Why the self-adjointness condition for positivity of an element of a C*-algebra?

Question

A positive element x of a C*-algebra A is a self-adjoint element whose spectrum is contained in the non-negative reals. If there's a faithful finite-dimensional representation of A where the involution is conjugate transposition, I think the second condition just means that x can be thought of as a matrix with positive eigenvalues, so it is self-adjoint*. Are there examples of C*-algebras with elements that have non-negative real spectra but that are not self-adjoint? What is the reason for not counting such elements as positive?

*This isn't true, but I'm leaving it in in case other people make the same mistake.

Answer 1

Sure. For example, any $n \times n$ nilpotent matrix has all eigenvalues zero, so has non-negative real spectrum as an element of $M_n(\mathbb{C})$, but no nonzero nilpotent matrix can be self-adjoint by the spectral theorem. In fact the first reason that occurs to me to not consider these positive is precisely that we don't have the benefit of various nice tools if we drop this condition, such as the spectral theorem and the functional calculus.

The functional calculus implies, for example, that we can treat a positive element $A$ more or less as if it were a positive real number, and in particular we can do things like take its square root $\sqrt{A}$. That's not an operation that makes sense on nilpotent matrices, and in fact not all nilpotent matrices even have square roots.

Answer 2

You can also have an infinite dimensional example. Take $x$ and $y$ be two distinct non-null elements in a Hilbert space $\mathcal{H}$ with dimension at least $2$ such that $\langle x,y\rangle\ge 0$. Then the rank-one operator $x\otimes y$ (defined as $z\mapsto\langle z,y\rangle x$) has the spectrum $\{0,\langle x,y\rangle\}\subset [0,\infty)$ and it is not selfadjoint as $(x\otimes y)^*=y\otimes x$. The $C^*$-algebra can be taken, in this case, $B(\mathcal{H})$.