Solving the matrix equation












2












$begingroup$


How can I solve the matrix equation of the form



$$
mathbf{SXK} + mathbf{X} = mathbf{Y}
$$



Here $mathbf{S}$ and $mathbf{K}$ are symmetric matrices, in addition $mathbf{K}$ is a sparse symmetric matrix. $X$ is the variable. Though $mathbf{S}$ and $mathbf{K}$ are symmetric, it is not invertible in general and $mathbf{X}$ is not symmetric. Is it possible to find a closed-form solution for $mathbf{X}$ ?. Is there any relevant literature study about solving such equations ?






linear-algebra matrices matrix-equations







share|cite|improve this question











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  • 1




    $begingroup$
    Are all square matrices? $S,K$ are invertible?
    $endgroup$
    – Exodd
    Dec 23 '18 at 19:45










  • $begingroup$
    @Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
    $endgroup$
    – Shew
    Dec 23 '18 at 20:05
















2












$begingroup$


How can I solve the matrix equation of the form



$$
mathbf{SXK} + mathbf{X} = mathbf{Y}
$$



Here $mathbf{S}$ and $mathbf{K}$ are symmetric matrices, in addition $mathbf{K}$ is a sparse symmetric matrix. $X$ is the variable. Though $mathbf{S}$ and $mathbf{K}$ are symmetric, it is not invertible in general and $mathbf{X}$ is not symmetric. Is it possible to find a closed-form solution for $mathbf{X}$ ?. Is there any relevant literature study about solving such equations ?






linear-algebra matrices matrix-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are all square matrices? $S,K$ are invertible?
    $endgroup$
    – Exodd
    Dec 23 '18 at 19:45










  • $begingroup$
    @Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
    $endgroup$
    – Shew
    Dec 23 '18 at 20:05














2












2








2





$begingroup$


How can I solve the matrix equation of the form



$$
mathbf{SXK} + mathbf{X} = mathbf{Y}
$$



Here $mathbf{S}$ and $mathbf{K}$ are symmetric matrices, in addition $mathbf{K}$ is a sparse symmetric matrix. $X$ is the variable. Though $mathbf{S}$ and $mathbf{K}$ are symmetric, it is not invertible in general and $mathbf{X}$ is not symmetric. Is it possible to find a closed-form solution for $mathbf{X}$ ?. Is there any relevant literature study about solving such equations ?






linear-algebra matrices matrix-equations







share|cite|improve this question











$endgroup$




How can I solve the matrix equation of the form



$$
mathbf{SXK} + mathbf{X} = mathbf{Y}
$$



Here $mathbf{S}$ and $mathbf{K}$ are symmetric matrices, in addition $mathbf{K}$ is a sparse symmetric matrix. $X$ is the variable. Though $mathbf{S}$ and $mathbf{K}$ are symmetric, it is not invertible in general and $mathbf{X}$ is not symmetric. Is it possible to find a closed-form solution for $mathbf{X}$ ?. Is there any relevant literature study about solving such equations ?






linear-algebra matrices matrix-equations




linear-algebra matrices matrix-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 20:08







Shew

















asked Dec 23 '18 at 19:33









ShewShew

596413




596413








  • 1




    $begingroup$
    Are all square matrices? $S,K$ are invertible?
    $endgroup$
    – Exodd
    Dec 23 '18 at 19:45










  • $begingroup$
    @Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
    $endgroup$
    – Shew
    Dec 23 '18 at 20:05














  • 1




    $begingroup$
    Are all square matrices? $S,K$ are invertible?
    $endgroup$
    – Exodd
    Dec 23 '18 at 19:45










  • $begingroup$
    @Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
    $endgroup$
    – Shew
    Dec 23 '18 at 20:05








1




1




$begingroup$
Are all square matrices? $S,K$ are invertible?
$endgroup$
– Exodd
Dec 23 '18 at 19:45




$begingroup$
Are all square matrices? $S,K$ are invertible?
$endgroup$
– Exodd
Dec 23 '18 at 19:45












$begingroup$
@Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
$endgroup$
– Shew
Dec 23 '18 at 20:05




$begingroup$
@Exodd, S and K are not invertible in general. though S and K are symmetric, X is not symmetric.
$endgroup$
– Shew
Dec 23 '18 at 20:05










