Cartesian Metric is Rotation Invariant
Jump to navigation
Jump to search
Theorem
The cartesian metric does not change under rotation.
Proof
Let the cartesian metric be $\delta _{ij} = \langle e_i, e_j \rangle$.
Also, let $\delta_{ij}^\prime$ be the metric of the coordinate system of $\delta _{ij}$ rotated by a rotation matrix $A$.
Then, $\delta_{ij}^\prime = \langle A e_i, A e_j \rangle$.
$\langle A x, y \rangle = \langle x, A^T y \rangle$, see Factor Matrix in the Inner Product, so
$\delta_{ij}^\prime = \langle A e_i, A e_j \rangle = \langle e_i, A^T A e_j \rangle$.
For rotation matrices, we have $A^T = A^{-1}$, so $\delta_{ij}^\prime$ reduces to:
- $\delta_{ij}^\prime = \langle e_i, A^T A e_j \rangle = \langle e_i, A^{-1} A e_j \rangle = \langle e_i, I e_j \rangle = \langle e_i, e_j \rangle$
where $I$ is the unit matrix.
Thus:
- $\delta_{ij}^\prime = \delta_{ij}$
The result follows.
$\blacksquare$
![]() | This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
Notes
This proof holds for any finite-dimensional Euclidian space.