Linear Isometry on Positive Elements of a $C^{\ast}$-Algebra

Question

I am trying to complete problem 2.5 from Murphy's $\textit{$C^{\ast}$-Algebras and Operator Theory}$, which states the following

Let $\varphi : A \rightarrow B$ be a linear isometry between unital $C^{\ast}$-Algebras $A$ and $B$ such that $\varphi(a^\ast) = \varphi(a)^\ast$ ($a \in A$) and $\varphi(1) = 1$. Show that $\varphi(A^+) \subseteq B^+$.

Here, the notation $A^+$ denotes the set of positive elements of $A$.

I have made some progress on this problem. Let $a \in A^+$. Evidently, $\varphi(a)$ is hermitian, so it suffices to show that every element of $\sigma(\varphi(a))$ is a positive real number. To this end, let $\lambda \in \mathbb{C}$. Then,

$$\varphi(a) - \lambda 1 \not \in \text{Inv}(B) \text{ iff } \varphi(a - \lambda 1) \not \in \text{Inv}(B).$$

My goal is to eventually use positivity of $a$ to show that $\lambda \in \mathbb{R}_{\geq 0}$. I also haven't used the isometry property of $\varphi$, and I'm unsure how to relate this property to some fact about spectrum to get what I want. Could someone point me in the right direction?

Answer 1

Since $\varphi(1)=1$ and $\|\varphi(a)\|=\|a\|$, using Lemma 2.2.2 from Murphy's book we have $$\|\varphi(a)-\|\varphi(a)\|\|=\|a-\|a\|\|\leq\|a\|=\|\varphi(a)\|.$$ Now using Lemma 2.2.2 again, we see that $\varphi(a)$ is positive.