Center of Simple $C^*$-Algebras Given a simple $C^*$-Algebra $A$, consider its center $C = A \cap A'$, i.e. the set of elements in $A$ commuting with every other element in $A$.
I have shown that if $A$ is unital, its center is trivial, i.e. $C = \mathbb{C} \cdot I$. It should be the case that if $A$ is non-unital, we have $C = 0$, but I don't know how to approach showing this.
In the unital case I considered the spectrum of an arbitrary element $a \in C$, which has to be nonempty, and showed that for some $\lambda \in \sigma(a)$ the set $\overline{(\lambda - c)A}$ is an ideal not containing $1$, so it must be $\{0\}$. However, I cannot do that here.
 A: Here's what I would do. You might be able to make this argument more elegant. I'm a bit out of practice on C*-stuffs!
Proposition: Let $A$ be a simple, nonunital C*-algebra. Then, the centre of $A$ is zero.
Proof. Suppose for contradiction that $x$ is a nonzero central element of $A$. Without loss of generality, $x$ is self-adjoint (or else replace it with $x^*x$ which is also central and nonzero). It still makes sense to talk about $\operatorname{spec}(x)$ simply by thinking of $x$ as an element of the minimal unitization $\widetilde A$ (better yet, by spectral permanence, the spectrum of $x$ is the same in any ambient unital C*-algebra). Notice that $0 \in \operatorname{spec}(x)$ necessarily, because $x$ is noninvertible in $\widetilde A$. Let's consider a few different cases:

*

*There exist two distinct nonzero elements $\lambda_1, \lambda_2 \in \operatorname{spec}(a)$. Then, we may choose continuous functions $f_1,f_2:\mathbb{R} \to [0,1]$, both vanishing at $0$, such that $f_1(\lambda_1)=f_2(\lambda_2)=1$ and $f_1\cdot f_2 =0$. Define $x_i=f_i(x)$ for $i=1,2$. Then, $x_1,x_2$ are nonzero central elements of $A$. The principal ideals $x_1 A$ and $x_2 A$ are nonzero (because they contain $x_1$ and $x_2$ respectively) and proper (because their product is zero), contradicting the assumption that $A$ is simple.

*Suppose that $\operatorname{spec}(x)=\{0,\lambda\}$ for some $\lambda \neq 0$. Then, $p=\lambda^{-1}x$ has $\operatorname{spec}(p)=\{0,1\}$ which implies $p$ is a projection: $p^2=p=p^*$. The principal ideal $pA$ (sometimes called a "corner") is nonzero (it contains $p$). Since $A$ is simple, we then have $pA=A$. However, this implies $p$ is a multiplicative identity for $A$, contradicting the assumption that $A$ is nonunital.

*The last remaining case $\operatorname{spec}(x)=\{0\}$ cannot occur because that implies $x=0$, contrary to assumption.

