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Right now I'm trying to find a derivative that's stumping me:

Let $A, B$ be $m \times n$ matrices and $W$ be a $p\times m$ matrix.

$f = W \bullet (A \circ B)$

$\frac{\partial f}{\partial A} = ?$

(The $\bullet$ represents matrix multiplication and the $\circ$ represents taking Hadamard product.)

From the post Derivative of Hadamard product, I've seen that for $g = A \circ B$, $\frac{\partial g}{\partial A} = B:M$, where $M$ is a 6th-order tensor with $M_{ijklmn}=1$ if $(i=k=m)$ and $(j=l=n)$, $0$ otherwise. However, I'm not sure how to deal with the $W$? Thanks.

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The gradient is a fourth-order tensor $\big(\Gamma\big)$, which is calculated as follows. $$\eqalign{ F &= W\cdot(B\circ A) \\ &= W\cdot (B:{\mathbb M}:A) \\ &= W\cdot (B:{\mathbb M}):A \\ &= (W\cdot{\mathbb M}:B):A \\ dF &= (W\cdot{\mathbb M}:B):dA \\ \Gamma = \frac{\partial F}{\partial A} &= W\cdot{\mathbb M}:B \\ }$$ Here are those last few lines in index notation. $$\eqalign{ dF_{pq} &= W_{pk}B_{ij}{\mathbb M}_{ijkqmn}\,dA_{mn} \\ \Gamma_{pqmn} = \frac{\partial F_{pq}}{\partial A_{mn}} &= W_{pk}B_{ij}{\mathbb M}_{ijkqmn} \\ &= W_{pk}{\mathbb M}_{kqmnij}B_{ij} \\ }$$ The three index-pairs on ${\mathbb M}$ can be rearranged as needed, e.g. $$\eqalign{ {\mathbb M}_{\,ij\,kq\,mn} &= {\mathbb M}_{\,ij\,kq\,mn} \\ &= {\mathbb M}_{\,kq\,mn\,ij} \\ &= {\mathbb M}_{\,ij\,mn\,kq} \\ &= etc. \\ }$$


The problem can also be approached by vectorizing the matrices. $$\eqalign{ a &= \operatorname{vec}(A),\quad b = \operatorname{vec}(B),\quad f = \operatorname{vec}(F) \\ {\cal B} &= \operatorname{Diag}(b) \\ F &= W\cdot(B\circ A)\cdot I\\ f &= (I\otimes W)\cdot(b\circ a) \\ df &= (I\otimes W)\cdot(b\circ da) \\ &= (I\otimes W)\cdot({\cal B}\cdot da) \\ \frac{\partial f}{\partial a} &= (I\otimes W)\cdot{\cal B} \\ }$$ where $\otimes$ is the Kronecker product and $I$ is the identity matrix.

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  • $\begingroup$ I have seen the notation you use in the second part in Neudecker and Magnus. Do you have any reference for the notation you use in the first part? Just trying to learn what it means. $\endgroup$
    – Ted Black
    Commented Jan 4 at 17:26

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