Cartesian Metric is Rotation Invariant

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Theorem

The cartesian metric does not change under rotation.

Proof

Let the cartesian metric be $\delta_{ij} = \innerprod {e_i} {e_j}$.

Also, let $\delta_{ij}'$ be the metric of the coordinate system of $\delta_{ij}$ rotated by a rotation matrix $A$.

Then, $\delta_{ij}' = \innerprod {A e_i} {A e_j}$.

$\innerprod {A x} y = \innerprod x {A^T y}$, see Factor Matrix in the Inner Product, so

$\delta_{ij}' = \innerprod {A e_i} {A e_j} = \innerprod {e_i} {A^T A e_j}$.

For rotation matrices, we have $A^T = A^{-1}$, so $\delta_{ij}'$ reduces to:

$\delta_{ij}' = \innerprod {e_i} {A^T A e_j} = \innerprod {e_i} {A^{-1} A e_j} = \innerprod {e_i} {I e_j} = \innerprod {e_i} {e_j}$

where $I$ is the unit matrix.

Thus:

$\delta_{ij}' = \delta_{ij}$

The result follows.

$\blacksquare$

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This proof holds for any finite-dimensional Euclidean space.