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Given two side matrices $P$ and $Q$, extract (find) the diagonal scaling matrix $Sigma$ of a singular value decomposition



The 2019 Stack Overflow Developer Survey Results Are InSingular Value Decomposition for zero-diagonal symmetric matrixDecomposing a stochastic matrix into a product of stochastic matricesDiagonal and anti-diagonal integral matrices: a special decompositionSingular Value Decomposition and Square matricesSingular Value Decomposition of Commuting MatricesSingular value decomposition of a matrix multiplicationIs it possible to compute derivative of truncated SVD without computing a full SVD?Unique decomposition of a positive definite matrix into a sum of outer products $bf x_kbf x_k^rm T$ and a diagonal matrix?Singular Value Decomposition for Rectangular MatricesSingular value decomposition of Block Diagonal matrix.










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$begingroup$


I have an application where I have already approximated a given matrix $R$ of size $m times n$ by multiplying two matrices $P$ and $Q^mathrm T$: $hat R=PQ^mathrm T$. $P$ is size $m times k$ and $Q$ is size $n times k$ and $Q^mathrm T$ is size $k times n$. I desire now to use these two matrices to efficiently as possible find a proper singular value decomposition, which has three matrices as you know.



What gives me great hope is that Simon Funk said here that "The end result, it's worth noting, is exactly an SVD if the training set perfectly covers the matrix. Call it what you will when it doesn't. [If you're wondering where the diagonal scaling matrix is, it gets arbitrarily rolled in to the two side matrices, but could be trivially extracted if needed.]"



Can someone describe and detail the trivial extraction process he talked about which I can use to find that third matrix $Sigma$ in the famous SVD equation $hat R = U Sigma V^mathrm T$?



Never mind FunkSVD, as I am not using that algorithm currently, but I do have a pretty well-estimated pair of matrices $P$ and $Q$ as my starting point. I used a gradient descent and machine learning to get $P$ and $Q$ already.



I am required to NOT run SVD from scratch -- instead I must do something very efficient to "trivially extract" the sigma matrix, when given "two side matrices", which dear Mr. Funk said is possible.



Thanks for contributions if any!










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    I have an application where I have already approximated a given matrix $R$ of size $m times n$ by multiplying two matrices $P$ and $Q^mathrm T$: $hat R=PQ^mathrm T$. $P$ is size $m times k$ and $Q$ is size $n times k$ and $Q^mathrm T$ is size $k times n$. I desire now to use these two matrices to efficiently as possible find a proper singular value decomposition, which has three matrices as you know.



    What gives me great hope is that Simon Funk said here that "The end result, it's worth noting, is exactly an SVD if the training set perfectly covers the matrix. Call it what you will when it doesn't. [If you're wondering where the diagonal scaling matrix is, it gets arbitrarily rolled in to the two side matrices, but could be trivially extracted if needed.]"



    Can someone describe and detail the trivial extraction process he talked about which I can use to find that third matrix $Sigma$ in the famous SVD equation $hat R = U Sigma V^mathrm T$?



    Never mind FunkSVD, as I am not using that algorithm currently, but I do have a pretty well-estimated pair of matrices $P$ and $Q$ as my starting point. I used a gradient descent and machine learning to get $P$ and $Q$ already.



    I am required to NOT run SVD from scratch -- instead I must do something very efficient to "trivially extract" the sigma matrix, when given "two side matrices", which dear Mr. Funk said is possible.



    Thanks for contributions if any!










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      I have an application where I have already approximated a given matrix $R$ of size $m times n$ by multiplying two matrices $P$ and $Q^mathrm T$: $hat R=PQ^mathrm T$. $P$ is size $m times k$ and $Q$ is size $n times k$ and $Q^mathrm T$ is size $k times n$. I desire now to use these two matrices to efficiently as possible find a proper singular value decomposition, which has three matrices as you know.



      What gives me great hope is that Simon Funk said here that "The end result, it's worth noting, is exactly an SVD if the training set perfectly covers the matrix. Call it what you will when it doesn't. [If you're wondering where the diagonal scaling matrix is, it gets arbitrarily rolled in to the two side matrices, but could be trivially extracted if needed.]"



      Can someone describe and detail the trivial extraction process he talked about which I can use to find that third matrix $Sigma$ in the famous SVD equation $hat R = U Sigma V^mathrm T$?



      Never mind FunkSVD, as I am not using that algorithm currently, but I do have a pretty well-estimated pair of matrices $P$ and $Q$ as my starting point. I used a gradient descent and machine learning to get $P$ and $Q$ already.



      I am required to NOT run SVD from scratch -- instead I must do something very efficient to "trivially extract" the sigma matrix, when given "two side matrices", which dear Mr. Funk said is possible.



      Thanks for contributions if any!










      share|cite|improve this question











      $endgroup$




      I have an application where I have already approximated a given matrix $R$ of size $m times n$ by multiplying two matrices $P$ and $Q^mathrm T$: $hat R=PQ^mathrm T$. $P$ is size $m times k$ and $Q$ is size $n times k$ and $Q^mathrm T$ is size $k times n$. I desire now to use these two matrices to efficiently as possible find a proper singular value decomposition, which has three matrices as you know.



      What gives me great hope is that Simon Funk said here that "The end result, it's worth noting, is exactly an SVD if the training set perfectly covers the matrix. Call it what you will when it doesn't. [If you're wondering where the diagonal scaling matrix is, it gets arbitrarily rolled in to the two side matrices, but could be trivially extracted if needed.]"



      Can someone describe and detail the trivial extraction process he talked about which I can use to find that third matrix $Sigma$ in the famous SVD equation $hat R = U Sigma V^mathrm T$?



      Never mind FunkSVD, as I am not using that algorithm currently, but I do have a pretty well-estimated pair of matrices $P$ and $Q$ as my starting point. I used a gradient descent and machine learning to get $P$ and $Q$ already.



      I am required to NOT run SVD from scratch -- instead I must do something very efficient to "trivially extract" the sigma matrix, when given "two side matrices", which dear Mr. Funk said is possible.



      Thanks for contributions if any!







      matrix-decomposition






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 23 at 17:26









      Rócherz

      3,0263823




      3,0263823










      asked Aug 2 '18 at 19:30









      Geoffrey AndersonGeoffrey Anderson

      1012




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