# polynomial functional calculus

Let $\mathcal{A}$ be an unital associative algebra over $\mathbb{C}$ with identity element  $e$ and let $a\in\mathcal{A}$.

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

 $\displaystyle p(a)$

to make sense as an element of $\mathcal{A}$, for any polynomial    $p:\mathbb{C}\longrightarrow\mathbb{C}$.

This is achieved in the following natural way: for any polynomial $p(\lambda):=\sum c_{n}\,\lambda^{n}$ we the element $p(a):=\sum c_{n}\,a^{n}\in\mathcal{A}$.

## 1 Definition

Recall that the set of polynomial functions in $\mathbb{C}$, denoted by $\mathbb{C}[\lambda]$, is an associative algebra over $\mathbb{C}$ under pointwise operations and is generated by the constant polynomial $1$ and the variable  $\lambda$ (corresponding to the identity function in $\mathbb{C}$).

Consider the algebra homomorphism $\pi:\mathbb{C}[\lambda]\longrightarrow\mathcal{A}$ such that $\pi(1)=e$ and $\pi(\lambda)=a$. This homomorphism is denoted by

 $\displaystyle p\longmapsto p(a)$

and it is called the polynomial functional calculus for $a$.

It is clear that for any polynomial $p(\lambda):=\sum c_{n}\lambda^{n}$ we have $p(a)=\sum c_{n}\,a^{n}$.

## 2 Spectral Properties

We will denote by $\sigma(x)$ the spectrum (http://planetmath.org/Spectrum) of an element $x\in\mathcal{A}$.

Let $\mathcal{A}$ be an unital associative algebra over $\mathbb{C}$ and $a$ an element in $\mathcal{A}$. For any polynomial $p$ we have that

 $\displaystyle\sigma(p(a))=p(\sigma(a))$

: Let us first prove that $\sigma(p(a))\subseteq p(\sigma(a))$. Suppose $\widetilde{\lambda}\in\sigma(p(a))$, which means that $p(a)-\widetilde{\lambda}e$ is not invertible  . Now consider the polynomial in $\mathbb{C}$ given by $q:=p-\widetilde{\lambda}$. It is clear that $q(a)=p(a)-\widetilde{\lambda}e$, and therefore $q(a)$ is not invertible. Since $\mathbb{C}$ is algebraically closed  (http://planetmath.org/FundamentalTheoremOfAlgebra), we have that

 $\displaystyle q(\lambda)=(\lambda-\lambda_{1})^{n_{1}}\cdots(\lambda-\lambda_{% k})^{n_{k}}$

for some $\lambda_{1},\dots,\lambda_{k}\in\mathbb{C}$ and $n_{1},\dots,n_{k}\in\mathbb{N}$. Thus, we can also write a similar product   for $q(a)$ as

 $\displaystyle q(a)=(a-\lambda_{1}e)^{n_{1}}\cdots(a-\lambda_{k}e)^{n_{k}}$

Now, since $q(a)$ is not invertible we must have that at least one of the factors $(a-\lambda_{i}e)$ is not invertible, which means that for that particular $\lambda_{i}$ we have $\lambda_{i}\in\sigma(a)$. But we also have that $q(\lambda_{i})=0$, i.e. $p(\lambda_{i})=\widetilde{\lambda}$, and hence $\widetilde{\lambda}\in p(\sigma(a))$.

We now prove the inclusion $\sigma(p(a))\supseteq p(\sigma(a))$. Suppose $\widetilde{\lambda}\in p(\sigma(a))$, which means that $\widetilde{\lambda}=p(\lambda_{0})$ for some $\lambda_{0}\in\sigma(a)$. The polynomial $p-\widetilde{\lambda}$ has a zero at $\lambda_{0}$, hence there is a polynomial $d$ such that

 $\displaystyle p(\lambda)-\widetilde{\lambda}=d(\lambda)(\lambda-\lambda_{0})\,% ,\qquad\qquad\lambda\in\mathbb{C}$

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

 $\displaystyle p(a)-\widetilde{\lambda}e=d(a)(a-\lambda_{0}e)$

If $p(a)-\widetilde{\lambda}e$ was invertible, then we would see that $a-\lambda_{0}e$ had a left (http://planetmath.org/InversesInRings) and a right inverse  (http://planetmath.org/InversesInRings), thus being invertible. But we know that $\lambda_{0}\in\sigma(a)$, hence we conclude that $p(a)-\widetilde{\lambda}e$ cannot be invertible, i.e. $\widetilde{\lambda}\in\sigma(p(a))$. $\square$

Title polynomial functional calculus PolynomialFunctionalCalculus 2013-03-22 18:48:23 2013-03-22 18:48:23 asteroid (17536) asteroid (17536) 8 asteroid (17536) Feature msc 46H30 msc 47A60 FunctionalCalculus ContinuousFunctionalCalculus2 BorelFunctionalCalculus polynomial spectral mapping theorem