How to find $A^{100}$ where $A$ is a matrix with trigonometric entries? 
How to find $A^{100}$ where $A=\left(\begin{matrix}\cos \theta &  -\sin \theta\\  \sin \theta & \cos \theta\end{matrix}\right)$. ? (Given$\theta=\frac{2\pi}{7}$)

I am a highschool student and new in linear algebra and seeking for only problem solving strategies. I found somewhere the trick to solve this kind of problem is find characteristic equation and then for some of the cases it is possible to find any power of matrix just by exponentiation of both sides. But here this is not applicable.
The characteristics equation found to be $\lambda^2-2\lambda \cos \theta+1=0$
I
how to proceed now ?
 A: Standard Techniques should apply. 
Diagonalize the system first. I assume you had learned how (find eigenvalues and vectors). So you have $A = P D P^{-1}$ where $D$ is diagonal and has your eigenvalues along the main diagonal. P is the stack of your eigenvectors.
Thus $A^{100} = PD^{100}P^{-1}$. $D^{100}$ is easy to compute just raise diagonal elements to the 100 th power. 
A: There are numerous ways to tackle this problem. You could immediately recognise that its a rotation matrix, so squaring it will just rotate you through twice that angle. Then you'd see that raising the matrix to the $100$th power will give the same matrix with $\theta=100\times\frac{2\pi}7$. 
If you don't immediately see this though, you could test to see what happens when you square the matrix. You'll find it gives $$A^2=\left(\begin{matrix}\cos 2\theta &  -\sin 2\theta\\  \sin 2\theta & \cos 2\theta\end{matrix}\right)$$ (after using some double angle formulae). Then you could spot a pattern when you find that $$A^3=\left(\begin{matrix}\cos 3\theta &  -\sin 3\theta\\  \sin 3\theta & \cos 3\theta\end{matrix}\right)$$You could prove this pattern by induction, and thus you've solved it. 
Finally, you could diagonalize the matrix, then raise it to the $100$th power. However that's probably overkill for this problem.
A: From the characteristic equation,
$$\begin{align*}
\lambda^2 - 2\lambda \cos \theta + 1 &= 0\\
\lambda &= \frac{2\cos \theta\pm \sqrt{4\cos^2\theta-4}}{2}\\
&= \cos\theta \pm i\sin\theta
\end{align*}$$
For $\lambda_0 = \cos\theta + i\sin\theta$,
$$\begin{align*}
\pmatrix{\cos \theta-\lambda_0 &-\sin\theta\\\sin\theta &\cos\theta-\lambda_0}\mathbf v &= 0\\
\pmatrix{-i\sin\theta &-\sin\theta\\\sin\theta &-i\sin\theta}\mathbf v &= 0\\
\pmatrix{-i&-1\\1&-i}\mathbf v &= 0\\
\mathbf v &= k\pmatrix{i\\1}\\
A\pmatrix{i\\1} &= (\cos\theta + i\sin\theta)\pmatrix{i\\1} \tag 1
\end{align*}$$
Similarly, for $\lambda_1 = \cos \theta - i\sin\theta$,
$$A\pmatrix{1\\i} = (\cos\theta - i\sin\theta)\pmatrix{1\\i}\tag2$$
$(1)\pmatrix{1 &0} + (2)\pmatrix{0&1}$,
$$\begin{align*}
A\pmatrix{i\\1}\pmatrix{1 &0} + A\pmatrix{1\\i}\pmatrix{0&1} &= \lambda_0\pmatrix{i\\1}\pmatrix{1 &0} + \lambda_1\pmatrix{1\\i}\pmatrix{0&1}\\
&= \pmatrix{i\\1}\pmatrix{\lambda_0 &0} + \pmatrix{1\\i}\pmatrix{0&\lambda_1}\\
A\pmatrix{i&1\\1&i} &= \pmatrix{i&1\\1&i}\pmatrix{\lambda_0 &0\\0&0} + \pmatrix{i&1\\1&i}\pmatrix{0 &0\\0&\lambda_1}\\
&= \pmatrix{i&1\\1&i}\pmatrix{\lambda_0 &0\\0&\lambda_1}\\
A &= \pmatrix{i&1\\1&i}\pmatrix{\lambda_0 &0\\0&\lambda_1}\pmatrix{i&1\\1&i}^{-1}\\
A^{n} &= \pmatrix{i&1\\1&i}\pmatrix{\lambda_0 &0\\0&\lambda_1}^{n}\pmatrix{i&1\\1&i}^{-1}\quad(n\in\mathbb N)\\
&= \pmatrix{i&1\\1&i}\pmatrix{\lambda_0^{n} &0\\0&\lambda_1^{n}}\pmatrix{i&1\\1&i}^{-1}\\
&=\frac1{-2} \pmatrix{i&1\\1&i}\pmatrix{\lambda_0^{n} &0\\0&\lambda_1^{n}}\pmatrix{i&-1\\-1&i}\\
&= \frac1{-2} \pmatrix{i&1\\1&i}\pmatrix{\cos n\theta + i\sin n\theta &0\\0&\cos n \theta - i\sin n\theta}\pmatrix{i&-1\\-1&i}\\
&= \frac1{-2} \pmatrix{i\cos n\theta -\sin n\theta &\cos n\theta - i\sin n\theta\\\cos n\theta + i\sin n\theta &i\cos n\theta +\sin n\theta}\pmatrix{i&-1\\-1&i}\\
&= \frac1{-2} \pmatrix{-2\cos n\theta&2\sin n\theta\\-2\sin n\theta &-2\cos n\theta}\\
&= \pmatrix{\cos n\theta&-\sin n\theta\\\sin n\theta &\cos n\theta}\\
\end{align*}$$
A: It might be helpful to realize, by multiplying it out and applying some trigonometric identities,
$$\left(\begin{matrix}\cos \theta &  -\sin \theta\\  \sin \theta & \cos \theta\end{matrix}\right)^2 = \left(\begin{matrix}\cos 2\theta &  -\sin 2\theta\\  \sin 2\theta & \cos 2\theta\end{matrix}\right)$$
More generally,
$$\left(\begin{matrix}\cos \theta &  -\sin \theta\\  \sin \theta & \cos \theta\end{matrix}\right)\left(\begin{matrix}\cos \varphi&  -\sin \varphi\\  \sin \varphi & \cos\varphi\end{matrix}\right)=\left(\begin{matrix}\cos \theta + \varphi &  -\sin \theta+ \varphi\\  \sin \theta + \varphi& \cos \theta+ \varphi\end{matrix}\right)$$
These can be verified by doing the multiplication as normal, and using the identities
$$\begin{align}
\sin(\alpha + \beta) &= \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)\\
\cos(\alpha + \beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)
\end{align}$$
and the original identity is derived for $\alpha = \beta = \theta$. Proceeding inductively, you can see
$$\left(\begin{matrix}\cos \theta &  -\sin \theta\\  \sin \theta & \cos \theta\end{matrix}\right)^n = \left(\begin{matrix}\cos n\theta &  -\sin n\theta\\  \sin n\theta & \cos n\theta\end{matrix}\right)$$
This is particularly noteworthy since the original matrix is a rotation matrix for two dimensions, and you would multiply it by a vector $[ x \; y ]^T$ on the right to rotate that vector by an angle $\theta$. What if you wanted to rotate it twice by the same angle? You could multiply by the same matrix twice (rotating the vector, and then rotating the rotated vector), or just double the angle in the original matrix. Want to rotate it five times by $\theta$ for whatever reason? Either multiply by the fifth power of the matrix, or just replace the angle with $5\theta$. So this formula seems to agree with our intuition.
