Computing the product $A_1 A_2 \cdots A_n$ with $A_n =\left(\displaystyle\begin{smallmatrix} n^2&1\\-1&n^2\end{smallmatrix}\right)$ 
Let $n \in \Bbb N$ and consider the square matrix $$A_n =\begin{pmatrix} n^2 & 1\\-1 & n^2\end{pmatrix}.$$
  
  
*
  
*Prove that there are sequence, $x_n,y_n$ such that 
  $$A_1A_2\cdots A_n =\begin{pmatrix} x_n&y_n\\-y_n& x_n\end{pmatrix}.$$
  
*Find the explicit expression of $x_n$ and $y_n$.
  
*What can we say about the convergence of  $x_n$ and $y_n$?

I have shown the existence of  $x_n$ and $y_n$ by induction and it turn out after identification that they satisfy the relations, 
$$ x_{n+1} =(n+1)^2x_n-y_n, \qquad\qquad y_{n+1} = x_n +(n+1)^2y_n$$ 

Can someone help to solve this recursive formula in other to get into the two last questions? Is there a more elegant way to overcome this problem?

 A: Similarly like here we identify 
$$ 1 \equiv  \left(\begin{matrix}  1& 0\\0&1 \end{matrix}\right)~~~\text{and}~~~~ i \equiv  \left(\begin{matrix}  0& 1\\-1&0 \end{matrix}\right).$$
$$A_n =\left(\begin{smallmatrix} n^2&1\\-1&n^2\end{smallmatrix}\right)\equiv n^2+i := z_n$$
$z_n$ is a complex number hence product of $z_n$ is also a complex number. this literally prove the existence of $x_n$ and $y_n$  and we have, $$A_1A_2\cdots A_n =\left(\begin{smallmatrix} x_n&y_n\\-y_n& x_n\end{smallmatrix}\right) \equiv z_1z_2\cdots z_n =x_n+iy_n.$$
Moreover, we have utilizing the polar form  we get,
 $$z_n = n^2+i =\sqrt{n^4+1}\exp\left(i\sum_{k=1}^{n}\arctan(\frac{1}{k^2})\right)$$
Hence $$x_n+iy_n = z_1z_2\cdots z_n = \left(\prod_{k= 1}^{n}\sqrt{k^4+1}\right)\exp\left(i\sum_{k=1}^{n}\arctan(\frac{1}{k^2})\right) $$
By identification we have 
$$x_n = \left(\prod_{k= 1}^{n}\sqrt{k^4+1}\right)\cos\left(\sum_{k=1}^{n}\arctan(\frac{1}{k^2})\right)$$
$$y_n = \left(\prod_{k= 1}^{n}\sqrt{k^4+1}\right)\sin\left(\sum_{k=1}^{n}\arctan(\frac{1}{k^2})\right)$$
A: This is in the same spirit as Guy Fsone's answer, but uses only linear algebra with no knowledge of complex numbers required. Let $S$ be the set of matrices of the form $A=\alpha R_\theta$ for some $\alpha>0$ and $\theta\in\mathbb R$, where
$$R_\theta=\begin{pmatrix}
\cos\theta&\sin\theta\\
-\sin\theta&\cos\theta
\end{pmatrix}.
$$
Observe that any matrix $A$ of the form
$$A=\begin{pmatrix}
a&b\\
-b&a
\end{pmatrix}$$
lies in $S$ by taking $\alpha=\sqrt{a^2+b^2}$ and $\theta=\arctan(\frac ba)$. To solve part (1), it suffices to show $S$ is closed under multiplication, which is a straightforward consequence of the easily proven identity $R_\theta R_\phi=R_{\theta+\phi}$. This identity also gives an easy solution for (2), since by taking $\theta_k=\arctan(\frac1{k^2})$ we have
$$A_1A_2\ldots A_n=\prod_{k=1}^n\Big(\sqrt{k^4+1}\,R_{\theta_k}\Big)=\left(\prod_{k=1}^n(k^4+1)\right)^{1/2}R_{\sum_{k=1}^n\theta_k},$$
from which we can read off the values of $x_n$ and $y_n$. While it should be fairly obvious that $x_n,y_n$ do not converge, observe that since $\arctan(x)\le x$ for all $x\ge0$, the series
$$\sum_{k=1}^\infty\theta_k=\sum_{k=1}^\infty\arctan\big(\tfrac1{k^2}\big)$$
converges; hence, one has that both $\lim_{n\to\infty}\frac{x_n}{n^2}$ and $\lim_{n\to\infty}\frac{y_n}{n^2}$ exist.
