[proofplan]
We pass from $A$ to the unital commutative $C^*$-subalgebra generated by $a$ and $1_A$. Continuous functional calculus identifies this subalgebra with $C(\sigma(a))$, sending $a$ to the coordinate function. Since $a$ is positive and nonzero, its spectrum contains a strictly positive point, and evaluation at that point gives a state on the subalgebra that is positive on $a$. The state [extension theorem](/theorems/59) then extends this state to all of $A$ without changing its value on $a$.
[/proofplan]
[step:Represent the generated subalgebra by continuous functions on the spectrum]
Let $B:=C^*(a,1_A)\subset A$ denote the unital $C^*$-subalgebra generated by $a$ and $1_A$. Since $a\in A_+$, the element $a$ is self-adjoint, so $B$ is commutative.
Let $\sigma_A(a)\subset\mathbb C$ denote the spectrum of $a$ computed in $A$. By continuous functional calculus for one self-adjoint element, there is a unital $*$-isomorphism
\begin{align*}
\Gamma:C(\sigma_A(a))\to B
\end{align*}
such that, for the coordinate function
\begin{align*}
\operatorname{id}_{\sigma_A(a)}:\sigma_A(a)\to\mathbb C
\end{align*}
defined by $\operatorname{id}_{\sigma_A(a)}(\lambda)=\lambda$, one has
\begin{align*}
\Gamma(\operatorname{id}_{\sigma_A(a)})=a.
\end{align*}
This use of functional calculus is compatible with [Basic Rules of Functional Calculus][citetheorem:8553], applied to the normal element $a$.
[/step]
[step:Choose a strictly positive spectral value]
Let
\begin{align*}
r(a):=\sup\{|\lambda|:\lambda\in\sigma_A(a)\}
\end{align*}
denote the spectral radius of $a$ in $A$. Because $a\in A_+$, the spectrum of $a$ is contained in $[0,\infty)$. Since $a\neq 0$, the spectral radius formula for positive elements in a $C^*$-algebra gives
\begin{align*}
r(a)=\|a\|_A>0.
\end{align*}
The spectrum $\sigma_A(a)$ is a nonempty compact subset of $\mathbb C$, and the [continuous function](/page/Continuous%20Function) $\lambda\mapsto |\lambda|$ attains its maximum on $\sigma_A(a)$. Hence there exists $\lambda_0\in\sigma_A(a)$ such that
\begin{align*}
|\lambda_0|=r(a)>0.
\end{align*}
Since $\sigma_A(a)\subset[0,\infty)$, this implies
\begin{align*}
\lambda_0>0.
\end{align*}
[guided]
We need a point of the spectrum at which the coordinate function has positive value. Define the spectral radius of $a$ in $A$ by
\begin{align*}
r(a):=\sup\{|\lambda|:\lambda\in\sigma_A(a)\}.
\end{align*}
Positivity gives the first half of the argument: for a positive element $a$, the spectrum satisfies
\begin{align*}
\sigma_A(a)\subset[0,\infty).
\end{align*}
It remains to rule out the possibility that the spectrum is only $\{0\}$. Since $a\neq 0$, its $C^*$-norm satisfies $\|a\|_A>0$. For a positive element in a $C^*$-algebra, the spectral radius agrees with the norm, so
\begin{align*}
r(a)=\|a\|_A>0.
\end{align*}
The spectrum $\sigma_A(a)$ is compact and nonempty, so the supremum is attained. Therefore there is some $\lambda_0\in\sigma_A(a)$ with
\begin{align*}
|\lambda_0|=r(a)>0.
\end{align*}
Because every spectral value of $a$ is real and nonnegative, this equality implies $\lambda_0>0$. This is the point at which evaluation will detect $a$ positively.
[/guided]
[/step]
[step:Evaluate at the positive spectral value to obtain a state on the subalgebra]
Define the evaluation functional
\begin{align*}
\varepsilon_{\lambda_0}:C(\sigma_A(a))\to\mathbb C
\end{align*}
by
\begin{align*}
\varepsilon_{\lambda_0}(f)=f(\lambda_0)
\end{align*}
for every $f\in C(\sigma_A(a))$.
This functional is linear. If $f\in C(\sigma_A(a))$ satisfies $f\ge 0$ pointwise, then
\begin{align*}
\varepsilon_{\lambda_0}(f)=f(\lambda_0)\ge 0,
\end{align*}
so $\varepsilon_{\lambda_0}$ is positive. Also
\begin{align*}
\varepsilon_{\lambda_0}(1_{\sigma_A(a)})=1,
\end{align*}
where $1_{\sigma_A(a)}$ is the constant function with value $1$. Hence $\varepsilon_{\lambda_0}$ is a state on $C(\sigma_A(a))$.
Define
\begin{align*}
\psi:B\to\mathbb C
\end{align*}
by
\begin{align*}
\psi(b)=\varepsilon_{\lambda_0}(\Gamma^{-1}(b))
\end{align*}
for every $b\in B$. Since $\Gamma$ is a unital $*$-isomorphism and $\varepsilon_{\lambda_0}$ is a state, $\psi$ is a state on $B$. Moreover,
\begin{align*}
\psi(a)=\varepsilon_{\lambda_0}(\operatorname{id}_{\sigma_A(a)})=\lambda_0>0.
\end{align*}
[/step]
[step:Extend the subalgebra state to a state on $A$]
The subalgebra $B\subset A$ is a unital $C^*$-subalgebra with the same unit $1_A$. By the [State Extension Theorem][citetheorem:8561], the state $\psi:B\to\mathbb C$ extends to a state
\begin{align*}
\phi:A\to\mathbb C
\end{align*}
such that $\phi|_B=\psi$.
Since $a\in B$, the restriction identity gives
\begin{align*}
\phi(a)=\psi(a)=\lambda_0>0.
\end{align*}
Thus there exists a state $\phi$ on $A$ satisfying $\phi(a)>0$, as required.
[/step]
