N-th power of a matrix

Question

I have to find the n-th power for the following matrix $$A=\begin{pmatrix} 1+\sqrt3 & 1-\sqrt3\\ \sqrt3 - 1 & \sqrt3 +1\end{pmatrix} $$ My thoughts is that here could be the same situation as for $B=\begin{pmatrix} \sin x & - \cos x\\ \cos x & \sin x\end{pmatrix} $ which give $$B^n=\begin{pmatrix} \sin (nx) & - \cos (nx) \\ \cos (nx) & \sin (nx) \end{pmatrix}. $$ So I think I have to make a connection between A and B, however in A I dont have only $1$ term per place so I dont know how to proceed. $$\frac12 A=\begin{pmatrix} \sin(\pi /6) +\cos(\pi /6) &\sin(\pi /6) - \cos(\pi /6) \\ \cos(\pi /6) - \sin(\pi /6) &\sin(\pi /6) +\cos(\pi /6) \end{pmatrix} $$ Couls you help me?

Answer 1

As can be easily verified $$ \left(\begin{array}{cc} a & -b\\ b & a \end{array}\right)=\left(\begin{array}{cc} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{array}\right)\left(\begin{array}{cc} b\cos\theta+a\sin\theta & a\cos\theta-b\sin\theta\\ -a\cos\theta+b\sin\theta & b\cos\theta+a\sin\theta \end{array}\right) $$

but

$$ b\cos\theta+a\sin\theta=\sqrt{a^{2}+b^{2}}\left(\frac{b}{\sqrt{a^{2}+b^{2}}}\cos\theta+\frac{a}{\sqrt{a^{2}+b^{2}}}\sin\theta\right)=\rho\cos\left(\theta_{0}-\theta\right) $$

and then

$$ \left(\begin{array}{cc} b\cos\theta+a\sin\theta & a\cos\theta-b\sin\theta\\ -a\cos\theta+b\sin\theta & b\cos\theta+a\sin\theta \end{array}\right)=\rho\left(\begin{array}{cc} \cos\left(\theta_{0}-\theta\right) & \sin\left(\theta_{0}-\theta\right)\\ -\sin\left(\theta_{0}-\theta\right) & \cos\left(\theta_{0}-\theta\right) \end{array}\right) $$

so

$$ \left(\begin{array}{cc} a & -b\\ b & a \end{array}\right)=\rho\left(\begin{array}{cc} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{array}\right)\left(\begin{array}{cc} \cos\left(\theta_{0}-\theta\right) & \sin\left(\theta_{0}-\theta\right)\\ -\sin\left(\theta_{0}-\theta\right) & \cos\left(\theta_{0}-\theta\right) \end{array}\right)=\rho\left(\begin{array}{cc} \sin\theta_{0} & -\cos\theta_{0}\\ \cos\theta_{0} & \sin\theta_{0} \end{array}\right) $$

hence

$$ \left(\begin{array}{cc} a & -b\\ b & a \end{array}\right)^{n}=\rho^{n}\left(\begin{array}{cc} \sin\left(n\theta_{0}\right) & -\cos\left(n\theta_{0}\right)\\ \cos\left(n\theta_{0}\right) & \sin\left(n\theta_{0}\right) \end{array}\right) $$

with

$$ \theta_0 = \arctan\left(\frac ab\right)\\ \rho = \sqrt{a^2+b^2} $$

Answer 2

Hint: \begin{align} \sin(\pi/6)+\cos(\pi/6) =& \sqrt{2}\left(\frac{\sin(\pi/6)}{\sqrt{2}}+\frac{\cos(\pi/6)}{\sqrt{2}} \right)\\ =&\sqrt{2}\left(\sin(\pi/6)\sin(\pi/4)+\cos(\pi/6)\cos(\pi/4)\right)=\ldots \end{align}

Answer 3

You’re on the right track by trying to pull out a common scalar factor. You just have to use a different one. Hint: A key feature of $B$ is that its rows and columns are unit vectors, so see if both rows of $A$ have the same norm. You will then have $A = kB$, from which $A^n=k^nB^n$.

Answer 4

$$\frac12 A=\begin{pmatrix} \sin(\pi /6) +\cos(\pi /6) &\sin(\pi /6) - \cos(\pi /6) \\ \cos(\pi /6) - \sin(\pi /6) &\sin(\pi /6) +\cos(\pi /6) \end{pmatrix}$$ $$\frac12 A= \begin{pmatrix} \sin(\pi /6) &\cos(\pi /6) \\ \cos(\pi /6) &-\sin(\pi /6) \end{pmatrix}\begin{pmatrix} 1 &1 \\ 1 &-1 \end{pmatrix}$$