# 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?

• I guess $a\in A$ is positive iff $a$ is Hermitian and $\sigma(a)$ is subset of $[0, \infty)$. – Timon Apr 26 '15 at 17:14
• Does hermitian mean self-adjoint, ie. $a=a^*$? Then this is already implied by $\sigma(a)\subset\mathbb{R}$. – Kalua Apr 26 '15 at 17:21
• Yes. By Hermitian I mean Self adjoint. Remember $\sigma (a) \subset \mathbb{R}$ doesn't imply $a$ is self adjoint. For example Take take the nilpotent operator $T$ on $\mathbb{C} ^2$ defined by $T(e_1)= 0, T(e_2)= e_1$. Here $\sigma(T) =\{0\} \subset \mathbb{R}$ but T is not self adjoint. – Timon Apr 26 '15 at 17:26
• You're right, of course! I edited the question accordingly, thanks :) – Kalua Apr 26 '15 at 17:35
• For your question this may be a good hint. If $f$ is a positive linear functional on a $C^*$ algebra $A$ with norm 1 and if $a$ is a self-adjoint element then show that $f(a)$ lies in closed convex hull of the $\sigma (a)$. – Timon Apr 26 '15 at 17:37

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.
• Of course you're right, how could I miss something so obvious. That explains why an alternative characterization for strictly positive elements other than $\sigma(a)\subset(0,\infty)$ is needed in the non-unital case. So the answer to my question is no, and with timons comment i may find an easier proof for the unital case. I may add the easier proof later, when i find it. Thanks everyone :) – Kalua Apr 26 '15 at 17:57
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.