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Lame matrix calculus question: Suppose you have that $y$ is an $n \times 1$ vector function of $t$ and $G$ is an $n \times n$ symmetric matrix function of $t$. Is it ok to write

$\frac{d}{dt}(y^T G y) = 2 \frac{d(y^T)}{dt} G y + y^T \frac{dG}{dt}y$,

where superscript $T$ denotes the transpose?

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    Where does $G$ live? Is it symmetric? What about the $y$? Is it real?2012-07-19

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Use $\frac{d}{dt}(y^T G y) =(\frac{dy^T}{dt} G y)+(y^T \frac{dG}{dt} y)+(y^T G \frac{dy}{dt})$. So if $(\frac{dy^T}{dt} G y)=(y^T G \frac{dy}{dt})$ your result is correct.

In a quantum mechanical setup this is related to Ehrenfest's Theorem. It says that $ \frac{d}{dt}\langle A \rangle_\psi = \frac{d}{dt}\langle \psi | A | \psi \rangle = \frac{d}{dt}\langle A\rangle = \frac{1}{i\hbar}\langle [A,H] \rangle+ \left\langle \frac{\partial A}{\partial t}\right\rangle $

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    ... with $\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar}H\psi $ and $H$ being the Hamiltionian...2012-07-19