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If $ \begin{bmatrix} \cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7} \\ \end{bmatrix}^k = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $, then the least positive integral value of $k$ is?

Actually, I got no idea how to solve this.
I did trial and error method and got 7 as the answer.
how do i solve this?
Can you please offer your assistance? Thank you

chndn
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    Can you think of a geometrical interpretation of that matrix? What, geometrically, happens to a point $(x,y)$ when you multiply it by that matrix? – Gerry Myerson May 24 '13 at 12:37
  • I need to prove it step by step. – chndn May 24 '13 at 12:40
  • A not illuminating step-by-step proof would be to compute $A^k$ by induction using the appropriate trignometric formulae. But it would be better to realize, as Gerry Myerson suggests, that this is the matrix of a rotation. – Julien May 24 '13 at 12:44
  • Calculate the eigenvalues of $[\dots ]$ and use what you've learned here... – draks ... May 24 '13 at 12:46
  • So there is no other method? how do i calculate eigenvalues? istn there any simpler method? – chndn May 24 '13 at 12:47
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    There are thousands of methods, but I think I've given you the best one. Is there something you don't like about it? – Gerry Myerson May 24 '13 at 12:48
  • I dont know what is eigenvalues – chndn May 24 '13 at 12:50
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    Well, I wrote my answer before seeing your last comment… But it is really strange that you are doing exercices on powers of matrices without having seen what eigenvalues are. –  May 24 '13 at 13:04
  • i am studying in +2 ( http://en.wikipedia.org/wiki/10%2B2 ) – chndn May 24 '13 at 13:05
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    @chndn: if you don't know what eigenvalues are, reading some basic linear algebra book would be good. This is a problem related to Cayley-Hamilton theorem, which is common in the linear system theory. – Stan May 24 '13 at 13:07
  • @Stan: While Cayley-Hamilton's theorem is important, it is not needed to solve this problem. –  May 24 '13 at 13:10
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    chndn, my suggestion has nothing to do with eigenvalues. Take a point $(x,y)$ in the plane. Think of it as a vector, and multiply it by your matrix, and then think of that resulting vector as a point in the plane again. How does the new point relate to the original, geometrically? Even look at some simple cases; what happens to the point $(1,0)$? what happens to $(0,1)$? Please try to take these ideas seriously, instead of just shrugging them off. – Gerry Myerson May 24 '13 at 13:19
  • @chndn: Really consider Gerry's answer because my answer is useless if you haven't seen the concepts I use. –  May 24 '13 at 13:33
  • "I have to write the steps in the exams." What is that supposed to mean? How does that engage, in any way, with what we have been saying to you? You are really getting on my nerves! – Gerry Myerson May 25 '13 at 04:03

4 Answers4

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We start with polar coordinates of a point $P=(r,\theta)$ vs its Cartesian form $(x,y)$. You know $x=r\cos\theta, y=r\sin\theta$. Now rotate the point $P$ by an angle of $\alpha$ with respect to origin so its angle with respect to $x$ axis becomes $\theta+\alpha$. Its new polar coordinates is $P'=(r,\theta+\alpha)$ and its new Cartesian coordinates $(x',y')$, where $x'=r\cos(\theta+\alpha), y'=r\sin(\theta+\alpha)$. Use trig identities to expand these

$x'=r(\cos \theta \cos \alpha -\sin \theta \sin \alpha)=x\cos \alpha -y \sin\alpha$, and

$ y'=r(\sin\theta\cos\alpha+\cos\theta\sin\alpha)=y\cos\alpha+x\sin\alpha$.

If you write these as a matrix equation you can finish your problem. $ \left[ \begin{array}{c} x' \\ y' \end{array} \right] = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{bmatrix} \times \left[ \begin{array}{c} x \\ y \end{array} \right] $

If you have several rotation $\alpha_1, \cdots, \alpha_n$ then the effect of multiplying all the related matrices is the same as adding the given angles. In particular if you have $n$ rotations by the same angle $\alpha$ the result will be same as one rotation of $n\alpha$. In short $ \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{bmatrix}^n= \begin{bmatrix} \cos n\alpha & -\sin n\alpha \\ \sin n\alpha & \cos n\alpha \\ \end{bmatrix}$

