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Let us work in $\mathbb{R}^2$. Consider the following rotation matrix:

$$ R = \begin{bmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} $$

I agree that this represents a rotation of angle $\theta$ in $\mathbb{R}^2$ under the standard basis of $\beta = \{(1,0), (0,1)\}$. But would it still represent a rotation if we changed the basis to something non-standard (in both the domain and the codomain), say $\beta_1 = \{1, 1), (1, -1) \}$?

That is, would we have that

$$ \mathbf{\vec{b}}^t R \mathbf{c} $$

represents a rotation if $\mathbf{\vec{b}}$ represents our non-standard basis and $\mathbf{c}$ represents an arbitrary element in $\mathbb{R}^2$ with respect to the non-standard basis?

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  • $\begingroup$ No in general. Fix one of the basis vector. Almost any random choice of a second basis vector will cause the resulting linear transformation to fail to be orthogonal. $\endgroup$ – Matt Samuel Sep 8 '16 at 23:12
  • $\begingroup$ Even if we use the same basis for both the domain and the codomain? $\endgroup$ – user1770201 Sep 8 '16 at 23:38
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Yes, it's possible. In particular, $R=\pm I_2$ is invariant under every change of basis.

If you need a nontrivial example, note that when $R\ne\pm I_2$, the product $P^{-1}RP$ is a $2\times2$ rotation matrix if and only if $P$ is a nonzero multiple of any $2\times2$ real orthogonal matrix. The "if" part can be easily verified if you multiply out $P^{-1}RP$ directly; to prove the "only if" part, consider $(P^{-1}RP)^T(P^{-1}RP)$.

So, in your case, when the change-of-basis matrix is $P=\pmatrix{1&1\\ 1&-1}$, a rotation will always remain a rotation in the new basis, because $P$ is a nonzero multiple of the real orthogonal matrix $\frac1{\sqrt{2}}\pmatrix{1&1\\ 1&-1}$.

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  • $\begingroup$ So the upshot is that it is possible to find a rotation matrix that is invariant under change in basis, but by no means guaranteed? Do you know if the rotation matrix I used in my example remains a rotation under any change in basis? $\endgroup$ – user1770201 Sep 9 '16 at 12:21
  • $\begingroup$ @user1770201 As stated in my answer, a nontrivial rotation remains a rotation in the new basis if and only if the change-of-basis matrix is a nonzero multiple of a real orthogonal matrix. It follows that in your example, a rotation will always remain a rotation in the new basis. $\endgroup$ – user1551 Sep 9 '16 at 13:50

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