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Question: Let $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$, then su...

Let A=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}, then sum of elements in lim⁡n→∞(Ann)\lim_{n\to\infty} \left( \frac{A^n}{n} \right), where θ∈R(n∈N)\theta \in R (n \in N)

A

0

B

1

C

2

D

3

Answer

0

Explanation

Solution

The given matrix is A=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}. This matrix represents a rotation in the 2D plane by an angle θ\theta counterclockwise. The nn-th power of a rotation matrix representing rotation by θ\theta is a rotation matrix representing rotation by nθn\theta. Thus, An=[cos⁡(nθ)sin⁡(nθ)−sin⁡(nθ)cos⁡(nθ)]A^n = \begin{bmatrix} \cos(n\theta) & \sin(n\theta) \\ -\sin(n\theta) & \cos(n\theta) \end{bmatrix}.

We are asked to find the sum of elements in the matrix lim⁡n→∞(Ann)\lim_{n\to\infty} \left( \frac{A^n}{n} \right). First, let's find the matrix Ann\frac{A^n}{n}:

Ann=1n[cos⁡(nθ)sin⁡(nθ)−sin⁡(nθ)cos⁡(nθ)]=[cos⁡(nθ)nsin⁡(nθ)n−sin⁡(nθ)ncos⁡(nθ)n]\frac{A^n}{n} = \frac{1}{n} \begin{bmatrix} \cos(n\theta) & \sin(n\theta) \\ -\sin(n\theta) & \cos(n\theta) \end{bmatrix} = \begin{bmatrix} \frac{\cos(n\theta)}{n} & \frac{\sin(n\theta)}{n} \\ -\frac{\sin(n\theta)}{n} & \frac{\cos(n\theta)}{n} \end{bmatrix}.

Now, we need to find the limit of this matrix as n→∞n\to\infty. The limit of a matrix is the matrix of the limits of its elements, provided the limits of the elements exist. Let's evaluate the limit of each element:

For the element cos⁡(nθ)n\frac{\cos(n\theta)}{n}: We know that −1≤cos⁡(nθ)≤1-1 \le \cos(n\theta) \le 1 for all real θ\theta and natural numbers nn. Dividing by nn (which is positive), we get −1n≤cos⁡(nθ)n≤1n-\frac{1}{n} \le \frac{\cos(n\theta)}{n} \le \frac{1}{n}. As n→∞n \to \infty, 1n→0\frac{1}{n} \to 0 and −1n→0-\frac{1}{n} \to 0. By the Squeeze Theorem, lim⁡n→∞cos⁡(nθ)n=0\lim_{n\to\infty} \frac{\cos(n\theta)}{n} = 0.

For the element sin⁡(nθ)n\frac{\sin(n\theta)}{n}: We know that −1≤sin⁡(nθ)≤1-1 \le \sin(n\theta) \le 1 for all real θ\theta and natural numbers nn. Dividing by nn (which is positive), we get −1n≤sin⁡(nθ)n≤1n-\frac{1}{n} \le \frac{\sin(n\theta)}{n} \le \frac{1}{n}. As n→∞n \to \infty, 1n→0\frac{1}{n} \to 0 and −1n→0-\frac{1}{n} \to 0. By the Squeeze Theorem, lim⁡n→∞sin⁡(nθ)n=0\lim_{n\to\infty} \frac{\sin(n\theta)}{n} = 0.

Now we can find the limit of the matrix:

lim⁡n→∞(Ann)=[lim⁡n→∞cos⁡(nθ)nlim⁡n→∞sin⁡(nθ)n−lim⁡n→∞sin⁡(nθ)nlim⁡n→∞cos⁡(nθ)n]=[00−00]=[0000]\lim_{n\to\infty} \left( \frac{A^n}{n} \right) = \begin{bmatrix} \lim_{n\to\infty} \frac{\cos(n\theta)}{n} & \lim_{n\to\infty} \frac{\sin(n\theta)}{n} \\ -\lim_{n\to\infty} \frac{\sin(n\theta)}{n} & \lim_{n\to\infty} \frac{\cos(n\theta)}{n} \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ -0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.

The limit matrix is the zero matrix of size 2×22 \times 2. The question asks for the sum of the elements in this limit matrix. The sum of the elements is 0+0+0+0=00 + 0 + 0 + 0 = 0.