Polynomial Calculus: well defined?

Question

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Problem

Given a C*-algebra $\mathcal{A}$.

Consider an element $A\in\mathcal{A}$.

Introduce an abstract polynomial calculus: $$A=A^*:\quad p(A):=\sum_ka_kA^k\quad(p\in\mathbb{C}[X])$$ This induces the concret polynomial calculus: $$A=A^*:\quad p(A):=\sum_ka_kA^k\quad(p\in\mathcal{P}[\sigma(A)])$$ Why is the concret polynomial calculus well defined: $$p(x)\equiv q(x)\quad x\in\sigma(A)\implies p(A)=q(A)$$

(Besides, what can happen for nonselfadjoint ones?)

Answer 1

Selfadjoints

By the C*-property one has: $\|A\|^2=r(A^*A)$

By the pre-spectral theorem also: $\sigma(p(A))=p(\sigma(A))$

And for continuous functions especially: $\sigma(f)=f(\Omega)$

Thus the concrete polynomial calculus is well-defined since: $$A=A^*:\quad\|p(A)\|^2=r(p(A)^*p(A))=r(\overline{p}p(A))=\|p\|_{\sigma(A)}^2$$

Normals

For merely normal ones the second equality is not valid. It requires another method!

Nilpotents

Consider a nilpotent one: $N\neq0,N^2=0$

So the spectrum is trivial: $\sigma(N)=\{0\}$

Thus the concret polynomial calculus is ill-defined: $\mathrm{id}(0)=0,N\neq0$