polynomials in an operator commute
Operators in general do not commute: AB and BA can be wildly different. But there is one safe haven. Any two polynomials in the SAME operator T commute with each other: p(T) q(T) = q(T) p(T), always. The reason is simply that powers of a single operator commute (T^i T^j = T^(i+j) = T^j T^i), and polynomials are built from those powers.
Collect all polynomials in T into the set F[T] = { p(T) : p in F[x] }. This is a commutative subalgebra of End(V) — closed under addition, scalar multiplication, and (commutative) multiplication, and containing I. It is a small, well-behaved island of commutativity sitting inside the generally noncommutative ocean of all operators, and its dimension is the degree of the minimal polynomial.
Commuting operators share structure. Every p(T) leaves invariant each eigenspace and each generalized eigenspace of T, because (T - lambda I) and p(T) commute, so p(T) maps ker(T - lambda I)^j into itself. In particular the spectral projections, the nilpotent and semisimple parts, and T^-1 (when it exists) all live in F[T] and respect T's eigenstructure automatically.
This is why polynomial methods are so powerful for a single operator: you never leave a friendly commutative world, every construction you build out of T inherits T's invariant subspaces, and functional calculus, spectral projections, and the inverse all sit in one tidy algebra. The flip side: to relate two DIFFERENT operators you need them to commute first, which is a separate and stronger hypothesis.
All polynomials in a single operator commute and form a finite-dimensional commutative algebra that automatically preserves the operator's invariant subspaces.
F[T] is the smallest subalgebra of End(V) containing T, and it is exactly the image of the evaluation homomorphism. It need not be all operators that commute with T (the full commutant can be larger), but it always sits inside that commutant.