If A= + &2 -2 & - , prove that A^2= + &2 -2 & - for all n .

Question

If $\text{A}=\begin{bmatrix}\cos\alpha+\sin\alpha&\sqrt{2}\sin\alpha\\-\sqrt{2}\sin\alpha&\cos\alpha-\sin\alpha\end{bmatrix},$ prove that
$ \text{A}^2=\begin{bmatrix}\cos\text{n}\alpha+\sin\text{n}\alpha&\sqrt{2}\sin\text{n}\alpha\\-\sqrt{2}\sin\text{n}\alpha&\cos\text{n}\alpha-\sin\text{n}\alpha\end{bmatrix}$ for all $\text{n}\in\text{N}.$

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Answer

Given,
$\text{A}=\begin{bmatrix}\cos\alpha+\sin\alpha&\sqrt{2}\sin\alpha\\-\sqrt{2}\sin\alpha&\cos\alpha-\sin\alpha\end{bmatrix}$
To prove $P(n) : \text{A}^2=\begin{bmatrix}\cos\text{n}\alpha+\sin\text{n}\alpha&\sqrt{2}\sin\text{n}\alpha\\-\sqrt{2}\sin\text{n}\alpha&\cos\text{n}\alpha-\sin\text{n}\alpha\end{bmatrix}$ we use mathematical induction.
Step $1$ : To show $P(1)$ is true.
$A^n$ is true for $n = 1$
Step $2$ : Let $P(k)$ be true,
So
$\text{A}^\text{k}=\begin{bmatrix}\cos\text{k}\alpha+\sin\text{k}\alpha&\sqrt{2}\sin\text{k}\alpha\\-\sqrt{2}\sin\text{k}\alpha&\cos\text{k}\alpha-\sin\text{k}\alpha\end{bmatrix}$
Step $3$ : Let $P(k)$ is true.
Now, we have to show that
$ \text{A}^\text{k+1}=\begin{bmatrix}\cos(\text{k}+1)\alpha+\sin(\text{k}+1)\alpha&\sqrt{2}\sin(\text{k}+1)\alpha\\-\sqrt{2}\sin(\text{k}+1)\alpha&\cos(\text{k}++1)\alpha-\sin(\text{k}+1)\alpha\end{bmatrix}$
Now,
$\text{A}^{\text{k}+1}=\text{A}^\text{k}\times\text{A}$
$=\begin{bmatrix}\cos\text{k}\alpha+\sin\text{k}\alpha&\sqrt{2}\sin\text{k}\alpha\\-\sqrt{2}\sin\text{k}\alpha&\cos\text{k}\alpha\sin\text{k}\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha+\sin\alpha&\sqrt{2}\sin\alpha\\-\sqrt{2}\sin\alpha&\cos\alpha-\sin\alpha\end{bmatrix}$
$ =\begin{bmatrix}(\cos\text{k}\alpha+\sin\text{k}\alpha)(\cos\alpha+\sin\alpha)-2\sin\alpha\sin\text{k}\alpha&(\cos\text{k}\alpha+\sin\text{k}\alpha)\sqrt{2}\sin\alpha+\sqrt{2}\sin\text{k}\alpha(\sin\alpha-\cos\alpha)\ \cos\alpha+\sin\alpha)(-\sqrt{2}\sin\text{k}\alpha)-\sqrt{2}\sin\alpha(\cos\text{k}\alpha-\sin\text{k}\alpha)&-2\sin\text{k}\alpha\sin\alpha+(\cos\text{k}\alpha-\sin\text{k}\alpha)(\cos\alpha-\sin\alpha)\end{bmatrix}$
$ =\begin{bmatrix}\cos\text{k}\alpha\cos\alpha+\sin\text{k}\alpha\cos\alpha+\cos\text{k}\alpha\sin\alpha+\sin\alpha\sin\text{k}\alpha-2\sin\alpha\sin\text{k}\alpha&\sqrt{2}\cos\text{k}\alpha\sin\alpha+\sqrt{2}\sin\alpha\sin\text{k}\alpha+\sqrt{2}\sin\text{k}\alpha\cos\alpha-\sqrt{2}\sin\text{k}\alpha\sin\alpha\\-\sqrt{2}\cos\alpha\sin\alpha-\sqrt{2}\sin\alpha\sin\text{k}\alpha-\sqrt{2}\sin\alpha\cos\text{k}\alpha+\sqrt{2}\sin\alpha\sin\text{k}\alpha&-2\sin\text{k}\alpha\sin\alpha+\cos\text{k}\alpha\cos\alpha-\cos\alpha\sin\text{k}\alpha-\sin\alpha\cos\text{k}\alpha\sin\alpha\sin\text{k}\alpha\end{bmatrix}$
$ =\begin{bmatrix}\cos\alpha\cos\text{k}\alpha+\sin\alpha\sin\text{k}\alpha+\sin\alpha\cos\text{k}\alpha+\sin\text{k}\alpha\cos\alpha&\sqrt{2}(\sin\text{k}\alpha\cos\alpha+\cos\text{k}\alpha\sin\alpha)\\-\sqrt{2}(\sin\text{k}\alpha\cos\alpha+\cos\text{k}\alpha\sin\alpha)&\cos\text{k}\alpha\cos\alpha-\sin\text{k}\alpha\sin\alpha-(\sin\text{k}\alpha\cos\alpha+\sin\alpha\cos\text{k}\alpha)\end{bmatrix}$
$ =\begin{bmatrix}\cos(\text{k}+1)\alpha+\sin(\text{k}+1)\alpha&\sqrt{2}\sin(\text{k}+1)\alpha\\-\sqrt{2}\sin(\text{k}+1)\alpha&\cos(\text{k}+1)\alpha-\sin(\text{k}+1)\alpha\end{bmatrix}$
So $, P(k + 1)$ is true whenever $P(k)$ is true.
Hence, by principle of mathematical induction $P(n)$ is true for all positive in teger.