Find $P$ such $P^{-1}AP = kR$.

Question

Let $ A=\begin{bmatrix} 3 & -5 \\ 1 & -1 \\ \end{bmatrix}$

Find P such that $P^{-1}AP = k R$ where $k \neq 0$ is a scalar and R an element of $SO(2)$ = {R element of $R^{2x2}$ | $R^{T} R = I, det(R) = 1$}.

I have no idea what to do.

Answer 1

If you have complex roots $\alpha \pm i \beta$, you can find corresponding eigenvectors of the form $\mathbf{v}_1 \pm i\mathbf{v}_2$. Then if $P = [\mathbf{v}_1 \; \mathbf{v}_2]$, you get:

$$A = PCP^{-1}$$

With

$$C = \begin{bmatrix} \alpha & \beta \\ -\beta & \alpha \end{bmatrix} = k\begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix}$$

Where $k = \sqrt{\alpha^2 + \beta^2}$ and $\theta = \tan^{-1}(\beta/\alpha)$