2 Answers
2






active

oldest

votes


















2












$begingroup$

The symmetry of $S$ and $K$ allow for you to write them as diagonalized matrices. Suppose we have



$$S=PDP^{-1},;;;;K=QCQ^{-1}$$



for diagonal matrices $C$ and $D$. We also may write $X$ as



$$X=PP^{-1}XQQ^{-1}$$



which gives us



$$PDP^{-1}XQCQ^{-1}+PP^{-1}XQQ^{-1}=Y$$



Letting $Z=P^{-1}XQ$, we have



$$PDZCQ^{-1}+PQ^{-1}Z=Y$$



or



$$DZC+Z=P^{-1}YQ$$



The matrix $DZC$ is the matrix $Z$ with rows scaled by elements of $D$ and columns scaled by elements of $C$. This leads to a system of equations that look like



$$[D]_i[Z]_{ij}[C]_j+[Z]_{ij}=[P^{-1}YQ]_{ij}$$



The existence and uniqueness of a solution can now be seen, as it is dependent upon the value of $[D]_i[C]_j$. Provided that this is not $-1$, then



$$[Z]_{ij}=frac{[P^{-1}YQ]_{ij}}{[D]_i[C]_j+1}$$



The matrix $X$ may the be recovered by $X=PZQ^{-1}$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    This equation is similar to the discrete Lyapunov equation and can be solved in a similar way. Using the equality
    $$ operatorname{vec}(ABC)=(C^{T} otimes A)operatorname{vec}(B) $$
    one obtains the system of linear equations
    $$
    left( K^T otimes S+I_{n^2} right)operatorname{vec}(X)=operatorname{vec}(Y).
    $$






    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









      2












      $begingroup$

      The symmetry of $S$ and $K$ allow for you to write them as diagonalized matrices. Suppose we have



      $$S=PDP^{-1},;;;;K=QCQ^{-1}$$



      for diagonal matrices $C$ and $D$. We also may write $X$ as



      $$X=PP^{-1}XQQ^{-1}$$



      which gives us



      $$PDP^{-1}XQCQ^{-1}+PP^{-1}XQQ^{-1}=Y$$



      Letting $Z=P^{-1}XQ$, we have



      $$PDZCQ^{-1}+PQ^{-1}Z=Y$$



      or



      $$DZC+Z=P^{-1}YQ$$



      The matrix $DZC$ is the matrix $Z$ with rows scaled by elements of $D$ and columns scaled by elements of $C$. This leads to a system of equations that look like



      $$[D]_i[Z]_{ij}[C]_j+[Z]_{ij}=[P^{-1}YQ]_{ij}$$



      The existence and uniqueness of a solution can now be seen, as it is dependent upon the value of $[D]_i[C]_j$. Provided that this is not $-1$, then



      $$[Z]_{ij}=frac{[P^{-1}YQ]_{ij}}{[D]_i[C]_j+1}$$



      The matrix $X$ may the be recovered by $X=PZQ^{-1}$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        The symmetry of $S$ and $K$ allow for you to write them as diagonalized matrices. Suppose we have



        $$S=PDP^{-1},;;;;K=QCQ^{-1}$$



        for diagonal matrices $C$ and $D$. We also may write $X$ as



        $$X=PP^{-1}XQQ^{-1}$$



        which gives us



        $$PDP^{-1}XQCQ^{-1}+PP^{-1}XQQ^{-1}=Y$$



        Letting $Z=P^{-1}XQ$, we have



        $$PDZCQ^{-1}+PQ^{-1}Z=Y$$



        or



        $$DZC+Z=P^{-1}YQ$$



        The matrix $DZC$ is the matrix $Z$ with rows scaled by elements of $D$ and columns scaled by elements of $C$. This leads to a system of equations that look like



        $$[D]_i[Z]_{ij}[C]_j+[Z]_{ij}=[P^{-1}YQ]_{ij}$$



        The existence and uniqueness of a solution can now be seen, as it is dependent upon the value of $[D]_i[C]_j$. Provided that this is not $-1$, then



        $$[Z]_{ij}=frac{[P^{-1}YQ]_{ij}}{[D]_i[C]_j+1}$$



        The matrix $X$ may the be recovered by $X=PZQ^{-1}$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          The symmetry of $S$ and $K$ allow for you to write them as diagonalized matrices. Suppose we have



