Can somebody help me to show:
$$\boldsymbol{a\cdot}\frac{\partial \boldsymbol{F}}{\partial \boldsymbol{q}}\boldsymbol{b}=\left(\frac{\partial \boldsymbol{F}}{\partial \boldsymbol{q}}\right)^T \boldsymbol{a\cdot}\boldsymbol{b}.$$
What I have done:
$\boldsymbol{a\cdot}\frac{\partial \boldsymbol{F}}{\partial \boldsymbol{q}}\boldsymbol{b}=a_k \boldsymbol{e}_k \boldsymbol{\cdot}\frac{\partial F_i}{\partial q_j} \boldsymbol{e}_i \boldsymbol{e}_j \:b_l \boldsymbol{e}_l=a_k \frac{\partial F_i}{\partial q_j}b_l \delta_{ki} \boldsymbol{e}_j \boldsymbol{e}_l= a_i \frac{\partial F_i}{\partial q_j}b_l \boldsymbol{e}_j \boldsymbol{e}_l=\color{red}{\frac{\partial F_j}{\partial q_i}a_j} b_l \boldsymbol{e}_i \boldsymbol{e}_l=\left(\frac{\partial F_i}{\partial q_j}\right)^T a_j b_l \boldsymbol{e}_i \boldsymbol{e}_l=\left(\frac{\partial F_i}{\partial q_j}\right)^T a_j \boldsymbol{e}_j\boldsymbol{\cdot} \boldsymbol{e}_j b_l \boldsymbol{e}_i \boldsymbol{e}_l=?$
In the third last step, I swapped the indices $i$ and $j$ to create $\left(\frac{\partial F_i}{\partial q_j}\right)^T$. But I'm not sure whether it was correct to interchange the Jacobi $\frac{\partial F_j}{\partial q_i}$ and $a_j$, as highlighted with $\color{red}{\text{red}}$. And in the last step, I multiplied $\boldsymbol{e}_j\boldsymbol{\cdot}\boldsymbol{e}_j(=1)$ to construct $\boldsymbol{a \cdot}$. But how should I continue to obtain the right hand side?
Thank you in advance.
Let us use Einstein notation and denote $\boldsymbol{A} = \partial\boldsymbol{F}/\partial\boldsymbol{q}$, which coordinates are $A_{ij} = \partial F_i/\partial q_j$. Thus, \begin{aligned} (\boldsymbol{A}\cdot\boldsymbol{b})\cdot\boldsymbol{a} = \boldsymbol{a}\cdot(\boldsymbol{A}\cdot\boldsymbol{b}) &= {a}_i A_{ij} {b}_j \\ &= A_{ij} {a}_i {b}_j\\ &= A_{ji}^\top {a}_i {b}_j = (\boldsymbol{A}^\top\!\cdot \boldsymbol{a})\cdot \boldsymbol{b} = \boldsymbol{b}\cdot(\boldsymbol{A}^\top\!\cdot \boldsymbol{a}) \end{aligned}