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Let $A$ be a $C^*$-algebra and consider it's unitization $A_1$ whose underlying vector space is the direct sum $A\oplus \mathbb{C}$. I want to know how does positive elements in $A_1$ look like.

I tried to find out it with the definition (maybe an other criterion is better) but I'm stuck : $a+\lambda 1\in A_1$ positive means, 1. $(a+\lambda 1\in A_1)^*=a+\lambda 1$, and 2. $\sigma(a+\lambda 1)\subseteq [0,\infty)$.

From 1. follows: $a^*=a$ and $\lambda\in\mathbb{R}$. My problem is 2.: Consider $\mu\in \mathbb{C}\setminus \sigma(a+\lambda 1)$. This means $a+\lambda 1-\mu$ is invertible in $A_1$. There exists an element $b+\eta 1\in A_1$ such that $(a+\lambda 1-\mu)(b+\eta 1)=ab+(\lambda-\mu)b+\eta a+(\lambda-\mu)\eta =0+1$. It follows $ab+(\lambda-\mu)b+\eta a=0$ and $(\lambda-\mu)\eta=1$. But it's pointless.

Could you help me to find out, how does positive elements in $A_1$ look like?

tau
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1 Answers1

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Suppose that $a+\lambda 1 \in A_1$ is positive, so that $\sigma(a+\lambda 1) \subseteq [0,\infty)$. Then one can show that $\sigma(a+\lambda 1)=\{ \mu+\lambda: \mu \in \sigma(a) \}$ (when thinking of $\sigma(a)$ as its spectrum in the unitization). Then if $\sigma(a) \subseteq [c,d]$ for some $c,d \in \mathbb{R}$, we have $\sigma(a+\lambda 1) \subseteq [c+\lambda,d+\lambda]$. That is, $a+\lambda 1$ is positive iff $a=a^*$, $\lambda \in \mathbb{R}$ and the function $f(z)=z+\lambda$ is positive (non-negative) on $\sigma(a)$.

sharris
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  • Thank you! I think the equality $\sigma(a+\lambda 1)={\mu+\lambda; \mu\in\sigma_{A_1}(a)}$ follows from the continuous functional calculus and the spectral mapping theorem $f(\sigma(a))=\sigma(a)+\lambda=\sigma(f(a))=\sigma(a+\lambda)$, is it correct? The function f is positive an $\sigma(a)$, if $f(z)=z+\lambda\ge 0$ for all $z\in \sigma(a)$, but has $\lambda$ to be nonnegative too? – tau Jun 13 '15 at 14:47
  • Meanwhile, the questions are clear now, thank you very much!I will vote up your answer later (I'm still unregistered) – tau Jun 13 '15 at 15:10
  • Yes, that's where it comes from. The constant on 1 does not need to be positive (for example, if your spectrum is contained in [5,10], then the constant -4 will still work), but the constant must be real. – sharris Jun 13 '15 at 23:02