Matrix problem invovling orthogonal matrices

Question

If $$P=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}\text{ and } A=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$$ And $Q=PAP'$ then $P'Q^{2009}P=?$

I noticed that $P$ is an orthogonal matrix so $$P'=P^{-1}$$ and $P^2(\theta)=P(2\theta)$

$$Q^{2009}=P^{2009}A^{2009}(P^{-1})^{2009}$$ I am stuck at this step. Is there any property of orthogonal matrix that I can use here?

Answer 1

Hint Since $P' = P^{-1}$, we have $P P' = I$ and hence $$P' Q^k P = P'QPP'QP\cdots P'QPP'QP = \underbrace{(P' Q P) \cdots (P' Q P)}_k = (P' Q P)^k .$$

Answer 2

Hint: $$A^\color{red}2 = \begin{pmatrix}1 & \color{red}2\\0 & 1\end{pmatrix}$$

$$A^\color{red}3 = \begin{pmatrix}1 & \color{red}3\\0 & 1\end{pmatrix}$$

$$A^\color{red}4 = \begin{pmatrix}1 & \color{red}4\\0 & 1\end{pmatrix}$$ $$\dots$$

And $$Q^{2009} = PA^{2009}P^{-1}$$