# There are 3 points in the spectrum of some self-adjoint element of a non-unital C*-algebra.

Let $A$ be a non-unital C*-algebra.

I would like to know a simple way to show that $A$ contains a self-adjoint element whose spectrum has at least $3$ elements.

Note that the spectrum of an element of $A$ is by definition the spectrum inside the unitization of $A$, and $0$ is in each spectrum. Thus the problem is to have 2 nonzero elements of the spectrum of some self-adjoint element. A self-adjoint element with only 1 nonzero element in its spectrum is a scalar multiple of a projection, so the problem is equivalent to showing that there are self-adjoint elements that are not scalar multiples of projections.

Every finite-dimensional C*-algebra is unital, hence $A$ is infinite dimensional. In every infinite-dimensional C*-algebra, there are self-adjoint elements with infinite spectrum. But I am looking to avoid such a strong result, and to learn a much simpler proof for a much simpler fact.

I realize this is partly subjective, but I hope that the goal is clear enough. This is idle curiosity. It is related to this question.

## 2 Answers

Suppose that every non-invertible positive element in $A$ is a scalar multiple of a projection. Fix projections $p,q$. Then $p+q=\lambda r$ for some projection $r$. If $\lambda=1$, then $pq=0$. Otherwise, $\lambda^2r=p+q+pq+qp$, or $(\lambda^2-\lambda)r=pq+qp$. So $(1-p)r(1-p)=0$. This implies $(1-p)q(1-p)=0$, i.e. $q(1-p)=0$, $q\leq p$.

In either case, $pq=qp$. As every positive element is a multiple of a projection, $A$ is abelian. If it contains a projection $p$ with a proper subprojection $q$, then we can construct $2p-q$ with three-point spectrum (since $2p-q=q+2(p-q)$). In other words, the only C$^*$-algebra where every element has one or two-point spectrum is $\mathbb C^2$.

Let $A$ be a nonunital C*-algebra. Then $A$ has dimension greater than $1$, and because self-adjoint elements span $A$, there are linearly independent self-adjoint $a$ and $b$ in $A$.

Suppose to reach a contradiction that $A$ contains no self adjoint element whose spectrum has more than $2$ elements. Then for each self-adjoint $x\in A$, $\sigma(x)=\{0,\|x\|\}$ or $\sigma(x)=\{0,-\|x\|\}$. In particular, $x\geq 0$ or $x\leq 0$.

Thus for each real $t$, $a+tb\geq 0$ or $a+tb\leq 0$, and because $a$ and $b$ are linearly independent, $a+tb\neq 0$. The set of positive elements of $A$ is closed and $t\mapsto a+tb$ is continuous on $\mathbb R$, so this implies that either $a+tb\geq 0$ for all $t$, or $a+tb\leq 0$ for all $t$. We could replace $a$ and $b$ with $-a$ and $-b$ if needed, so without loss of generality suppose $a+tb\geq 0$ for all real $t$.

In the unitization of $A$, $a\leq \|a\|1$. Therefore $0\leq a+tb\leq \|a\|1+tb$ for all real $t$. Thus $\|a\|+t\sigma(b)\subseteq [0,\infty)$ for all real $t$, which implies $b=0$, contradicting the fact that $\{a,b\}$ is linearly independent.