What is the scalar derivative?












1












$begingroup$


I quote paragraph 2.5 of The Matrix Cookbook document: Assume $F(X)$ to be a differentiable function of each of the elements of $X$... $f(cdot)$ is the scalar derivative of $F(cdot)$. $X$ is here a matrix.



What is the scalar derivative? It is not defined in this document and I have issues to find a definition using Mister Google.



But the way, I'm puzzled by formula (100) of that document:
$$frac{partial}{partial X} mathsf{Tr}(XA) = A^T$$



$X mapsto {Tr}(XA)$ is a linear form defined on the matrices vector space and therefore it's derivative is itself everywhere



$$frac{partial}{partial X} mathsf{Tr}(XA).H = mathsf{Tr}(HA)$$



What is the link with $A^T$?






matrices derivatives







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  • $begingroup$
    Mister Wikipedia provides some help.
    $endgroup$
    – Paul Sinclair
    Jan 10 '18 at 0:33
















1












$begingroup$


I quote paragraph 2.5 of The Matrix Cookbook document: Assume $F(X)$ to be a differentiable function of each of the elements of $X$... $f(cdot)$ is the scalar derivative of $F(cdot)$. $X$ is here a matrix.



What is the scalar derivative? It is not defined in this document and I have issues to find a definition using Mister Google.



But the way, I'm puzzled by formula (100) of that document:
$$frac{partial}{partial X} mathsf{Tr}(XA) = A^T$$



$X mapsto {Tr}(XA)$ is a linear form defined on the matrices vector space and therefore it's derivative is itself everywhere



$$frac{partial}{partial X} mathsf{Tr}(XA).H = mathsf{Tr}(HA)$$



What is the link with $A^T$?






matrices derivatives







share|cite|improve this question









$endgroup$












  • $begingroup$
    Mister Wikipedia provides some help.
    $endgroup$
    – Paul Sinclair
    Jan 10 '18 at 0:33














1












1








1





$begingroup$


I quote paragraph 2.5 of The Matrix Cookbook document: Assume $F(X)$ to be a differentiable function of each of the elements of $X$... $f(cdot)$ is the scalar derivative of $F(cdot)$. $X$ is here a matrix.



What is the scalar derivative? It is not defined in this document and I have issues to find a definition using Mister Google.



But the way, I'm puzzled by formula (100) of that document:
$$frac{partial}{partial X} mathsf{Tr}(XA) = A^T$$



$X mapsto {Tr}(XA)$ is a linear form defined on the matrices vector space and therefore it's derivative is itself everywhere



$$frac{partial}{partial X} mathsf{Tr}(XA).H = mathsf{Tr}(HA)$$



What is the link with $A^T$?






matrices derivatives







share|cite|improve this question









$endgroup$




I quote paragraph 2.5 of The Matrix Cookbook document: Assume $F(X)$ to be a differentiable function of each of the elements of $X$... $f(cdot)$ is the scalar derivative of $F(cdot)$. $X$ is here a matrix.



What is the scalar derivative? It is not defined in this document and I have issues to find a definition using Mister Google.



But the way, I'm puzzled by formula (100) of that document:
$$frac{partial}{partial X} mathsf{Tr}(XA) = A^T$$



$X mapsto {Tr}(XA)$ is a linear form defined on the matrices vector space and therefore it's derivative is itself everywhere



$$frac{partial}{partial X} mathsf{Tr}(XA).H = mathsf{Tr}(HA)$$



What is the link with $A^T$?






matrices derivatives




matrices derivatives






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 '18 at 18:35









mathcounterexamples.netmathcounterexamples.net

26.9k22157




26.9k22157












  • $begingroup$
    Mister Wikipedia provides some help.
    $endgroup$
    – Paul Sinclair
    Jan 10 '18 at 0:33


















  • $begingroup$
    Mister Wikipedia provides some help.
    $endgroup$
    – Paul Sinclair
    Jan 10 '18 at 0:33
















$begingroup$
Mister Wikipedia provides some help.
$endgroup$
– Paul Sinclair
Jan 10 '18 at 0:33




$begingroup$
Mister Wikipedia provides some help.
$endgroup$
– Paul Sinclair
Jan 10 '18 at 0:33










2 Answers
2






active

oldest

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0












$begingroup$

For applied matrix calculus in deep learning the term 'scalar derivative' is used to explicitly confirm that the output of the partial derivative of the function with respect to a variable is a scalar and not a vector.



