We can use the scalars of a polynomial to build a new operator, which scales copies of an operator with the coefficients a_{j} of the polynomial. constituents p(z) = a_{0} + a_{1}z + a_{2}z^{2} + \cdots + a_{m}z^{m}, a polynomial for z \in \mathbb{F} T \in \mathcal{L}(V) requirements p(T) is an operator refined by:

where, T^{m} is the power of operator additional information p(z) \to p(T) is a linear function additivity: (p_{1} + p_2)T = (a_{0}+b_{0})I … = a_{0} I + b_{0} I … = p_{1}(T) + p_{2}(T) homogeneity: (\lambda p)T = (\lambda a_{0})I … = \lambda (a_{0} I \dots) = \lambda p(T) polynomial of operator is commutative (pq)T = p(T) q(T) p(T)q(T) = q(T)p(T) The first result can be shown because the product of polynomials are a result of rote algebra, and when you come across pq trying to combine z^{j+k} at each FOIL part, you just swap that into T^{j+k} = T^{j}T^{k}. Then, you re-split the constants towards either side (i.e. if the FOIL gave a_{j} b_{k} T^{j+k} \implies a_{j} T^{j} b_{k} T^{k}), then you factor the sums out into two separate pieces to get to p(T) and q(T). The second result: p(T) q(T) = (pq)(T) = (qp)T = q(T) p(T), with the middle commutativity because \mathbb{F} commutes.