# Is a Rotation Matrix Still a Rotation Matrix under a Non-Standard Basis?

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?

• 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. – Matt Samuel Sep 8 '16 at 23:12
• Even if we use the same basis for both the domain and the codomain? – user1770201 Sep 8 '16 at 23:38

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 $\frac1{\sqrt{2}}\pmatrix{1&1\\ 1&-1}$.