Question

\[A = \begin{bmatrix}\cos \alpha + \sin \alpha & \sqrt{2}\sin \alpha \\ - \sqrt{2}\sin \alpha & \cos \alpha - \sin \alpha\end{bmatrix}\] ,prove that

\[A^n = \begin{bmatrix}\text{cos n α} + \text{sin n α}  & \sqrt{2}\text{sin n  α} \\ - \sqrt{2}\text{sin n α} & \text{cos n α} - \text{sin  n  α} \end{bmatrix}\] for all n ∈ N.

 

Answer 1

We shall prove the result by the principle of mathematical induction on n.

Step 1:  If n = 1, by definition of integral power of a matrix, we have

\[A^1 = \begin{bmatrix}\cos 1\alpha + \sin 1\alpha & \sqrt{2}\sin 1\alpha \\ - \sqrt{2} \sin 1\alpha & \cos 1\alpha - \sin 1\alpha\end{bmatrix} = \begin{bmatrix}\cos \alpha + \sin \alpha & \sqrt{2}\sin \alpha \\ - \sqrt{2}\sin \alpha & \cos \alpha - \sin \alpha\end{bmatrix} = A\]So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,
\[\]\[A^m = \begin{bmatrix}\cos m\alpha + \sin m\alpha & \sqrt{2}\sin m\alpha \\ - \sqrt{2}\sin m\alpha & \cos m\alpha - \sin m\alpha\end{bmatrix}\]

Now we shall show that the result is true for

\[n = m + 1\]

Here,

 

\[A^{m + 1} = \begin{bmatrix}\cos \left( m + 1 \right)\alpha + \sin \left( m + 1 \right)\alpha & \sqrt{2}\sin \left( m + 1 \right)\alpha \\ - \sqrt{2}\sin \left( m + 1 \right)\alpha & \cos \left( m + 1 \right)\alpha - \sin \left( m + 1 \right)\alpha\end{bmatrix}\]

By definition of integral power of matrix, we have

\[A^{m + 1} = A^m . A\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}\cos m\alpha + \sin m\alpha & \sqrt{2}\sin m\alpha \\ - \sqrt{2}\sin m\alpha & \cos m\alpha - \sin m\alpha\end{bmatrix}\begin{bmatrix}\cos \alpha + \sin \alpha & \sqrt{2}\sin \alpha \\ - \sqrt{2}\sin \alpha & \cos \alpha - \sin \alpha\end{bmatrix} \left[ From eq . \left( 1 \right) \right]\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}\left( \cos m\alpha + \sin m\alpha \right)\left( \cos \alpha + \sin \alpha \right) - \sqrt{2}\sin m\alpha\left( \sqrt{2}\sin \alpha \right) & \left( \cos m\alpha + \sin m\alpha \right)\left( \sqrt{2}\sin \alpha \right) + \sqrt{2}\sin m\alpha\left( \cos \alpha - \sin \alpha \right) \\ - \sqrt{2}\sin m\alpha\left( \cos \alpha + \sin \alpha \right) - \left( \cos m\alpha - \sin m\alpha \right)\left( \sqrt{2}\sin \alpha \right) & - \sqrt{2}\sin m\alpha\left( \sqrt{2}\sin \alpha \right) + \left( \cos m\alpha - \sin m\alpha \right)\left( \cos \alpha - \sin \alpha \right)\end{bmatrix}\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}\cos m\alpha \cos\alpha + \sin m\alpha \cos\alpha + \cos m\alpha \sin\alpha + \sin m\alpha \sin\alpha - 2\sin m\alpha \sin\alpha & \sqrt{2}\sin \alpha \cos m\alpha + \sqrt{2}\sin \alpha \sin m\alpha + \sqrt{2}\sin m\alpha \cos\alpha - \sqrt{2}\sin ma \sin\alpha \\ - \sqrt{2}\sin ma \cos\alpha - \sqrt{2}\sin ma \sin\alpha - \sqrt{2}\sin \alpha \cos m\alpha + \sqrt{2}\sin \alpha \sin m\alpha & - 2\sin \alpha \sin m\alpha + \cos m\alpha \cos\alpha - \sin m\alpha \cos\alpha - \cos m\alpha \sin\alpha + \sin m\alpha \sin\alpha\end{bmatrix}\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}\cos\left( m\alpha - \alpha \right) + \sin\left( m\alpha + \alpha \right) - \cos\left( m\alpha - \alpha \right) + \cos\left( m\alpha + \alpha \right) & \sqrt{2}\sin\left( m\alpha + \alpha \right) \\ - \sqrt{2}\sin\left( m\alpha + \alpha \right) & \cos\left( m\alpha + \alpha \right) - \sin\left( m\alpha + \alpha \right)\end{bmatrix}\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}\cos\left( m\alpha + \alpha \right) + \sin\left( m\alpha + \alpha \right) & \sqrt{2}\sin\left( m\alpha + a \right) \\ - \sqrt{2}\sin\left( m\alpha + \alpha \right) & \cos\left( m\alpha + \alpha \right) - \sin\left( m\alpha + \alpha \right)\end{bmatrix}\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}\cos\left( m + 1 \right)\alpha + \sin\left( m + 1 \right)\alpha & \sqrt{2}\sin\left( m + 1 \right)\alpha \\ - \sqrt{2}\sin\left( m + 1 \right)\alpha & \cos\left( m + 1 \right)\alpha - \sin\left( m + 1 \right)\alpha\end{bmatrix}\]
\[\]This show that when the result is true for n = m, it is also true for n = m +1. 

Hence, by the principle of mathematical induction, the result is valid for all n

\[\in N\]