The transition matrix P used in the last two posts can be factored into a rotation, R, about the normal to the plane of the probability vectors followed by a radial contraction, V, in this plane. So we have P=V·R. To show that V is a contraction we can choose a set of unit vectors, the basis B, that are perpendicular to each other with two in the plane of the probability vectors and one perpendicular to it. The choice of the vectors in the plane is somewhat arbitrary. B

^{T}·B shows the projection of each of the column vectors of B onto each other. The unit diagonals indicate that the magnitude of each is 1 and the 0 components tell us that the vectors are normal to each other. Multiplying B by V we see that the third vector of B which is normal to the plane is left unchanged. (V·B)

^{T}·B indicates that the directions of the vectors in B are also left unchanged but the new vectors in the plane have their components parallel to the plane reduced in magnitude by the same factor, 0.854 showing that it is a contraction.

We see that R has the some of the properties of a rotation with its determinant |R|=1, its column vectors being unit vectors and perpendicular to each other and that they form a right handed set of vectors. One can see that R is a rotation in the plane of the probability vectors by computing R·B and comparing its column vectors with those of B by computing (R·B)

^{T}·B. We see that the column vector B

^{<0>}moves towards B

^{<1>}and B

^{<1>}moves towards -B

^{<0>}. The angle of rotation is θ = -0.1 radians. By computing V·R we see that it is indeed equal to P.

The rotation preserves the relative position of points in the plane of the probability vectors and the contraction preserves the relative directions of the points. The result is that the images of the border's triangle is always a triangle similar to it. The paths that the points follow are somewhat like those of a flow field.

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