@mathcounterexamples.net See:




  • http://cs231n.stanford.edu/vecDerivs.pdf

  • https://arxiv.org/pdf/1802.01528.pdf






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The simplest explanation is that the word $scalar$ is a typo.



    The formula itself seem correct. For instance, let
    $$eqalign{
    F(x) &= sin(x) cr
    f(x) &= frac{dF}{dx} = cos(x) cr
    }$$
    Then, for a matrix argument $A$, one has the result
    $$eqalign{
    frac{partial,{rm Tr}(sin(A))}{partial A} &= cos(A)^T cr
    }$$
    ...or $cos(A)$ depending on which layout convention you prefer.






    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









      0












      $begingroup$

      For applied matrix calculus in deep learning the term 'scalar derivative' is used to explicitly confirm that the output of the partial derivative of the function with respect to a variable is a scalar and not a vector.



      @mathcounterexamples.net See:




      • http://cs231n.stanford.edu/vecDerivs.pdf

      • https://arxiv.org/pdf/1802.01528.pdf






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        For applied matrix calculus in deep learning the term 'scalar derivative' is used to explicitly confirm that the output of the partial derivative of the function with respect to a variable is a scalar and not a vector.



        @mathcounterexamples.net See:




        • http://cs231n.stanford.edu/vecDerivs.pdf

        • https://arxiv.org/pdf/1802.01528.pdf






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          For applied matrix calculus in deep learning the term 'scalar derivative' is used to explicitly confirm that the output of the partial derivative of the function with respect to a variable is a scalar and not a vector.



          @mathcounterexamples.net See:




          • http://cs231n.stanford.edu/vecDerivs.pdf

          • https://arxiv.org/pdf/1802.01528.pdf






          share|cite|improve this answer









          $endgroup$



          For applied matrix calculus in deep learning the term 'scalar derivative' is used to explicitly confirm that the output of the partial derivative of the function with respect to a variable is a scalar and not a vector.



          @mathcounterexamples.net See:




          • http://cs231n.stanford.edu/vecDerivs.pdf

          • https://arxiv.org/pdf/1802.01528.pdf







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 '18 at 14:30









          Matthew ArthurMatthew Arthur

          183




          183























              0












              $begingroup$

              The simplest explanation is that the word $scalar$ is a typo.



              The formula itself seem correct. For instance, let
              $$eqalign{
              F(x) &= sin(x) cr
              f(x) &= frac{dF}{dx} = cos(x) cr
              }$$
              Then, for a matrix argument $A$, one has the result
              $$eqalign{
              frac{partial,{rm Tr}(sin(A))}{partial A} &= cos(A)^T cr
              }$$
              ...or $cos(A)$ depending on which layout convention you prefer.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The simplest explanation is that the word $scalar$ is a typo.



                The formula itself seem correct. For instance, let
                $$eqalign{
                F(x) &= sin(x) cr
                f(x) &= frac{dF}{dx} = cos(x) cr
                }$$
                Then, for a matrix argument $A$, one has the result
                $$eqalign{
                frac{partial,{rm Tr}(sin(A))}{partial A} &= cos(A)^T cr
                }$$
                ...or $cos(A)$ depending on which layout convention you prefer.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The simplest explanation is that the word $scalar$ is a typo.



                  The formula itself seem correct. For instance, let
                  $$eqalign{
                  F(x) &= sin(x) cr
                  f(x) &= frac{dF}{dx} = cos(x) cr
                  }$$
                  Then, for a matrix argument $A$, one has the result
                  $$eqalign{
                  frac{partial,{rm Tr}(sin(A))}{partial A} &= cos(A)^T cr
                  }$$
                  ...or $cos(A)$ depending on which layout convention you prefer.






                  share|cite|improve this answer









                  $endgroup$



                  The simplest explanation is that the word $scalar$ is a typo.



                  The formula itself seem correct. For instance, let
                  $$eqalign{
                  F(x) &= sin(x) cr
                  f(x) &= frac{dF}{dx} = cos(x) cr
                  }$$
                  Then, for a matrix argument $A$, one has the result
                  $$eqalign{
                  frac{partial,{rm Tr}(sin(A))}{partial A} &= cos(A)^T cr
                  }$$
                  ...or $cos(A)$ depending on which layout convention you prefer.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 '18 at 3:20









                  greggreg

                  8,5851823




                  8,5851823






























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