Let v ( k ) = ⎣ ⎡ x ( k ) y ( k ) z ( k ) ⎦ ⎤
If
v ( k + 1 ) = A v ( k )
where
A = ⎣ ⎡ 0 . 9 4 4 9 0 1 5 4 3 0 . 2 8 5 7 2 2 2 2 6 − 0 . 1 5 9 7 6 1 9 6 3 − 0 . 2 6 7 6 7 7 9 9 1 0 . 9 5 5 3 1 9 3 8 6 0 . 1 2 5 3 5 2 9 5 6 0 . 1 8 8 4 3 9 8 2 6 − 0 . 0 7 5 6 8 1 4 4 0 . 9 7 9 1 6 4 3 1 3 ⎦ ⎤
And v ( 0 ) = ⎣ ⎡ 3 4 5 ⎦ ⎤
The sequence { v ( k ) } is periodic. Find the period.
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It's easy enough to check that the first time we get v ( k ) = v ( 0 ) is when k = 1 8 . It's a bit more interesting to work out why.
What kind of geometric transformation in 3 D can be periodic? It must be the case that A is either a rotation or a rotation with a reflection.
We can verify that ∣ 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 ) , we have tr ( P − 1 A P ) = tr ( P − 1 ( A P ) ) = tr ( ( A P ) P − 1 ) = tr ( A )
Knowing that ⎝ ⎛ 1 0 0 0 cos θ sin θ 0 − sin θ cos θ ⎠ ⎞
is a rotation matrix, we find tr ( A ) = 1 + 2 cos θ
for any 3 D rotation matrix. In this case, tr ( A ) = 2 . 8 7 9 …
and solving we find θ = 0 . 3 4 9 … ≈ 9 π
so again we find the period is 1 8 .