Positive Elements in a C*-Algebra

**Nusc** · March 24th 2009, 05:47 PM

Let $\cal{A} = \cal{B}$ (l^2), let $a = the \;unilateral \;shift \;on \;l^2$ , and let $b=a^*$ . Show that $\sigma(ab) \neq \sigma(ba)$

Given $\cal{A} = \cal{B}$ (l^2) is a C* -Algebra where for each operator T in B(l^2), T* = is the adjoint of T is that true? or only for B(H)?

**Opalg** · March 25th 2009, 08:13 AM

Originally Posted by Nusc

Let $\mathcal{A} = \mathcal{B}(l^2)$ , let $a = \text{the unilateral shift on }l^2$ , and let $b=a^*$ . Show that $\sigma(ab) \neq \sigma(ba)$

The adjoint of the (forwards) unilateral shift is the backwards unilateral shift. So the product $ba$ is the identity (if you shift forwards and then backwards you get back to where you started). But $ab$ is not invertible because the backwards shift kills off the first basis vector. Thus $0\in\sigma(ab)$ but $0\notin\sigma(ba)$ .

In fact, 0 is the only number that can be in the spectrum of st but not in the spectrum of ts (where s, t are elements of a C*-algebra). There is a theorem which says that $\sigma(st)\cup\{0\} = \sigma(ts)\cup\{0\}$ .

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