Question:

If \[ A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \] prove that \[ A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix}, \] where \( n \in \mathbb{N} \).

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The power of a rotation matrix follows the identity \( A^n = \begin{bmatrix} \cos n\theta & \sin n\theta
-\sin n\theta & \cos n\theta \end{bmatrix} \).
Updated On: Feb 27, 2025
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Solution and Explanation

Step 1: Prove by induction. \[ \text{For } n = 1, \quad A^1 = A \] \[ \text{Assume } A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix}. \] \[ \text{Multiply by } A: \quad A^{n+1} = A^n \cdot A \] \[ A^{n+1} = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \] \[ = \begin{bmatrix} \cos (n+1)\theta & \sin (n+1)\theta \\ -\sin (n+1)\theta & \cos (n+1)\theta \end{bmatrix}. \]

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