# Shouldn't work be the same in all coordinates?

We know that the work done by a force $\mathbf{F}$, along a path $\mathbf{x}$, is given by: $$W = \mathbf{F}^T \cdot \mathbf{x}$$ However, suppose that i apply some change of basis, given by a matrix $A$. So, $\mathbf{F}$ will become $A\mathbf{F}$ and $\mathbf{x}$ will become $A\mathbf{x}$. And so $$W = (A\mathbf{F})^T \cdot A\mathbf{x} = \mathbf{F}^TA^TA\mathbf{x}$$ Which may not be equal to $\mathbf{F} \cdot \mathbf{x}$. What am i missing? If the path and the force are both the same, shouldn't the work be the same in both cases, no matter what basis am i using for $\Bbb{R}^3$? I am supposing that $A$ is not necessarily orthonormal.

Let us write the dot product like this: $$\vec F^T \vec x$$ where the $T$ means transpose. If we now apply the change of coordinates we get: $$\vec F'^T \vec x'=(A\vec F)^T A\vec x$$ $$=\vec F^T A^T A \vec x$$ Now a change of coordinates matrix must be orthogonal (in this case) so $$A^TA=I$$ Hence we get: $$\vec F'^T \vec x'=\vec F^TI \vec x=\vec F^T \vec x$$ which is coordinate independent.