I'm working on my master thesis in computational quantum chemistry and need some math help.
I have the gradient of the dipole moment of a 3-atom molecule in one basis, but need it in another.
The dipole moment, $\mu : \mathbb{R}^9 \rightarrow \mathbb{R^3}$, is a vector in three dimensional space as a function of the position of all the 3 atoms, giving 3*3=9 coordinates.
The gradient $\nabla{\mu (x_1,...,x_9)}$ is then a 9x3 matrix.
I have the numerical values of $G$, as well as the change of basis vector, $T\in{M_{9x9}}$.
Had the gradient been of a scalar field instead of of a vector field, I could manage, but I'm uncertain how to proceed when the gradient is a matrix and not a vector.
I think your difficulty stems from how you have defined $\mu$. A more natural definition is $\mu \colon \mathbb{R}^3 \times \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$ so that the function maps the position vectors of the 3 atoms to the dipole moment vector.
Now, let $T$ be an invertible $3\times3$ matrix that defines the linear transformation from the original coordinates to the new coordinates. Let $\nu$ be the dipole moment function in the new coordinates. Then, we have $$ \nu( x_1, x_2 , x_3 ) = T \, \mu (T^{-1} x_1, T^{-1} x_2, T^{-1} x_3 ), $$ because $T^{-1}$ maps a position in the new coordinates to a position in the old coordinates and $T$ maps the dipole moment in the old coordinates to the dipole moment in the new coordinates.
Now, we can use the chain rule for differentiation to compute the Jacobian derivative of $\nu$ with respect to each position. This gives \begin{align*} \frac{\partial}{\partial x_i} \nu(x_1,x_2,x_3) &= \frac{\partial}{\partial x_i} T \, \mu (T^{-1} x_1, T^{-1} x_2, T^{-1} x_3 ) \\ &= T \frac{\partial \mu}{\partial x_i} (T^{-1} x_1, T^{-1} x_2, T^{-1} x_3 ) T^{-1}. \end{align*}
Basically, this implies that your $9\times 3$ matrix consists of three $3\times 3$ sub-blocks $S_1,S_2,S_3$ associated with $x_1,x_2,x_3$. After the change of basis, the $i$-th sub-block should be mapped to $TS_i T^{-1}$.