We know that the work done by a force $\mathbf{F}$, along a path $\mathbf{x}$, is given by: \begin{equation} W = \mathbf{F}^T \cdot \mathbf{x} \end{equation} 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.

  • $\begingroup$ See also this related answer to a related question. $\endgroup$
    – garyp
    Commented Aug 12, 2017 at 18:14

1 Answer 1


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.


Sorry I missed the statement about orthogonal matrices in the question. The point is that we actually only expect the work done to remain the same under orthogonal transformations. Orthogonal transformations correspond to rotations (and reflections) under which we do not on physical grounds expect the work to change. If the matrix is not orthogonal we then are doing things like stretches - these changes units and with the same scalar product we do not expect to get the same answer.

As a side note as ACuriousMind stated in the answers to one of my questions a proper calculation could be done but this would involve a change in the scalar product.

  • $\begingroup$ Thanks for the reconsideration about my scalar product. I will correct it. But this is not what i am asking. $\endgroup$ Commented Aug 12, 2017 at 17:53
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    $\begingroup$ @Dovah-king If your matrix is not orthogonal then you are doing something like a stretch in which case you have a change of units and there is no reason for them to be the same. Orthogonal matrices correspond to rotations - in which case you would expect the work to remain the same. $\endgroup$ Commented Aug 12, 2017 at 17:57

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