minimal polynomial and eigenvalues
The minimal polynomial and the characteristic polynomial look like they might carry different information, but on one question they agree exactly: their ROOTS are identical, and both root sets are precisely the eigenvalues of the operator. What they disagree about is multiplicity, not membership.
Here is why. If lambda is an eigenvalue with eigenvector v, then for ANY polynomial p we have p(T)v = p(lambda)v. So if p annihilates T, then p(lambda)v = 0, and since v is nonzero, p(lambda) = 0 — every eigenvalue is a root of every annihilator, the minimal polynomial included. Conversely, a root lambda of the minimal polynomial must be an eigenvalue, because (x - lambda) divides m and the corresponding factor of m(T) = 0 forces T - lambda I to be non-invertible.
The two polynomials therefore factor over the splitting field as chi(x) = prod (x - lambda_i)^a_i and m(x) = prod (x - lambda_i)^b_i over the SAME distinct eigenvalues lambda_i, with exponents satisfying 1 <= b_i <= a_i. Here a_i is the algebraic multiplicity and b_i is the size of the largest Jordan block at lambda_i — the minimal polynomial records the worst-case block, not the total count.
This is enormously useful: to find the minimal polynomial you first find the distinct eigenvalues (the roots), then determine each exponent b_i by testing the smallest power that makes (T - lambda_i I)^b_i vanish on the generalized eigenspace. The eigenvalues do half the work for you.
Both polynomials carry the eigenvalues 2 and 5; the characteristic exponents count all block sizes, the minimal exponents record only the largest block at each eigenvalue.
Same roots, possibly smaller exponents — that is the whole relationship in one line. The exponent of (x-lambda) in the minimal polynomial equals the index of lambda, i.e. the size of its largest Jordan block; in the characteristic polynomial it equals the algebraic multiplicity, i.e. the total Jordan-block size at lambda.