Question

If $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$ then ${\lim }_{n\to \infty }\frac{1}{n}{A}^n$ is

• A. a null matrix
• B. an identity matrix
• C. $\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$
• D. None of these

Answer 1

Answer: (A)

$A^n=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$
$\Rightarrow \frac{1}{n}{A}^n=\left[\begin{array}{cc}\frac{\cos n\theta }{n}& \frac{\sin n\theta }{n}\\ -\frac{\sin n\theta }{n}& \frac{\cos n\theta }{n}\end{array}\right]$
But $-1\le \cos n\theta \le 1$ and $-1\le \sin n\theta \le 1$
$\underset{n\to \infty }{\lim }\frac{\sin n\theta }{n}=0,\underset{n\to \infty }{\lim }\frac{\cos n\theta }{n}=0$
$\Rightarrow \underset{n\to \infty }{\lim }\frac{1}{n}{A}^n=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$