# Thread: [SOLVED] Tricky little matrix problem

1. ## [SOLVED] Tricky little matrix problem

This exercise has been troubling me. It seems like it should be simple, but I can't seem to crack it.

Let $\displaystyle A=\left[\begin{array}{cc} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta}\end{array} \right]$

Find $\displaystyle A^k$, where $\displaystyle k\in\mathbb{N}$

To solve this, I tried to find a pattern in the following:

$\displaystyle A^2=\left[\begin{array}{cc} \cos^2{\theta}-\sin^2{\theta} & 2\sin{\theta}\cos{\theta} \\ -2\sin{\theta}\cos{\theta} & \cos^2{\theta}-\sin^2{\theta}\end{array} \right]$

$\displaystyle A^3=\left[\begin{array}{cc} \cos^3{\theta}-3\sin^2{\theta}\cos{\theta} & 3\sin{\theta}\cos^2{\theta}-\sin^3{\theta} \\ -3\sin{\theta}\cos^2{\theta}+\sin^3{\theta} & \cos^3{\theta}-3\sin^2{\theta}\cos{\theta}\end{array} \right]$

$\displaystyle A^4=\left[\begin{array}{cc}\cos^4{\theta}-6\sin^2{\theta}\cos^2{\theta}+\sin^4{\theta}&4\sin {\theta}\cos^3{\theta}-4\sin^3{\theta}\cos{\theta}\\-4\sin{\theta}\cos^3{\theta}+4\sin^3{\theta}\cos{\t heta}&\cos^4{\theta}-6\sin^2{\theta}\cos^2{\theta}+\sin^4{\theta}\end{a rray}\right]$

From this I observe that

$\displaystyle A^k=\frac{1}{k^2}\left[\begin{array}{cc}kf'(\theta)&k^2f(\theta)\\f''(\th eta)&kf'(\theta)\end{array}\right]$

But I can't seem to generalize $\displaystyle f(\theta)$ in terms of $\displaystyle k$. Any ideas?

2. Did you notice that $\displaystyle A^2 = \left[ {\begin{array}{cc} {\cos (2x)} & {\sin (2x)} \\ { - \sin (2x)} & {\cos (2x)} \\ \end{array}} \right]$?
I don't know if that helps?

3. Originally Posted by hatsoff
This exercise has been troubling me. It seems like it should be simple, but I can't seem to crack it.

Let $\displaystyle A=\left[\begin{array}{cc} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta}\end{array} \right]$

Find $\displaystyle A^k$, where $\displaystyle k\in\mathbb{N}$
Think geometrically.
This is a rotation matrix. Meaning if $\displaystyle \bold{x}\in \mathbb{R}^2$ then $\displaystyle A\bold{x}$ represent rotation the vector $\displaystyle \bold{x}$ counterclockwise by $\displaystyle \theta$ (around the origin).

Now $\displaystyle A^2\bold{x} = A(A\bold{x})$ is rotating $\displaystyle \bold{x}$ twice by $\displaystyle \theta$ i.e. $\displaystyle 2\theta$. And in general $\displaystyle A^k$ is a rotation by $\displaystyle k\theta$. The matrix for the rotation is given by,
$\displaystyle A^k = \left[ \begin{array}{cc}\cos k\theta & \sin k\theta \\ - \sin k\theta & \cos k\theta \end{array} \right]$

4. Originally Posted by hatsoff
This exercise has been troubling me. It seems like it should be simple, but I can't seem to crack it.

Let $\displaystyle A=\left[\begin{array}{cc} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta}\end{array} \right]$

Find $\displaystyle A^k$, where $\displaystyle k\in\mathbb{N}$

To solve this, I tried to find a pattern in the following:

$\displaystyle A^2=\left[\begin{array}{cc} \cos^2{\theta}-\sin^2{\theta} & 2\sin{\theta}\cos{\theta} \\ -2\sin{\theta}\cos{\theta} & \cos^2{\theta}-\sin^2{\theta}\end{array} \right]$

$\displaystyle A^3=\left[\begin{array}{cc} \cos^3{\theta}-3\sin^2{\theta}\cos{\theta} & 3\sin{\theta}\cos^2{\theta}-\sin^3{\theta} \\ -3\sin{\theta}\cos^2{\theta}+\sin^3{\theta} & \cos^3{\theta}-3\sin^2{\theta}\cos{\theta}\end{array} \right]$

$\displaystyle A^4=\left[\begin{array}{cc}\cos^4{\theta}-6\sin^2{\theta}\cos^2{\theta}+\sin^4{\theta}&4\sin {\theta}\cos^3{\theta}-4\sin^3{\theta}\cos{\theta}\\-4\sin{\theta}\cos^3{\theta}+4\sin^3{\theta}\cos{\t heta}&\cos^4{\theta}-6\sin^2{\theta}\cos^2{\theta}+\sin^4{\theta}\end{a rray}\right]$

From this I observe that

$\displaystyle A^k=\frac{1}{k^2}\left[\begin{array}{cc}kf'(\theta)&k^2f(\theta)\\f''(\th eta)&kf'(\theta)\end{array}\right]$

But I can't seem to generalize $\displaystyle f(\theta)$ in terms of $\displaystyle k$. Any ideas?
A thread of related interest: http://www.mathhelpforum.com/math-he...wer-100-a.html