Solving a matrix least squares problem with a fixed Frobenius norm constraint












-1












$begingroup$


I am trying to solve for X an equation of the form :



Min ||AXB-CXD|| s.t. ||X||_F=1,



where , A, C are m-by-m matrices, and B, D are n-by-b matrices. Is there any effective algorithms?



Any hint is greatly appreciated :-)






numerical-linear-algebra







share|cite|improve this question









$endgroup$

















    -1












    $begingroup$


    I am trying to solve for X an equation of the form :



    Min ||AXB-CXD|| s.t. ||X||_F=1,



    where , A, C are m-by-m matrices, and B, D are n-by-b matrices. Is there any effective algorithms?



    Any hint is greatly appreciated :-)






    numerical-linear-algebra







    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1





      $begingroup$


      I am trying to solve for X an equation of the form :



      Min ||AXB-CXD|| s.t. ||X||_F=1,



      where , A, C are m-by-m matrices, and B, D are n-by-b matrices. Is there any effective algorithms?



      Any hint is greatly appreciated :-)






      numerical-linear-algebra







      share|cite|improve this question









      $endgroup$




      I am trying to solve for X an equation of the form :



      Min ||AXB-CXD|| s.t. ||X||_F=1,



      where , A, C are m-by-m matrices, and B, D are n-by-b matrices. Is there any effective algorithms?



      Any hint is greatly appreciated :-)






      numerical-linear-algebra




      numerical-linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 31 at 1:35









      Jiao-fen LiJiao-fen Li

      11




      11






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          Your problem is equivalent to



          $min | AXB - CXD |_{F}^{2} $



          subject to



          $| X |_{F}^{2} = 1$.



          Here, squaring the norms simplifies the calculations below without changing the optimal solutions.



          The key idea is that this problem is really a linear least squares problem in the entries of the matrix $X$. If you let the vector $u$ be defined by



          $u=mbox{vec}(X)=left[
          begin{array}{c}
          X_{1,1} \
          X_{2,1} \
          vdots \
          X_{m,1} \
          X_{1,2} \
          X_{2,2} \
          vdots \
          vdots \
          X_{m,n}
          end{array}
          right]$



          Then $mbox{vec}(AXB-CXD)$ can be written as



          $mbox{vec}(AXB-CXD)=Gu$



          where the entries of G can be computed by a tedious calculation. In terms of $G$ and $u$, your problem is now



          $min | Gu |_{2}^{2}$



          subject to



          $| u |_{2}^{2}=1$.



          This can be written as



          $min u^{T}G^{T}Gu$



          subject to



          $u^{T}u=1$.



          This is simply the problem of finding a normalized eigenvector of $G^{T}G$ corresponding to the smallest eigenvalue of $G^{T}G$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for your reply
            $endgroup$
            – Jiao-fen Li
            Jan 31 at 3:10












          Your Answer





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          1 Answer
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          active

          oldest

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          1












          $begingroup$

          Your problem is equivalent to



          $min | AXB - CXD |_{F}^{2} $



          subject to



          $| X |_{F}^{2} = 1$.



          Here, squaring the norms simplifies the calculations below without changing the optimal solutions.



          The key idea is that this problem is really a linear least squares problem in the entries of the matrix $X$. If you let the vector $u$ be defined by



          $u=mbox{vec}(X)=left[
          begin{array}{c}
          X_{1,1} \
          X_{2,1} \
          vdots \
          X_{m,1} \
          X_{1,2} \
          X_{2,2} \
          vdots \
          vdots \
          X_{m,n}
          end{array}
          right]$



          Then $mbox{vec}(AXB-CXD)$ can be written as



          $mbox{vec}(AXB-CXD)=Gu$



          where the entries of G can be computed by a tedious calculation. In terms of $G$ and $u$, your problem is now



          $min | Gu |_{2}^{2}$



          subject to



          $| u |_{2}^{2}=1$.



          This can be written as



          $min u^{T}G^{T}Gu$



          subject to



          $u^{T}u=1$.



          This is simply the problem of finding a normalized eigenvector of $G^{T}G$ corresponding to the smallest eigenvalue of $G^{T}G$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for your reply
            $endgroup$
            – Jiao-fen Li
            Jan 31 at 3:10
















          1












          $begingroup$

          Your problem is equivalent to



          $min | AXB - CXD |_{F}^{2} $



          subject to



          $| X |_{F}^{2} = 1$.



