|
(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. |