Inverse of identity minus matrix exponential












2












$begingroup$


I am trying to analytically find the inverse of a matrix given by:
begin{align}
W = left( I - alpha e^A right)^{-1},
end{align}
where $I$ is the identity matrix of appropriate size, $e^A$ denotes matrix exponential of $A$ and



begin{align}
A = begin{bmatrix}
frac{ (1-1)^2 }{ sigma^2 } & frac{ (2-1)^2 }{ sigma^2 } &
frac{ (3-1)^2 }{ sigma^2 } & dots & frac{ (n-1)^2 }{ sigma^2 } \
frac{ (1-2)^2 }{ sigma^2 } & frac{ (2-2)^2 }{ sigma^2 } &
frac{ (3-2)^2 }{ sigma^2 } & dots & frac{ (n-2)^2 }{ sigma^2 } \
vdots & vdots & vdots & ddots & vdots \
frac{ (1-N)^2 }{ sigma^2 } & frac{ (2-N)^2 }{ sigma^2 } &
frac{ (3-N)^2 }{ sigma^2 } & dots & frac{ (N-N)^2 }{ sigma^2 } \
end{bmatrix}
end{align}



Any help would be much appreciated!



Thank you very much,



Katie






matrices inverse matrix-exponential







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I am trying to analytically find the inverse of a matrix given by:
    begin{align}
    W = left( I - alpha e^A right)^{-1},
    end{align}
    where $I$ is the identity matrix of appropriate size, $e^A$ denotes matrix exponential of $A$ and



    begin{align}
    A = begin{bmatrix}
    frac{ (1-1)^2 }{ sigma^2 } & frac{ (2-1)^2 }{ sigma^2 } &
    frac{ (3-1)^2 }{ sigma^2 } & dots & frac{ (n-1)^2 }{ sigma^2 } \
    frac{ (1-2)^2 }{ sigma^2 } & frac{ (2-2)^2 }{ sigma^2 } &
    frac{ (3-2)^2 }{ sigma^2 } & dots & frac{ (n-2)^2 }{ sigma^2 } \
    vdots & vdots & vdots & ddots & vdots \
    frac{ (1-N)^2 }{ sigma^2 } & frac{ (2-N)^2 }{ sigma^2 } &
    frac{ (3-N)^2 }{ sigma^2 } & dots & frac{ (N-N)^2 }{ sigma^2 } \
    end{bmatrix}
    end{align}



    Any help would be much appreciated!



    Thank you very much,



    Katie






    matrices inverse matrix-exponential







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I am trying to analytically find the inverse of a matrix given by:
      begin{align}
      W = left( I - alpha e^A right)^{-1},
      end{align}
      where $I$ is the identity matrix of appropriate size, $e^A$ denotes matrix exponential of $A$ and



      begin{align}
      A = begin{bmatrix}
      frac{ (1-1)^2 }{ sigma^2 } & frac{ (2-1)^2 }{ sigma^2 } &
      frac{ (3-1)^2 }{ sigma^2 } & dots & frac{ (n-1)^2 }{ sigma^2 } \
      frac{ (1-2)^2 }{ sigma^2 } & frac{ (2-2)^2 }{ sigma^2 } &
      frac{ (3-2)^2 }{ sigma^2 } & dots & frac{ (n-2)^2 }{ sigma^2 } \
      vdots & vdots & vdots & ddots & vdots \
      frac{ (1-N)^2 }{ sigma^2 } & frac{ (2-N)^2 }{ sigma^2 } &
      frac{ (3-N)^2 }{ sigma^2 } & dots & frac{ (N-N)^2 }{ sigma^2 } \
      end{bmatrix}
      end{align}



      Any help would be much appreciated!



      Thank you very much,



      Katie






      matrices inverse matrix-exponential







      share|cite|improve this question









      $endgroup$




      I am trying to analytically find the inverse of a matrix given by:
      begin{align}
      W = left( I - alpha e^A right)^{-1},
      end{align}
      where $I$ is the identity matrix of appropriate size, $e^A$ denotes matrix exponential of $A$ and



      begin{align}
      A = begin{bmatrix}
      frac{ (1-1)^2 }{ sigma^2 } & frac{ (2-1)^2 }{ sigma^2 } &
      frac{ (3-1)^2 }{ sigma^2 } & dots & frac{ (n-1)^2 }{ sigma^2 } \
      frac{ (1-2)^2 }{ sigma^2 } & frac{ (2-2)^2 }{ sigma^2 } &
      frac{ (3-2)^2 }{ sigma^2 } & dots & frac{ (n-2)^2 }{ sigma^2 } \
      vdots & vdots & vdots & ddots & vdots \
      frac{ (1-N)^2 }{ sigma^2 } & frac{ (2-N)^2 }{ sigma^2 } &
      frac{ (3-N)^2 }{ sigma^2 } & dots & frac{ (N-N)^2 }{ sigma^2 } \
      end{bmatrix}
      end{align}



      Any help would be much appreciated!