Maesumi
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Powers of matrices should always be attacked with diagonalization, if feasible. Forget $2\pi/7$, for the moment, and look at $$ A=\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix} $$ whose characteristic polynomial is, easily, $p_A(X)=1-2X\cos\alpha+X^2$. The discriminant is $4(\cos^2\alpha-1)=4(i\sin\alpha)^2$, so the eigenvalues of $A$ are \begin{align} \lambda&=\cos\alpha+i\sin\alpha\\ \bar{\lambda}&=\cos\alpha-i\sin\alpha \end{align} Finding the eigenvectors is easy: $$ A-\lambda I= \begin{bmatrix} -i\sin\alpha & -\sin\alpha\\ i\sin\alpha & \sin\alpha \end{bmatrix} $$ and an eigenvector is $v=[-i\quad 1]^T$. Similarly, an eigenvector for $\bar{\lambda}$ is $w=[i\quad 1]^T$. If $$ S=\begin{bmatrix}-i & i\\1 & 1\end{bmatrix} $$ you get immediately that $$ S^{-1}=\frac{i}{2}\begin{bmatrix}1 & -i\\-1 & -i\end{bmatrix} $$ so, by well known rules, $$ A=SDS^{-1} $$ where $$ D= \begin{bmatrix} \cos\alpha+i\sin\alpha & 0 \\ 0 & \cos\alpha-i\sin\alpha \end{bmatrix}. $$ By De Moivre's formulas, you have $$ D^k= \begin{bmatrix} \cos(k\alpha)+i\sin(k\alpha) & 0 \\ 0 & \cos(k\alpha)-i\sin(k\alpha) \end{bmatrix}. $$ Since $A^k=S D^k S^{-1}$ your problem is now to find the minimum $k$ such that $\cos(k\alpha)+i\sin(k\alpha)=1$, that is, for $\alpha=2\pi/7$, $$ \begin{cases} \cos k(2\pi/7)=1\\ \sin k(2\pi/7)=0 \end{cases} $$ and you get $k=7$.

This should not be a surprise, after all: the effect of $A$ on vectors is exactly rotating them by the angle $\alpha$. If you think to the vector $v=[x\quad y]^T$ as the complex number $z=x+iy$, when you do $Av$ you get $$ Av= \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix} x\cos\alpha-y\sin\alpha\\x\sin\alpha+y\cos\alpha \end{bmatrix} $$ and $$ (x\cos\alpha-y\sin\alpha)+i(x\sin\alpha+y\cos\alpha)= (x+iy)(\cos\alpha+i\sin\alpha)=\lambda z $$ (notation as above). Thus $z$ is mapped to $\lambda z$, which is just $z$ rotated by an angle $\alpha$.

egreg
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Look if your matrix is diagonizable. If it is, write it's diagonal form. Let say that $A$ is the matrix you describe in your post, then we will write $B=S^{-1}AS$ its diagonal form where $S$ is the matrix that allows you to change from the canonical base $E$ to a base $G$ where $B$ is diagonal.

Now assume that $A$ is diagonizable you have a diagonal matrix so the problem is to find the smallest integer $k$ for which $\lambda^k_1=1$ and $\lambda^k_2=1$ where $\lambda_1$ and $\lambda_2$ are the eigenvalues of $A$. The answer you will find will be the one you are looking for. Indeed, let's say that $m$ is the smallest value for which $B^m=I_2$ then you have :

$$I_2=B^m=\underbrace{(S^{-1}AS)…(S^{-1}AS)}_{m \text{ times}}=S^{-1}A^mS \implies A^m=I_2$$

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    Judging from some of the comments, I doubt OP knows about diagonalizable, or about change of base, or eigenvalues. Maybe in another year or two, OP will find this answer helpful. – Gerry Myerson May 24 '13 at 13:21
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    Yes, I've seen his comments after writing my answer… So yes, what you proposed in the comments is clearly the easiest way to do it. –  May 24 '13 at 13:24
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I founded out that $$A^k= \begin{bmatrix} \cos \frac{k.2 \pi}{7} & -\sin \frac{k.2 \pi}{7} \\ \sin \frac{k.2 \pi}{7} & \cos \frac{k.2 \pi}{7} \\ \end{bmatrix} $$

using appropriate trigonometric formulae. Now for $A=I$, $k$ should be equal to $7$

chndn
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