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.