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Let A be a constant matrix. Define $$ Y:= \mathrm{tr}(X^\top A) A $$

I want to find \begin{align} \frac{\partial Y}{\partial X} \label{A} \end{align}

I know that $$ \frac{\partial\,\mathrm{tr}(X^\top A)}{\partial X} = A \label{B} $$

Of course, I should be able to extend this to the case I want. But I am a bit confused as to how I should notate the result. What I did to find \eqref{A} was to vectorize both sides and then use the fact in \eqref{B}.

$$ \mathrm{vec(Y)} = \mathrm{tr}(X^\top A)\mathrm{vec}(A) $$

I also knwo that I can write $$ \mathrm{tr}(X^\top A) = \mathrm{vec}(A)^\top \mathrm{vec}(X) $$ So the desired the derivative should be $$ \mathrm{vec}(A) \otimes \mathrm{vec}(A) $$ Is this correct?.

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  • $\begingroup$ The problem can be approached in either tensor or vector form $$\eqalign{&Y=(A\star A):X&\implies\,y=aa^Tx\cr &\frac{\partial Y}{\partial X}=(A\star A)&\implies\,\frac{\partial y}{\partial x}=aa^T}$$where $a={\rm vec}(A),\,x={\rm vec}(X),$ and $\star$ is the tensor product $${\mathcal A}=B\star C \implies {\mathcal A}_{ijkl}=B_{ij}C_{kl}$$ $\endgroup$ – greg Apr 29 '18 at 14:09
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Your result seems correct. You may verify it by using the entry-wise form.

Let the entry-wise form of the matrices $A$, $X$ and $Y$ be \begin{align} A&=\left(a_{jk}\right)_{jk},\\ X&=\left(x_{jk}\right)_{jk},\\ Y&=\left(y_{jk}\right)_{jk}. \end{align} Then $Y=\text{tr}\left(X^{\top}A\right)A$ reads $$ y_{jk}=\left(\sum_{i,r}x_{ir}a_{ir}\right)a_{jk}. $$

With this notation, \begin{align} \frac{\partial y_{jk}}{\partial x_{pq}}&=\frac{\partial}{\partial x_{pq}}\left[\left(\sum_{i,r}x_{ir}a_{ir}\right)a_{jk}\right]\\ &=\sum_{i,r}\frac{\partial x_{ir}}{\partial x_{pq}}a_{ir}a_{jk}\\ &=\sum_{i,r}\delta_{ip}\delta_{rq}a_{ir}a_{jk}\\ &=a_{pq}a_{jk}. \end{align} Thus if one regards $A$ as a tensor, one may write $$ \frac{\partial Y}{\partial X}=A\otimes A. $$

Alternatively, one may use the differential form $$ {\rm d}y_{jk}=\left(\sum_{p,q}a_{pq}{\rm d}x_{pq}\right)a_{jk}, $$ which, in its matrix form, reads $$ {\rm d}Y=\text{tr}\left(A^{\top}{\rm d}X\right)A. $$

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