Question
If \[ A = \begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix} \] prove that \[ A^n = \begin{bmatrix} \cos n\theta & i\sin n\theta \\ i\sin n\theta & \cos n\theta \end{bmatrix} \quad \forall n \in \mathbb{N}. \]
Solution (Mathematical Induction)
Step 1: Base Case (n = 1)
\[ A^1 = \begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos (1\theta) & i\sin (1\theta) \\ i\sin (1\theta) & \cos (1\theta) \end{bmatrix} \] ✔ True for \(n=1\)Step 2: Assume for \(n = k\)
\[ A^k = \begin{bmatrix} \cos k\theta & i\sin k\theta \\ i\sin k\theta & \cos k\theta \end{bmatrix} \]Step 3: Prove for \(n = k+1\)
\[ A^{k+1} = A^k \cdot A \] \[ = \begin{bmatrix} \cos k\theta & i\sin k\theta \\ i\sin k\theta & \cos k\theta \end{bmatrix} \begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix} \]Step 4: Multiply
\[ = \begin{bmatrix} \cos k\theta \cos\theta – \sin k\theta \sin\theta & i(\cos k\theta \sin\theta + \sin k\theta \cos\theta) \\ i(\sin k\theta \cos\theta + \cos k\theta \sin\theta) & \cos k\theta \cos\theta – \sin k\theta \sin\theta \end{bmatrix} \]Step 5: Use Identities
\[ \cos(k+1)\theta = \cos k\theta \cos\theta – \sin k\theta \sin\theta \] \[ \sin(k+1)\theta = \sin k\theta \cos\theta + \cos k\theta \sin\theta \]Step 6: Substitute
\[ A^{k+1} = \begin{bmatrix} \cos (k+1)\theta & i\sin (k+1)\theta \\ i\sin (k+1)\theta & \cos (k+1)\theta \end{bmatrix} \]Step 7: Conclusion
✔ True for \(k+1\) \[ \Rightarrow A^n = \begin{bmatrix} \cos n\theta & i\sin n\theta \\ i\sin n\theta & \cos n\theta \end{bmatrix} \]Final Result
\[
A^n =
\begin{bmatrix}
\cos n\theta & i\sin n\theta \\
i\sin n\theta & \cos n\theta
\end{bmatrix}
\]
Hence proved.