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 Identity Matrix.

Thus, $$\delta_{ij}^\prime = \delta_{ij}$$.

The result follows.