Derivative of $f = {rm tr} left[ U^T ; {rm unvec} left( B {rm vec}(X) right) right]$ w.r.t. $X$?











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Let a block diagonal matrix reads $$B := {rm blkdiag}left(A_1, cdots, A_i, cdots, A_N right) in mathbb{R}^{MN times KN} ,$$ where $A_i in mathbb{R}^{M times K}$.



How to take the derivative of $f = {rm tr} left[ U^T ; {rm unvec} left( B {rm vec}(X) right) right]$, where $U in mathbb{R}^{M times N}$ and $X in mathbb{R}^{K times N}$ w.r.t. $X$?










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  • Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
    – user550103
    Nov 25 at 19:24










  • @user550103. yes, that's correct.
    – learning
    Nov 25 at 19:25










  • What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
    – Did
    Nov 25 at 19:49















up vote
-1
down vote

favorite












Let a block diagonal matrix reads $$B := {rm blkdiag}left(A_1, cdots, A_i, cdots, A_N right) in mathbb{R}^{MN times KN} ,$$ where $A_i in mathbb{R}^{M times K}$.



How to take the derivative of $f = {rm tr} left[ U^T ; {rm unvec} left( B {rm vec}(X) right) right]$, where $U in mathbb{R}^{M times N}$ and $X in mathbb{R}^{K times N}$ w.r.t. $X$?










share|cite|improve this question






















  • Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
    – user550103
    Nov 25 at 19:24










  • @user550103. yes, that's correct.
    – learning
    Nov 25 at 19:25










  • What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
    – Did
    Nov 25 at 19:49













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Let a block diagonal matrix reads $$B := {rm blkdiag}left(A_1, cdots, A_i, cdots, A_N right) in mathbb{R}^{MN times KN} ,$$ where $A_i in mathbb{R}^{M times K}$.



How to take the derivative of $f = {rm tr} left[ U^T ; {rm unvec} left( B {rm vec}(X) right) right]$, where $U in mathbb{R}^{M times N}$ and $X in mathbb{R}^{K times N}$ w.r.t. $X$?










share|cite|improve this question













Let a block diagonal matrix reads $$B := {rm blkdiag}left(A_1, cdots, A_i, cdots, A_N right) in mathbb{R}^{MN times KN} ,$$ where $A_i in mathbb{R}^{M times K}$.



How to take the derivative of $f = {rm tr} left[ U^T ; {rm unvec} left( B {rm vec}(X) right) right]$, where $U in mathbb{R}^{M times N}$ and $X in mathbb{R}^{K times N}$ w.r.t. $X$?







multivariable-calculus matrix-calculus






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asked Nov 25 at 19:20









learning

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275












  • Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
    – user550103
    Nov 25 at 19:24










  • @user550103. yes, that's correct.
    – learning
    Nov 25 at 19:25










  • What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
    – Did
    Nov 25 at 19:49


















  • Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
    – user550103
    Nov 25 at 19:24










  • @user550103. yes, that's correct.
    – learning
    Nov 25 at 19:25










  • What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
    – Did
    Nov 25 at 19:49
















Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
– user550103
Nov 25 at 19:24




Just clarification. does "unvec" operation creates matrix, that is reverse operation of a "vec"?
– user550103
Nov 25 at 19:24












@user550103. yes, that's correct.
– learning
Nov 25 at 19:25




@user550103. yes, that's correct.
– learning
Nov 25 at 19:25












What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
– Did
Nov 25 at 19:49




What hapened when you evaluated $$f(X+H)-f(X)$$ with $H$ small?
– Did
Nov 25 at 19:49










1 Answer
1






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votes

















up vote
1
down vote



accepted










Define the vectors
$$eqalign{
x &= {rm vec}(X) cr
u &= {rm vec}(U) cr
}$$

Write your function in terms of the vectors. Then find the differential and gradient.
$$eqalign{
f &= u^TBx = (B^Tu)^Tx cr
df &= (B^Tu)^T,dx cr
frac{partial f}{partial x} &= B^Tu cr
}$$

Now de-vectorize this to obtain a matrix result.
$$eqalign{
frac{partial f}{partial X} &= {rm unvec}(B^Tu) cr
}$$






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Define the vectors
    $$eqalign{
    x &= {rm vec}(X) cr
    u &= {rm vec}(U) cr
    }$$

    Write your function in terms of the vectors. Then find the differential and gradient.
    $$eqalign{
    f &= u^TBx = (B^Tu)^Tx cr
    df &= (B^Tu)^T,dx cr
    frac{partial f}{partial x} &= B^Tu cr
    }$$

    Now de-vectorize this to obtain a matrix result.
    $$eqalign{
    frac{partial f}{partial X} &= {rm unvec}(B^Tu) cr
    }$$






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      Define the vectors
      $$eqalign{
      x &= {rm vec}(X) cr
      u &= {rm vec}(U) cr
      }$$

      Write your function in terms of the vectors. Then find the differential and gradient.
      $$eqalign{
      f &= u^TBx = (B^Tu)^Tx cr
      df &= (B^Tu)^T,dx cr
      frac{partial f}{partial x} &= B^Tu cr
      }$$

      Now de-vectorize this to obtain a matrix result.
      $$eqalign{
      frac{partial f}{partial X} &= {rm unvec}(B^Tu) cr
      }$$






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Define the vectors
        $$eqalign{
        x &= {rm vec}(X) cr
        u &= {rm vec}(U) cr
        }$$

        Write your function in terms of the vectors. Then find the differential and gradient.
        $$eqalign{
        f &= u^TBx = (B^Tu)^Tx cr
        df &= (B^Tu)^T,dx cr
        frac{partial f}{partial x} &= B^Tu cr
        }$$

        Now de-vectorize this to obtain a matrix result.
        $$eqalign{
        frac{partial f}{partial X} &= {rm unvec}(B^Tu) cr
        }$$






        share|cite|improve this answer












        Define the vectors
        $$eqalign{
        x &= {rm vec}(X) cr
        u &= {rm vec}(U) cr
        }$$

        Write your function in terms of the vectors. Then find the differential and gradient.
        $$eqalign{
        f &= u^TBx = (B^Tu)^Tx cr
        df &= (B^Tu)^T,dx cr
        frac{partial f}{partial x} &= B^Tu cr
        }$$

        Now de-vectorize this to obtain a matrix result.
        $$eqalign{
        frac{partial f}{partial X} &= {rm unvec}(B^Tu) cr
        }$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 at 20:19









        greg

        7,3251720




        7,3251720






























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