Recalling Vol I: the characteristic polynomial
In a first course you met the characteristic polynomial p(x) = det(xI - A) of a matrix, and you found eigenvalues as its roots. The variable x was a scalar: you substituted numbers to test which ones make xI - A singular. That viewpoint already mixes a scalar variable with a fixed matrix. The leap of Volume II is to let the variable BE the matrix.
Fix a linear operator T on a finite-dimensional space V (or a square matrix A). Powers of T make sense: T^0 = I, T^1 = T, T^2 = T compose T, and so on. Once you can take powers and add scalar multiples, you can evaluate ANY polynomial at T.
Evaluating a polynomial at the operator
Given f(x) = c_k x^k + ... + c_1 x + c_0, define f(T) = c_k T^k + ... + c_1 T + c_0 I. This substitution map f -> f(T) is the evaluation at T. It is a ring homomorphism: (f + g)(T) = f(T) + g(T) and (fg)(T) = f(T) g(T). The product rule works because powers of a SINGLE operator commute with each other.
A = [2, 1; 0, 2] # a 2x2 matrix f(x) = x^2 - 4x + 4 = (x - 2)^2 A^2 = [4, 4; 0, 4] -4A = [-8,-4; 0,-8] +4I = [ 4, 0; 0, 4] ----------------------------- f(A) = A^2 - 4A + 4I = [0, 0; 0, 0] So f(x) = (x-2)^2 ANNIHILATES A: f(A) = 0. Note (x - 2) alone does NOT: A - 2I = [0,1;0,0] != 0.
Annihilating polynomials and the smallest one
A polynomial f is an annihilating polynomial of T if f(T) = 0 (the zero operator). The matrix above is annihilated by (x-2)^2. Do annihilators always exist? Yes: on an n-dimensional space the operators I, T, T^2, ..., T^{n^2} live in the n^2-dimensional space of all operators, so these n^2 + 1 of them must be linearly dependent — that dependence is a nonzero annihilating polynomial of degree at most n^2.
Among all nonzero annihilators, there is a unique monic one of least degree: the minimal polynomial m_T(x). It is the single polynomial that captures everything a polynomial can know about T. The set of ALL annihilators forms an ideal — the subject of guide 3 — and m_T generates it. We will denote by p_T the characteristic polynomial det(xI - T); guide 2 proves it too annihilates T.