We consider 2D orthogonal rotation matrices $R$. We consider a real matrix
$$A = \begin{pmatrix} a & b\\ b & c \end{pmatrix}.$$
I write that $B = RAR^T$ for some diagonal matrix $B$.
I would like to know for which values of $\theta$ (angle of rotation of $R$) that this is possible. To determine all the possible values of $\theta$ I tried expanding out the equation $B = RAR^T$ and equating, but it was incredibly difficult to infer what the possible values of $\theta$ would be. What is a better way to approach this question?
Best Answer
The $2D$ rotation matrix $R$ is given by
$R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $
Diagonalizing a symmetric matrix $A$ means find a matrix of eigenvectors $R$ such that
$A R = R B $
It is known that for any symmetric matrix, the eigenvectors are orthogonal to each other, so by choosing them to be of unit length, matrix $R$ will be an orthgonal matrix, in which case $R^{-1} = R^T $, hence
$ B = R^{-1} A R = R^T A R $
Now explicitly compute $R^T A R $
$R^T A R = \begin{bmatrix} \cos \theta && \sin \theta \\ -\sin \theta && \cos \theta \end{bmatrix} \begin{bmatrix} a && b \\ b && c \end{bmatrix} \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $
Multiplying the two right most matrices,
$R^T A R = \begin{bmatrix} \cos \theta && \sin \theta \\ -\sin \theta && \cos \theta \end{bmatrix} \begin{bmatrix} a \cos \theta + b \sin \theta && -a \sin \theta + b \cos \theta \\ b \cos \theta + c \sin \theta && -b \sin \theta + c \cos \theta \end{bmatrix} $
Which multiplies to
$R^T A R = \begin{bmatrix} a \cos^2 \theta + c \sin^2 \theta + 2 b \sin \theta \cos \theta && b (\cos^2 \theta - \sin^2 \theta) + (c - a) \sin \theta \cos \theta \\ b (\cos^2 \theta - \sin^2 \theta) + (c - a) \sin \theta \cos \theta && a \sin^2 \theta + c \cos^2 \theta - 2 b \sin \theta \cos \theta \end{bmatrix} $
Now we want the off-diagonal entries to be $0$, hence,
$ b (\cos^2 \theta - \sin^2 \theta) + (c - a) \sin \theta \cos \theta = 0 $
But $\cos^2 \theta - \sin^2 \theta = \cos(2 \theta) $ and $ \sin \theta \cos \theta = \frac{1}{2} \sin(2 \theta) $
Therefore, the equation becomes
$ b \cos(2 \theta) + \dfrac{1}{2} (c - a) \sin(2 \theta) = 0 $
From which,
$ \tan(2 \theta) = \dfrac{ 2b }{ a - c} $
i.e.
$ \theta = \dfrac{1}{2} \tan^{-1} \left( \dfrac{2b}{a - c} \right) $