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Can anybody tell me why [itex]\nabla_{\textbf q} \langle \textbf q, P \textbf q \rangle = P \textbf q[/itex]?

Explanation of notations:

q is an n-dimensional vector: [itex]\textbf q = (q_1, q_2, \cdots, q_n)^T[/itex]

P is an nxn-dimensional, real, Hermitian matrix

[itex]\nabla_{\textbf q} := \left(\frac{\partial}{\partial q_1}, \cdots, \frac{\partial}{\partial q_n} \right)^T[/itex]

Background story: I know the (specific) Lagrangian is [itex]\mathcal L = \frac{m}{2} \langle \dot{\textbf q},\dot{\textbf q} \rangle - \frac{k}{2} \langle \textbf q, P \textbf q \rangle[/itex] and the book then tells me that [itex]\ddot{ \textbf q } = - \frac{k}{2m} P \textbf q[/itex] and I presume that they derived this from the Lagrangian equations [itex]\frac{\mathrm d}{\mathrm dt} \left( \nabla_{\dot{\textbf q}} \mathcal L \right) = \nabla_{\textbf q} \mathcal L[/itex] but to do that I need to have the above equality, right?

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# How to see this equality (Lagrangian mechanics)

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