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Let $F(X) = Tr(XBX^TA)$, where A,B are matrix.

What is the derivative $\frac{\partial F(X)}{\partial X}$?

Is it something like $A(XB^T+XB)$?

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1 Answer 1

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Coming back to the definition of the differential, one looks for a linear functional $L$ such that $F(X+H)=F(X)+L(H)+o(\|H\|)$ when $H\to O$, and then one knows that $D_XF=L$. Here, $$ F(X+H)=F(X)+\mathrm{tr}(HBX^TA)+\mathrm{tr}(XBH^TA)+\mathrm{tr}(HBH^TA), $$ hence $$ D_XF:H\mapsto\mathrm{tr}(HBX^TA)+\mathrm{tr}(XBH^TA). $$ Since $\mathrm{tr}(U)=\mathrm{tr}(U^T)$ for every square matrix $U$ and $\mathrm{tr}(VW)=\mathrm{tr}(WV)$ for every matrices $V$ and $W$, this is also $$ (D_XF)(H)=\mathrm{tr}(H^TC),\qquad C=AXB+A^TXB^T. $$ One sees that $C=AX(B+B^T)$ when $A$ is symmetric, but not in general.

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