Can a certain matrix $A$ be diagonalized by a rotation matrix?

Question

I have matrix $$\mathcal{A}=\pmatrix{a & c \\ c & b}$$ I can diagonalize this matrix by $\mathcal{A} = SDS^{-1} \Rightarrow D=S^{-1}\mathcal{A}S$, where $D$ is diagonalized matrix consists of eigenvalues and $S$ consists of eigenvectors. This is trivial approach I know. But can we discuss using rotation matrix $$\mathcal{R}=\pmatrix{\cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta}}$$ instead of $S$?

Answer 1

In general, no.

I write $R_\theta=\pmatrix{\cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta}}$ for the rotation of angle $\theta$. It is not difficult to show that

1. $R_\theta$ is invertible and $R_\theta^{-1}=R_{-\theta}$ for every $\theta$.
2. $R_\alpha R_\beta=R_{\alpha+\beta}$ for every $\alpha$ and $\beta$.

Consider the case where $A$ itself is a rotation $A=R_\alpha$. Then for every rotation matrix $S=R_{\theta}$, we have

$$ S^{-1}AS=R_{-\theta}R_{\alpha}R_{\theta}=R_{-\theta+\alpha+\theta}=R_\alpha = A $$

If we choose $\alpha$ such that $A$ is not diagonal (that is, $\sin\alpha\neq 0$), then $A$ is not "diagonalisable by rotation".