0
$\begingroup$

Disclaimer

This thread has been refreshed!

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?)

$\endgroup$
9
  • $\begingroup$ What do you mean by "well definition"? $\endgroup$ May 15, 2014 at 16:40
  • $\begingroup$ being well defined... $\endgroup$ May 15, 2014 at 16:42
  • 3
    $\begingroup$ Then say "Is this well defined?", since (perhaps unintuitively) saying "well definition" does not mean anything. Anyway, is this true? Say $E$ is $\mathbb C^2$, and $T$ is given by the $2\times 2$ matrix $\begin{pmatrix}1&1\\0&1\end{pmatrix}$, so $1$ is the only point in its spectrum. Now let $p_1(x)=x-1$ and $p_2(x)=(x-1)^2$, so $p_1$ and $p_2$ coincide on the spectrum (right?), but $T-I\ne 0=(T-I)^2$. $\endgroup$ May 15, 2014 at 16:45
  • $\begingroup$ Ah I got the problem! What I want is to extend the polynomial calculus to the continuous calculus. So what I need is an underlying compact domain so the corresponding polynomial functions become bounded that makes the space of polynomial functions a normed space... $\endgroup$ May 15, 2014 at 16:50
  • 1
    $\begingroup$ You're not going to extend the polynomial calculus to continuous functions, at least not for a general $T \in \mathcal{B}(E)$, even if you restrict to matrices $T$ on finite-dimensional spaces. $\endgroup$ May 16, 2014 at 3:20

1 Answer 1

1
$\begingroup$

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: $\mathrm{id}(0)=0,N\neq0$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .