Houjun Liu


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.


  • \(p(z) = a_{0} + a_{1}z + a_{2}z^{2} + \cdots + a_{m}z^{m}\), a polynomial for \(z \in \mathbb{F}\)
  • \(T \in \mathcal{L}(V)\)


\(p(T)\) is an operator refined by:

\begin{equation} p(T) = a_{0} I + a_{1} T + a_{2} T^{2} + \cdots + a_{m} T^{m} \end{equation}

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

  1. \((pq)T = p(T) q(T)\)
  2. \(p(T)q(T) = q(T)p(T)\)

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.