          Here, squaring the norms simplifies the calculations below without changing the optimal solutions.



          The key idea is that this problem is really a linear least squares problem in the entries of the matrix $X$. If you let the vector $u$ be defined by



          $u=mbox{vec}(X)=left[
          begin{array}{c}
          X_{1,1} \
          X_{2,1} \
          vdots \
          X_{m,1} \
          X_{1,2} \
          X_{2,2} \
          vdots \
          vdots \
          X_{m,n}
          end{array}
          right]$



          Then $mbox{vec}(AXB-CXD)$ can be written as



          $mbox{vec}(AXB-CXD)=Gu$



          where the entries of G can be computed by a tedious calculation. In terms of $G$ and $u$, your problem is now



          $min | Gu |_{2}^{2}$



          subject to



          $| u |_{2}^{2}=1$.



          This can be written as



          $min u^{T}G^{T}Gu$



          subject to



          $u^{T}u=1$.



          This is simply the problem of finding a normalized eigenvector of $G^{T}G$ corresponding to the smallest eigenvalue of $G^{T}G$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for your reply
            $endgroup$
            – Jiao-fen Li
            Jan 31 at 3:10














          1












          1








          1





          $begingroup$

          Your problem is equivalent to



          $min | AXB - CXD |_{F}^{2} $



          subject to



          $| X |_{F}^{2} = 1$.



          Here, squaring the norms simplifies the calculations below without changing the optimal solutions.



          The key idea is that this problem is really a linear least squares problem in the entries of the matrix $X$. If you let the vector $u$ be defined by



          $u=mbox{vec}(X)=left[
          begin{array}{c}
          X_{1,1} \
          X_{2,1} \
          vdots \
          X_{m,1} \
          X_{1,2} \
          X_{2,2} \
          vdots \
          vdots \
          X_{m,n}
          end{array}
          right]$



          Then $mbox{vec}(AXB-CXD)$ can be written as



          $mbox{vec}(AXB-CXD)=Gu$



          where the entries of G can be computed by a tedious calculation. In terms of $G$ and $u$, your problem is now



          $min | Gu |_{2}^{2}$



          subject to



          $| u |_{2}^{2}=1$.



          This can be written as



          $min u^{T}G^{T}Gu$



          subject to



          $u^{T}u=1$.



          This is simply the problem of finding a normalized eigenvector of $G^{T}G$ corresponding to the smallest eigenvalue of $G^{T}G$.






          share|cite|improve this answer









          $endgroup$



          Your problem is equivalent to



          $min | AXB - CXD |_{F}^{2} $



          subject to



          $| X |_{F}^{2} = 1$.



          Here, squaring the norms simplifies the calculations below without changing the optimal solutions.



          The key idea is that this problem is really a linear least squares problem in the entries of the matrix $X$. If you let the vector $u$ be defined by



          $u=mbox{vec}(X)=left[
          begin{array}{c}
          X_{1,1} \
          X_{2,1} \
          vdots \
          X_{m,1} \
          X_{1,2} \
          X_{2,2} \
          vdots \
          vdots \
          X_{m,n}
          end{array}
          right]$



          Then $mbox{vec}(AXB-CXD)$ can be written as



          $mbox{vec}(AXB-CXD)=Gu$



          where the entries of G can be computed by a tedious calculation. In terms of $G$ and $u$, your problem is now



          $min | Gu |_{2}^{2}$



          subject to



          $| u |_{2}^{2}=1$.



          This can be written as



          $min u^{T}G^{T}Gu$



          subject to



          $u^{T}u=1$.



          This is simply the problem of finding a normalized eigenvector of $G^{T}G$ corresponding to the smallest eigenvalue of $G^{T}G$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 31 at 2:07









          Brian BorchersBrian Borchers

          6,27611320




          6,27611320












          • $begingroup$
            Thank you very much for your reply
            $endgroup$
            – Jiao-fen Li
            Jan 31 at 3:10


















          • $begingroup$
            Thank you very much for your reply
            $endgroup$
            – Jiao-fen Li
            Jan 31 at 3:10
















          $begingroup$
          Thank you very much for your reply
          $endgroup$
          – Jiao-fen Li
          Jan 31 at 3:10




          $begingroup$
          Thank you very much for your reply
          $endgroup$
          – Jiao-fen Li
          Jan 31 at 3:10


















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