polynomial functional calculus

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

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

$p(a)$

to make sense as an element of $𝒜$, for any polynomial $p:ℂ⟶ℂ$.

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

1 Definition

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

Moreover, any homomorphism from the algebra $ℂ[\lambda ]$ is perfectly determined by the values of $1$ and $\lambda$.

Definition - Consider the algebra homomorphism $\pi \mathrm{:}\mathrm{C}\mathit{}\mathrm{[}\lambda \mathrm{]}\mathrm{⟶}\mathrm{A}$ such that $\pi \mathit{}\mathrm{(}\mathrm{1}\mathrm{)}\mathrm{=}e$ and $\pi \mathit{}\mathrm{(}\lambda \mathrm{)}\mathrm{=}a$. This homomorphism is denoted by

$p⟼p(a)$

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 $x\in 𝒜$.

Theorem - (polynomial spectral mapping theorem) - Let $\mathrm{A}$ be an unital associative algebra over $\mathrm{C}$ and $a$ an element in $\mathrm{A}$. For any polynomial $p$ we have that

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

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

$q(\lambda )={(\lambda -{\lambda }_1)}^{n_1}\mathrm{\cdots }{(\lambda -{\lambda }_k)}^{n_k}$

for some ${\lambda }_1,\mathrm{\dots },{\lambda }_k\in ℂ$ and $n_1,\mathrm{\dots },n_k\in \mathbb{N}$. Thus, we can also write a similar product for $q(a)$ as

$q(a)={(a-{\lambda }_{1}e)}^{n_1}\mathrm{\cdots }{(a-{\lambda }_{k}e)}^{n_k}$

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. $p({\lambda }_i)=\tilde{\lambda }$, and hence $\tilde{\lambda }\in p(\sigma (a))$.

We now prove the inclusion $\sigma (p(a))\supseteq p(\sigma (a))$. Suppose $\tilde{\lambda }\in p(\sigma (a))$, which means that $\tilde{\lambda }=p({\lambda }_0)$ for some ${\lambda }_0\in \sigma (a)$. The polynomial $p-\tilde{\lambda }$ has a zero at ${\lambda }_0$, hence there is a polynomial $d$ such that

$p(\lambda )-\tilde{\lambda }=d(\lambda )(\lambda -{\lambda }_0),\lambda \in ℂ$

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

$p(a)-\tilde{\lambda }e=d(a)(a-{\lambda }_{0}e)$

If $p(a)-\tilde{\lambda }e$ 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 $p(a)-\tilde{\lambda }e$ cannot be invertible, i.e. $\tilde{\lambda }\in \sigma (p(a))$. $\mathrm{\square }$