# 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 $$A_n =\begin{pmatrix} n^2 & 1\\-1 & n^2\end{pmatrix}.$$

1. Prove that there are sequence, $x_n,y_n$ such that $$A_1A_2\cdots A_n =\begin{pmatrix} x_n&y_n\\-y_n& x_n\end{pmatrix}.$$

2. Find the explicit expression of $x_n$ and $y_n$.

3. 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?

• One option is to observe that the matrices $A_n$ are diagonalizable and simultaneously diagonalizable, that is, there are diagonal matrices $\Lambda_n := \operatorname{diag}(\lambda_n, \mu_n)$ and a matrix $P$ such that $A_n = P \Lambda_n P^{-1}$, so that $A_1 \cdots A_n = P \operatorname{diag}(\prod_{a = 1}^n \lambda_a, \prod_{a = 1}^n \mu_a) P^{-1}$, which replaces the problem with the scalar problems of finding the sequences $(\lambda_a), (\mu_a)$ (in fact, these sequences are complex conjugates of one another, so it's really a single scalar problem). – Travis Feb 5 '18 at 18:20
• Please replace $y_{n+1} = -x_n +(n+1)^2y_n$ by $y_{n+1} = x_n +(n+1)^2y_n$. – Did Feb 5 '18 at 19:44

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 $$x_n = \left(\prod_{k= 1}^{n}\sqrt{k^4+1}\right)\cos\left(\sum_{k=1}^{n}\arctan(\frac{1}{k^2})\right)$$ $$y_n = \left(\prod_{k= 1}^{n}\sqrt{k^4+1}\right)\sin\left(\sum_{k=1}^{n}\arctan(\frac{1}{k^2})\right)$$

• You should also show it doesn’t converge... though maybe obvious I suppose... – Macavity Feb 5 '18 at 18:39
• @Macavity Yeaaa they heavily diverge as well I think OP can see that himself – Guy Fsone Feb 5 '18 at 19:57

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.