      Thank you very much,



      Katie






      matrices inverse matrix-exponential




      matrices inverse matrix-exponential






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 14 at 11:58









      KatieKatie

      636




      636






















          2 Answers
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          $begingroup$

          Define the variables
          $$eqalign{
          lambda &= log(alpha) &implies alpha = e^lambda cr
          X &= -tfrac{1}{2}(A+lambda I) cr
          }$$
          Then
          $$eqalign{
          W^{-1} &= Big(I-e^{-2X}Big) &= 2e^{-X}sinh(X) cr
          W &= Big(I-e^{-2X}Big)^{-1} &= tfrac{1}{2}e^{X}{,rm csch}(X) cr
          }$$






          share|cite|improve this answer











          $endgroup$





















            -1












            $begingroup$

            You can take out $frac{1}{sigma^2}$. Notice also that your matrix is symmetric, and of the form: Find rank of the matrix $a_{ij}=(i-j)^2$, $i,j=1,dots, n$
            You can make it upper triangular by a few transformations (after which it's easy to transform to identity).






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

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              active

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              2












              $begingroup$

              Define the variables
              $$eqalign{
              lambda &= log(alpha) &implies alpha = e^lambda cr
              X &= -tfrac{1}{2}(A+lambda I) cr
              }$$
              Then
              $$eqalign{
              W^{-1} &= Big(I-e^{-2X}Big) &= 2e^{-X}sinh(X) cr
              W &= Big(I-e^{-2X}Big)^{-1} &= tfrac{1}{2}e^{X}{,rm csch}(X) cr
              }$$






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Define the variables
                $$eqalign{
                lambda &= log(alpha) &implies alpha = e^lambda cr
                X &= -tfrac{1}{2}(A+lambda I) cr
                }$$
                Then
                $$eqalign{
                W^{-1} &= Big(I-e^{-2X}Big) &= 2e^{-X}sinh(X) cr
                W &= Big(I-e^{-2X}Big)^{-1} &= tfrac{1}{2}e^{X}{,rm csch}(X) cr
                }$$






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Define the variables
                  $$eqalign{
                  lambda &= log(alpha) &implies alpha = e^lambda cr
                  X &= -tfrac{1}{2}(A+lambda I) cr
                  }$$
                  Then
                  $$eqalign{
                  W^{-1} &= Big(I-e^{-2X}Big) &= 2e^{-X}sinh(X) cr
                  W &= Big(I-e^{-2X}Big)^{-1} &= tfrac{1}{2}e^{X}{,rm csch}(X) cr
                  }$$






                  share|cite|improve this answer











                  $endgroup$



                  Define the variables
                  $$eqalign{
                  lambda &= log(alpha) &implies alpha = e^lambda cr
                  X &= -tfrac{1}{2}(A+lambda I) cr
                  }$$
                  Then
                  $$eqalign{
                  W^{-1} &= Big(I-e^{-2X}Big) &= 2e^{-X}sinh(X) cr
                  W &= Big(I-e^{-2X}Big)^{-1} &= tfrac{1}{2}e^{X}{,rm csch}(X) cr
                  }$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 16 at 2:01

























                  answered Jan 16 at 1:44









                  greggreg

                  8,9551824




                  8,9551824























                      -1












                      $begingroup$

                      You can take out $frac{1}{sigma^2}$. Notice also that your matrix is symmetric, and of the form: Find rank of the matrix $a_{ij}=(i-j)^2$, $i,j=1,dots, n$
                      You can make it upper triangular by a few transformations (after which it's easy to transform to identity).






                      share|cite|improve this answer









                      $endgroup$


















                        -1












                        $begingroup$

                        You can take out $frac{1}{sigma^2}$. Notice also that your matrix is symmetric, and of the form: Find rank of the matrix $a_{ij}=(i-j)^2$, $i,j=1,dots, n$
                        You can make it upper triangular by a few transformations (after which it's easy to transform to identity).






                        share|cite|improve this answer









                        $endgroup$
















                          -1












                          -1








                          -1





                          $begingroup$

                          You can take out $frac{1}{sigma^2}$. Notice also that your matrix is symmetric, and of the form: Find rank of the matrix $a_{ij}=(i-j)^2$, $i,j=1,dots, n$
                          You can make it upper triangular by a few transformations (after which it's easy to transform to identity).






                          share|cite|improve this answer









                          $endgroup$



                          You can take out $frac{1}{sigma^2}$. Notice also that your matrix is symmetric, and of the form: Find rank of the matrix $a_{ij}=(i-j)^2$, $i,j=1,dots, n$
                          You can make it upper triangular by a few transformations (after which it's easy to transform to identity).







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 14 at 12:26









                          lightxbulblightxbulb

                          1,140311




                          1,140311






























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