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I have a function $F$, and $F=tr((X_1-f_1)(X_1-f_1)^T+(X_2-f_2)(X_2-f_2)^T+(X_1-X_2)(X_1-X_2)^T)$, where $X_1$, $X_2$, $f_1$ and $f_2$ are all $n\times n$ matrices.

So what is $\frac{dF}{dX_1}$ and $\frac{dF}{dX_2}$? It is a differentiation of a trace of a matrix. Hope you can give some details and a reference of where I can find the principle of this computation.

Thanks in advance!

(Is my solution right? for the first and last parts: $\frac{d}{dX_1}tr((X_1-f_1)(X_1-f_1)^T)=2(X_1-f_1)\frac{d}{dX_1}(X_1-f_1)=2(X_1-f_1)$

and

$\frac{d}{dX_1}tr((X_1-X_2)(X_1-X_2)^T)=2(X_1-X_2)\frac{d}{dX_1}(X_1-X_2)=2(X_1-X_2)$

by chain rule.

)

Is my solution correct?

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    $\begingroup$ Hint: $tr(A+B)=tr(A)+tr(B)$. So you only need to find out what the derivative of tr(AB) is. $\endgroup$
    – MrYouMath
    Commented Oct 27, 2016 at 21:58
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    $\begingroup$ and $(A,B)\mapsto tr(AB)$ is bilinear ... $\endgroup$
    – user251257
    Commented Oct 27, 2016 at 22:04

2 Answers 2

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Using the Frobenius (:) Inner Product the function can be written $$\eqalign{ F &= (X_1-f_1):(X_1-f_1)+ (X_2-f_2):(X_2-f_2) \cr &+\, (X_1-X_2):(X_1-X_2) }$$ Its differential is $$\eqalign{ dF &= 2(X_1-f_1):dX_1 + 2(X_2-f_2):dX_2 \cr &+\, 2(X_1-X_2):dX_1 + 2(X_2-X_1):dX_2 \cr\cr }$$ Setting $dX_2=0\,\,$ yields the gradient wrt $X_1$ $$\eqalign{ \frac{\partial F}{\partial X_1} &= 2(X_1-f_1) + 2(X_1-X_2) &= 4X_1 - 2X_2 - 2f_1 \cr\cr }$$ Setting $dX_1=0\,\,$ yields the gradient wrt $X_2$ $$\eqalign{ \frac{\partial F}{\partial X_2} &= 4X_2 - 2X_1 - 2f_2 \cr }$$

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Since you ask for "principle of this computation", I recommend the differential operator technique for Jacobian identification in p. 199, Table 2 of this book. Working out matrix derivatives of trace of matrix functions appear therein p. 200 (Section 9) as a special case of this principle. It is instructive to master this disciplined way to approach such computation, especially for more complicated functions.

Your conclusions for the derivative of the first and last terms are correct.

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