Cartesian Metric is Rotation Invariant

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.

$\blacksquare$

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Notes

This proof holds for any finite-dimensional Euclidian space.