Since the matrix A satisfies the polynomial x³ + x² - 1, it follows that the minimal polynomial m(x) of A divides x³ + x² - 1. Note that x³ + x² - 1 is irreducible (over Q) by the rational root test and therefore, m(x) = x³ + x² - 1. Since zero is not a root of m(x), it follows that zero is not an eigenvalue of A. Consequently, det(A) is nonzero.

Now, the characteristic polynomial p(x) of A is given by the product of invariant factors:

p(x) = q₁(x)q₂(x) ^{. . .} q_r(x),

where q₁(x) | q₂(x) | ^{. . .} | q_r(x).

Moreover, q_r(x) is the minimal polynomial of A. In this case, the minimal polynomial m(x) is an irreducible polynomial of degree three. Hence, all the invariant factors are equal to m(x) and so deg p(x) = 3r. Thus n [= deg p(x)] is a multiple of 3.