Searching about strictly positive elements, I found this exercise. I tried to solve it, and the following is my attempt. Please check my proof. Is it correct?
Suppose $a$ is strictly positive. By definition, $\overline {aAa} =A$. Also $\overline {aAa}$ is generated by $a$. $1\in A$, so there is a sequence of polynomials $p_n(a)$ such that $1=\lim p_n(a)$. Also $0\not\in \sigma(1)$, and $\sigma(1)=\lim\sigma(p_n(a)) =\lim p_n(\sigma(a))$. Thus $0\not\in \sigma(a^n)$ for every $n$, which implies that $a$ is invertible.
Thanks in advance.
It is almost never true that $\overline{aAa}$ is generated by $a$. Note that $a$ is positive, so the algebra it generates is abelian.
For the second part of the argument (which is irrelevant to the exercise, but let me address it for your benefit): your write "$\lim\sigma(p_n(a))$"; what do you mean by the limit of a sequence of sets? Also, when you consider polynomials, you need to be careful in cases like this because you want polynomials with no constant term. Any such polynomial $p_n(a)$ is of the form $aq_n(a)$, so none can be invertible if $a$ is not invertible; and the set of invertible operators is open, so an invertible cannot be a limit of non-invertibles. Thus if $aq_n(a)\to 1$, then eventually $aq_n(a)$ is invertible, which implies that $a$ is invertible.