polynomial functional calculus

Let 𝒜 be an unital associative algebra over with identity elementMathworldPlanetmath e and let a𝒜.

The polynomial functional calculus is the most basic form of a functional calculus. It allows the expression


to make sense as an element of 𝒜, for any polynomialMathworldPlanetmathPlanetmathPlanetmath p:.

This is achieved in the following natural way: for any polynomial p(λ):=cnλn we the element p(a):=cnan𝒜.

1 Definition

Recall that the set of polynomial functions in , denoted by [λ], is an associative algebra over under pointwise operations and is generated by the constant polynomial 1 and the variableMathworldPlanetmath λ (corresponding to the identity function in ).

Moreover, any homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath from the algebraMathworldPlanetmathPlanetmathPlanetmathPlanetmath [λ] is perfectly determined by the values of 1 and λ.

Definition - Consider the algebra homomorphism π:C[λ]A such that π(1)=e and π(λ)=a. This homomorphism is denoted by


and it is called the polynomial functional calculus for a.

It is clear that for any polynomial p(λ):=cnλn we have p(a)=cnan.

2 Spectral Properties

We will denote by σ(x) the spectrum (http://planetmath.org/Spectrum) of an element x𝒜.

TheoremMathworldPlanetmath - (polynomial spectral mapping theorem) - Let A be an unital associative algebra over C and a an element in A. For any polynomial p we have that


: Let us first prove that σ(p(a))p(σ(a)). Suppose λ~σ(p(a)), which means that p(a)-λ~e is not invertiblePlanetmathPlanetmath. Now consider the polynomial in given by q:=p-λ~. It is clear that q(a)=p(a)-λ~e, and therefore q(a) is not invertible. Since is algebraically closedMathworldPlanetmath (http://planetmath.org/FundamentalTheoremOfAlgebra), we have that


for some λ1,,λk and n1,,nk. Thus, we can also write a similar productMathworldPlanetmathPlanetmath for q(a) as


Now, since q(a) is not invertible we must have that at least one of the factors (a-λie) is not invertible, which means that for that particular λi we have λiσ(a). But we also have that q(λi)=0, i.e. p(λi)=λ~, and hence λ~p(σ(a)).

We now prove the inclusion σ(p(a))p(σ(a)). Suppose λ~p(σ(a)), which means that λ~=p(λ0) for some λ0σ(a). The polynomial p-λ~ has a zero at λ0, hence there is a polynomial d such that


Thus, we can also write a similar product for q(a) as


If p(a)-λ~e was invertible, then we would see that a-λ0e had a left (http://planetmath.org/InversesInRings) and a right inverseMathworldPlanetmath (http://planetmath.org/InversesInRings), thus being invertible. But we know that λ0σ(a), hence we conclude that p(a)-λ~e cannot be invertible, i.e. λ~σ(p(a)).

Title polynomial functional calculus
Canonical name PolynomialFunctionalCalculus
Date of creation 2013-03-22 18:48:23
Last modified on 2013-03-22 18:48:23
Owner asteroid (17536)
Last modified by asteroid (17536)
Numerical id 8
Author asteroid (17536)
Entry type Feature
Classification msc 46H30
Classification msc 47A60
Related topic FunctionalCalculus
Related topic ContinuousFunctionalCalculus2
Related topic BorelFunctionalCalculus
Defines polynomial spectral mapping theorem