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M · hu gua over F2

The nuclear operator as a matrix

The hu gua is a linear map over F2

Choose a hexagram and check that multiplying it by the matrix M gives its nuclear hexagram.

The nuclear hexagram (hu gua) takes the inner lines: the output is (l2, l3, l4, l3, l4, l5) of the input hexagram. Since every output line is a copy of an input line, hu gua is a linear map over F2: there is a matrix M of 6×6 such that M·x = hu gua(x) for the 64 hexagrams. And out of that matrix comes, with nothing else, the whole dynamics of the nuclear forest.

The matrix M (6×6 over F2)
l1′l2′l3′l4′l5′l6′
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
l1l2l3l4l5l6

row = output line (l′), column = input line. A 1 at (i, j) means: output line i copies input line j.

M · x = hu gua(x)
x · Cuì
0
0
0
1
1
0
→
0
0
1
0
1
1
=
hu gua = Jiàn

M·x agrees with hu gua(x) ✓

The powers of M and the collapse of the rank
M
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
rank 4
image 16
M²
0
0
1
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
1
0
0
rank 2
image 4
M³
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
rank 2
image 4
M⁴
0
0
1
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
1
0
0
rank 2
image 4

The rank falls 6 → 4 → 2 and stabilises there: they are the images 64 → 16 → 4 of the nuclear forest, now as powers of a matrix. Moreover M⁴ = M² (verified): from the second step onwards, the system merely oscillates.

The image of M² is the 4 attractors
2. Kūn
fixed point
64. Wèi Jì
cycle of 2
63. Jì Jì
cycle of 2
1. Qián
fixed point
0, 21, 42, 63
values of the image of M² = Kūn, Wèi Jì, Jì Jì, Qián

On those 4, M fixes Kūn and Qián and swaps Wèi Jì and Jì Jì: the only cycle of the forest comes from the fact that M, restricted to the image, is an involution with 2 fixed points. Linear algebra tells the same story as the arrows.