“…Hence, by considering that y (1) y (2) y (3) y (4) y (5) , y (1) y (2) y (3) y (4) y (5) , and that the matrices on the right-hand side of the expressions above have full rank, by the same reasoning used in the proof of Theorem 1, one has that ⟨g 1, 1 , g 1, 2 , g 2, 1 , g 2, 2 , g 2, 3 , g 2, 4 ⟩ = ⟨y (0) , y (1) , …, y (5) ⟩, ⟨g 1, 3 , g 1, 4 , g 3, 1 , g 3, 2 , g 3, 3 , g 3, 4 ⟩ = ⟨y (0) , y (1) , …, y (5) ⟩ .…”