fact used in C*-algebra proofs.

Recently I have seen this argument being used a lot in the book I am reading. Namely the fact that if $\mathcal{A}$ is a unital $C^{*}$-algebra, let $\mathcal{A}_{+}$ be the collection of positive elements, and let $\mathcal{A}_{-} = \{ A: -A \in \mathcal{A}_{+} \}$ then $\mathcal{A}_{+} \cap \mathcal{A}_{-} = \{0 \}$, how can we just assume that?

Solution 1:

Assume $x \in \mathcal{A}_+\cap \mathcal{A}_-$. Then $x \in \mathcal{A}_+$, so $\sigma_\mathcal{A}(x)\subseteq [0, \infty[$. On the other hand, $-x \in \mathcal{A}_+$, so $\sigma_\mathcal{A}(-x)\subseteq [0,\infty[$, and thus $\sigma_\mathcal{A}(x)\subseteq ]-\infty,0]$. Hence, $$\sigma_\mathcal{A}(x)\subseteq ]-\infty,0]\cap [0, \infty[ = \{0\}$$ and thus $\sigma_\mathcal{A}(x)=0$. By assumption, $x$ is positive, so self-adjoint, so its norm equals its spectral radius. Hence, $x=0$, since the spectral radius is $0$.