not a good idea because you will get overly complex formula without any practical value (with nested inverses etc...).



                Beside this comment, observe that:
                $$
                M^{-1}=left(
                begin{array}{cc}
                A^{-1}& 0 \
                0 & b^{-1}I
                end{array}
                right)
                $$
                with
                $$
                A=left(
                begin{array}{ccc}
                aI & A_{12} & A_{13} \
                A_{21} & aI & A_{23} \
                A_{31} & A_{32} & aI
                end{array}
                right)
                $$
                which is certainly the first simplification to considere.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 17 at 10:14









                Picaud VincentPicaud Vincent

                1,529310




                1,529310






























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