The Annihilating Ideal and the Minimal Polynomial

The kernel of evaluation is an ideal

Reconsider evaluation at T as a ring homomorphism ev_T: F[x] -> End(V), f |-> f(T). Its kernel is precisely the set of annihilating polynomials. The kernel of a ring homomorphism is always an ideal: it is closed under addition, and if f annihilates T then so does g f for ANY g, because (gf)(T) = g(T) f(T) = g(T) 0 = 0.

F[x] is a principal ideal domain: every ideal is generated by a single polynomial, found by Euclidean division. The monic generator of the annihilating ideal is the minimal polynomial m_T. So 'f annihilates T' is EXACTLY 'm_T divides f'. Uniqueness from guide 1 and divisibility from guide 2 are now one statement.

Minimal polynomial and eigenvalues share roots

Key theorem: m_T and p_T have exactly the same roots (they differ only in multiplicities). In particular, the roots of m_T are precisely the eigenvalues of T. This is the content of minimal polynomial and eigenvalues.

1. If lambda is an eigenvalue with eigenvector v, then m_T(T) v = m_T(lambda) v (apply each power: T^k v = lambda^k v). But m_T(T) = 0, so m_T(lambda) v = 0, and v != 0 forces m_T(lambda) = 0. Every eigenvalue is a root of m_T.
2. Conversely, if m_T(lambda) = 0, factor m_T(x) = (x - lambda) g(x). Then 0 = m_T(T) = (T - lambda I) g(T). If T - lambda I were invertible, g(T) = 0 would make g a smaller annihilator, contradicting minimality. So T - lambda I is singular: lambda is an eigenvalue.

Computing m_T and reading off nilpotency

Since m_T divides p_T and has the same roots, m_T is built from the same irreducible factors as p_T, each raised to AT MOST its multiplicity in p_T. To find m_T over an algebraically closed field: factor p_T(x) = prod (x - lambda_i)^{a_i}, then for each lambda_i find the smallest power e_i with (T - lambda_i I)^{e_i} acting as zero on the corresponding generalized eigenspace. Then m_T = prod (x - lambda_i)^{e_i}.

N = [0, 1, 0;
     0, 0, 1;
     0, 0, 0]      # a nilpotent Jordan block, eigenvalue 0

p_N(x) = x^3              (char poly, all eigenvalues 0)

N^1 = [0,1,0; 0,0,1; 0,0,0]  != 0
N^2 = [0,0,1; 0,0,0; 0,0,0]  != 0
N^3 = [0,0,0; 0,0,0; 0,0,0]  == 0   <- first zero power

m_N(x) = x^3.   Index of nilpotency = 3 = degree of m_N.