Periodic sequence

Algebra Level 4

Let v ( k ) = [ x ( k ) y ( k ) z ( k ) ] \mathbf{v}(k) = \begin{bmatrix} x(k) \\ y(k) \\ z(k) \end{bmatrix}

If

v ( k + 1 ) = A v ( k ) \mathbf{v}(k+1) =A \mathbf{v}(k)

where

A = [ 0.944901543 0.267677991 0.188439826 0.285722226 0.955319386 0.07568144 0.159761963 0.125352956 0.979164313 ] A =\begin{bmatrix} 0.944901543 &&-0.267677991 && 0.188439826 \\ 0.285722226 && 0.955319386&&-0.07568144 \\ -0.159761963&&0.125352956&&0.979164313\end{bmatrix}

And v ( 0 ) = [ 3 4 5 ] \mathbf{v}(0) = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix}

The sequence { v ( k ) } \{ \mathbf{v}(k) \} is periodic. Find the period.


The answer is 18.

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1 solution

Chris Lewis
Sep 23, 2020

It's easy enough to check that the first time we get v ( k ) = v ( 0 ) \bold{v}(k)=\bold{v}(0) is when k = 18 k=\boxed{18} . It's a bit more interesting to work out why.

What kind of geometric transformation in 3 3 D can be periodic? It must be the case that A A is either a rotation or a rotation with a reflection.

We can verify that A = 1 |A|=1 ; so it is a rotation.

The trace of a rotation matrix depends only on the angle of rotation (ie not the axis of rotation). This is because the trace is invariant under change of basis; using the property that tr ( U V ) = tr ( V U ) \text{tr}(UV)=\text{tr}(VU) , we have tr ( P 1 A P ) = tr ( P 1 ( A P ) ) = tr ( ( A P ) P 1 ) = tr ( A ) \text{tr}(P^{-1}AP) =\text{tr}(P^{-1}(AP))=\text{tr}((AP)P^{-1})=\text{tr}(A)

Knowing that ( 1 0 0 0 cos θ sin θ 0 sin θ cos θ ) \begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\theta & -\sin\theta\\ 0 & \sin\theta & \cos\theta \end{pmatrix}

is a rotation matrix, we find tr ( A ) = 1 + 2 cos θ \text{tr}(A)=1+2\cos\theta

for any 3 3 D rotation matrix. In this case, tr ( A ) = 2.879 \text{tr}(A)=2.879\ldots

and solving we find θ = 0.349 π 9 \theta=0.349\ldots \approx \frac{\pi}{9}

so again we find the period is 18 \boxed{18} .

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