Unital $C^\ast$-algebra $\mathcal{A}$ with positive elements $\mathcal{A}$.

Question

We are given a unital $C^\ast$-algebra $\mathcal{A}$ where the positive elements of $\mathcal{A}$ are defined to be self-adjoint $A\in\mathcal{A}$ with $\sigma(A)\subseteq[0;+\infty[$. Denoting the collection of positive elements $\mathcal{A}_+$ and set $\mathcal{A}_-=\{A|-A\in\mathcal{A}_+\}$.

In the following I want to show that $\mathcal{A}_{-} \cap \mathcal{A}_{+}=\{0\}$.

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Idea: We have $\sigma(A)\subseteq[0;+\infty[$ and also $\sigma(-A)\subseteq[0;+\infty[$. Now this implies that $\sigma(A)=\{0\}$ which holds if $A=0$ and therefore one can conclude that $\mathcal{A}_+\cap\mathcal{A}_-=\{0\}$.

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Am I on the right track? If no, then how can one actually show $\mathcal{A}_{-} \cap \mathcal{A}_{+}=\{0\}$?

Answer 1

You have reduced the problem to showing that if $A$ is selfadjoint and $\sigma(A)=\{0\}$, then $A=0$. This can be done in several ways, depending on what tools you have available. Here are a couple.

• From Banach algebra theory one can show that the spectral radius of $A$ is $\lim_n\|A^n\|^{1/n}$. In the particular case where $A=A^*$, it is not too hard to show that this limit is $\|A\|$.

• If you have the Gelfand transform available, then you know that $C^*(A)\simeq C(\sigma(A))=C(\{0\})=\mathbb C$, and the isomorphism takes $A$ to the identity function $t\longmapsto t$ on the set $\{0\}$, which is the zero function.