Is the derivative of $Tr(A^T(\alpha)BA(\alpha))$ given by $\frac{d}{d\alpha}A(\alpha).2BA^T(\alpha)$, using the chain rule, or is it something else? I was expecting it to be a scalar valued function. This is where am stuck.
2026-04-11 12:56:31.1775912191
What is the derivative of $\frac{d}{d\alpha}Tr(A^T(\alpha)BA(\alpha))$?
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Assuming that the derivative $$G = \frac{dA}{d\alpha}$$ is known.
Write the function in terms of the Frobenius (:) Inner Product and take its differential $$\eqalign{ f &= A:BA \cr\cr df &= dA:BA + A:B\,dA \cr &= (B+B^T)A:dA \cr &= (B+B^T)A:G\,d\alpha \cr\cr \frac{df}{d\alpha} &= (B+B^T)A:G \cr &= {\rm tr}(G^T(B+B^T)A) \cr }$$ The result is scalar-valued, as you suspected.