N-th power of a matrix

linear algebramatrices

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?

Best Answer

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

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

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