[thaumasiotes]

I'd like to understand what you're doing, but I'm missing something.

1. Adjoining ε such that ε² = 0.

If the numbers have the form a + bε, isn't the action of ε to map (a, b) to (0, a)? Did you mean to say that ε = [0,1; 0,0]?

2. Adjoining φ = (1 + √5)/2 = [0,1; 1,1].

I'm fascinated by the idea of determining that a 2x2 matrix is equal to a real number.

I see that if you take successive powers of the real number φ, and express them in the form aφ + b, the coefficients a and b will take on values from the Fibonacci sequence. So far so good.

I don't follow the claim that the matrix [0,1; 1,1] actually represents φ. This would imply that the formula F_n = [1, 0] · φ^(n+1) · [1, 0] means that F_N equals the real number 1 (= 1 + 0φ), times φ^{n+1}, times 1 again. But this isn't true. It certainly is true that [1,0][0,1; 1,1]^{n+1}[1,0] is equal to F_n, but I don't follow the interpretation as adjoined numeric values (as opposed to as coefficients of φ).

3. Equivalence of representations [a,b] and aI + bM, where M is a 2x2 matrix representing any adjoined number.

This looks to me like the claim that if [a,b]M = [c,d], then (aI + bM)M = cI + dM. I tried to work this out algebraically and I'm pretty sure it isn't true in general. I'm open to being told that I'm wrong about this. Have I interpreted the idea correctly? What are the conditions under which the equivalence holds?

[crdrost]

(1) This gets a little into difficult notation but e.g. in Mathematica-style notation,

    {{0,0},{1,0}} {{a},{b}} = {{0}, {a}}.

I think that you are preferring to left multiply your matrices by your points so that your points remain horizontal, whereas I am just used to my vectors being column vectors that I write as points sometimes? So that is why we are getting transposes of each other's notation.

(2) So the case for φ is very similar to the case for ε: you want to start with the constitutive relation, in this case φ² = φ + 1, to build the matrix with first column {{0},{1}} (multiplying 1 by φ gives φ) and the second column {{1},{1}} (the above constitutive relation).

The isomorphism is then that if X represents this adjoined unit then the point {{a}, {b}} becomes a I + b X one way [or say if you have a cubic constitutive relation then {{a},{b},{c}} becomes a I + b X + c X² similarly] and M becomes M {{1},{0}} [or say M {{1},{0},{0}} etc].

I would definitely agree that this is probably a much narrower statement than your “if [a,b]M = [c,d], then (aI + bM)M = cI + dM” as the structure of X is very tight. It is always [e_2 e_3 ... e_n c] where e_i is a unit column vector with a 1 in the i’th place and 0s everywhere else, and c is the vector embodying the constitutive relation.

But if you want a condition, the isomorphism condition is probably the best place to go.

So the claim is that φ^n = F_{n-1} + F_n φ.