Are elements of a $C^*$-Algebra strictly positive iff their spectrum is strictly positive? Let $A$ be a $C^*$-Algebra. An element $a\in A$ is said to be positive iff $a=a^*$ and the spectrum $\sigma(a)$ is nonnegative, ie. $\sigma(a)\subset[0,\infty)$. This is equivalent to $\varphi(a)\ge 0$ for all positive linear functionals $\varphi:A\to\mathbb{C}$.
The standard definition for $a$ being strictly positive seems to be that $\varphi(a)>0$ for all nonzero positive linear functionals. Is this definition equivalent to (the more intuitive characterization) $a=a^*$ and $\sigma(a)\subset(0,\infty)$?
I know that this is true in the unital case (proof: An equivalent definition of $a$ being strictly positive is that $a$ is positive and $\overline{aAa}=A$. Assume $a$ is strictly positive, then $a$ is invertible, because $\|axa-1\|<\frac{1}{2}$ for some $x\in A$, hence
$axa$ is invertible, which means that $a$ has a left and right inverse, thus
$a$ is invertible. So $a$ is invertible and positive, ie. $\sigma(a)\subset(0,\infty)$. Conversely assume that $\sigma(a)\subset(0,\infty)$ and $a=a^*$, then $a$ is positive and invertible. Because $a$ is invertible we have $aAa=A$, so $a$ is strictly positive.)
What can be said about the non-unital case?
 A: If $ A $ is non-unital, then the spectrum of an element of $ A $ is defined via the unitization $ A^{+} $ of $ A $. Hence, $ \lambda \in \sigma(a) $ if and only if $ (a,- \lambda) \in A^{+} $ is not invertible. It follows readily that $ 0 \in \sigma(a) $ for any $ a \in A $. If $ 0 \notin \sigma(a) $, then $ (a,0_{\Bbb{C}}) $ would be invertible in $ A^{+} $, but multiplying $ (a,0_{\Bbb{C}}) $ by any $ (b,z) \in A^{+} $ can never give us $ (0_{A},1_{\Bbb{C}}) $, so we get a contradiction.
Therefore, self-adjoint elements of $ A $ can never have strictly positive spectra if $ A $ is non-unital.
A: As you note, the notion of strictly positive is irrelevant on unital C$^*$-algebras. But it is a different notion in the non-unital case. Consider the algebra of compact operators $K(H)$ on a separable Hilbert space. It has a strictly positive element, because every separable C$^*$-algebra has one; but no compact operator is invertible. 
Explicitly, in terms of matrix units you can consider $x=\sum_n\frac1n\,e_{nn}$. This is strictly positive because it is easy to check that $x K(H) x$ contains all rank-one operators. 
