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Prove that $$\nabla_{\mathrm X} \mbox{tr} (\mathrm A \mathrm X^{-1} \mathrm B) = - \mathrm X^{-\top} \mathrm A^\top \mathrm B^\top \mathrm X^{-\top}$$

My proof is below. I am interested in other proofs.

My proof

Let

$$f (\mathrm X) := \mbox{tr} (\mathrm A \mathrm X^{-1} \mathrm B)$$

Hence,

$$\begin{array}{rl} f (\mathrm X + h \mathrm V) &= \mbox{tr} (\mathrm A (\mathrm X + h \mathrm V)^{-1} \mathrm B)\\ &= \mbox{tr} (\mathrm A (\mathrm X ( \mathrm I + h \mathrm X^{-1} \mathrm V))^{-1} \mathrm B)\\ &= \mbox{tr} (\mathrm A ( \mathrm I + h \mathrm X^{-1} \mathrm V)^{-1} \mathrm X^{-1} \mathrm B)\\ &= \mbox{tr} (\mathrm A ( \mathrm I - h \mathrm X^{-1} \mathrm V + O (h^2)) \mathrm X^{-1} \mathrm B)\\ &= \mbox{tr} (\mathrm A \mathrm X^{-1} \mathrm B) - h \, \mbox{tr} ( \mathrm A \mathrm X^{-1} \mathrm V \mathrm X^{-1} \mathrm B) + O (h^2)\\ &= f (\mathrm X) - h \, \mbox{tr} ( \mathrm X^{-1} \mathrm B \mathrm A \mathrm X^{-1} \mathrm V ) + O (h^2)\end{array}$$

Thus, the directional derivative of $f$ in the direction of $\mathrm V$ at $\mathrm X$ is

$$D_{\mathrm V} f (\mathrm X) = - \mbox{tr} ( \mathrm X^{-1} \mathrm B \mathrm A \mathrm X^{-1} \mathrm V ) = - \mbox{tr} ( (\mathrm X^{-\top} \mathrm A^\top \mathrm B^\top \mathrm X^{-\top})^\top \mathrm V ) = - \langle \mathrm X^{-\top} \mathrm A^\top \mathrm B^\top \mathrm X^{-\top}, \mathrm V \rangle$$

and, lastly,

$$\nabla_{\mathrm X} \mbox{tr} (\mathrm A \mathrm X^{-1} \mathrm B) = - \mathrm X^{-\top} \mathrm A^\top \mathrm B^\top \mathrm X^{-\top}$$

Rodrigo de Azevedo
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1 Answers1

4

Using differentials and the Frobenius product $\big(\operatorname{tr}(A^TB)=A\!:\!B\,\big)$ yields a compact derivation $$\eqalign{ f &= A^TB^T:X^{-1}\cr\cr df &= A^TB^T:dX^{-1}\cr &= -A^TB^T:X^{-1}\,dX\,X^{-1}\cr &= -X^{-T}A^TB^TX^{-T}:dX \cr\cr \frac{\partial f}{\partial X} &= -X^{-T}A^TB^TX^{-T} \cr }$$

greg
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