          $$S=PDP^{-1},;;;;K=QCQ^{-1}$$



          for diagonal matrices $C$ and $D$. We also may write $X$ as



          $$X=PP^{-1}XQQ^{-1}$$



          which gives us



          $$PDP^{-1}XQCQ^{-1}+PP^{-1}XQQ^{-1}=Y$$



          Letting $Z=P^{-1}XQ$, we have



          $$PDZCQ^{-1}+PQ^{-1}Z=Y$$



          or



          $$DZC+Z=P^{-1}YQ$$



          The matrix $DZC$ is the matrix $Z$ with rows scaled by elements of $D$ and columns scaled by elements of $C$. This leads to a system of equations that look like



          $$[D]_i[Z]_{ij}[C]_j+[Z]_{ij}=[P^{-1}YQ]_{ij}$$



          The existence and uniqueness of a solution can now be seen, as it is dependent upon the value of $[D]_i[C]_j$. Provided that this is not $-1$, then



          $$[Z]_{ij}=frac{[P^{-1}YQ]_{ij}}{[D]_i[C]_j+1}$$



          The matrix $X$ may the be recovered by $X=PZQ^{-1}$.






          share|cite|improve this answer









          $endgroup$



          The symmetry of $S$ and $K$ allow for you to write them as diagonalized matrices. Suppose we have



          $$S=PDP^{-1},;;;;K=QCQ^{-1}$$



          for diagonal matrices $C$ and $D$. We also may write $X$ as



          $$X=PP^{-1}XQQ^{-1}$$



          which gives us



          $$PDP^{-1}XQCQ^{-1}+PP^{-1}XQQ^{-1}=Y$$



          Letting $Z=P^{-1}XQ$, we have



          $$PDZCQ^{-1}+PQ^{-1}Z=Y$$



          or



          $$DZC+Z=P^{-1}YQ$$



          The matrix $DZC$ is the matrix $Z$ with rows scaled by elements of $D$ and columns scaled by elements of $C$. This leads to a system of equations that look like



          $$[D]_i[Z]_{ij}[C]_j+[Z]_{ij}=[P^{-1}YQ]_{ij}$$



          The existence and uniqueness of a solution can now be seen, as it is dependent upon the value of $[D]_i[C]_j$. Provided that this is not $-1$, then



          $$[Z]_{ij}=frac{[P^{-1}YQ]_{ij}}{[D]_i[C]_j+1}$$



          The matrix $X$ may the be recovered by $X=PZQ^{-1}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 20:33









          Josh B.Josh B.

          1,0298




          1,0298























              2












              $begingroup$

              This equation is similar to the discrete Lyapunov equation and can be solved in a similar way. Using the equality
              $$ operatorname{vec}(ABC)=(C^{T} otimes A)operatorname{vec}(B) $$
              one obtains the system of linear equations
              $$
              left( K^T otimes S+I_{n^2} right)operatorname{vec}(X)=operatorname{vec}(Y).
              $$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                This equation is similar to the discrete Lyapunov equation and can be solved in a similar way. Using the equality
                $$ operatorname{vec}(ABC)=(C^{T} otimes A)operatorname{vec}(B) $$
                one obtains the system of linear equations
                $$
                left( K^T otimes S+I_{n^2} right)operatorname{vec}(X)=operatorname{vec}(Y).
                $$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  This equation is similar to the discrete Lyapunov equation and can be solved in a similar way. Using the equality
                  $$ operatorname{vec}(ABC)=(C^{T} otimes A)operatorname{vec}(B) $$
                  one obtains the system of linear equations
                  $$
                  left( K^T otimes S+I_{n^2} right)operatorname{vec}(X)=operatorname{vec}(Y).
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  This equation is similar to the discrete Lyapunov equation and can be solved in a similar way. Using the equality
                  $$ operatorname{vec}(ABC)=(C^{T} otimes A)operatorname{vec}(B) $$
                  one obtains the system of linear equations
                  $$
                  left( K^T otimes S+I_{n^2} right)operatorname{vec}(X)=operatorname{vec}(Y).
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 23 '18 at 20:42









                  AVKAVK

                  2,1561518




                  2,1561518






























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