Nonhermitian elements whose spectra are real? Let $\mathcal{A}$ be a complex C* algebra. It is well-known that $\sigma(a)\subset\mathbb{R}$ if $a$ is a hermitian.
I wonder whether the converse is true. That is, if $\sigma(a)\subset\mathbb{R}$, do we have that $a$ is hermitian?
At least in the algebra of continuous functions, this is true because the spectrum of a function lies on the real line only if the function only takes real values, or equivalently, is hermitian.
But for other algebras I am not quite sure.
 A: As azarel mentioned, if $a$ is normal and $\sigma(a)\subset \mathbb R$, then $a$ is self-adjoint.  For example, this can be seen from your remark about algebras of continuous functions, considering the Gelfand map on $C^*(1,a)$.  (Similary, $a$ is unitary if and only if $a$ is normal and $\sigma(a)\subseteq\mathbb T$, $a$ is positive if and only if $a$ is normal and $\sigma(a)\subseteq[0,\infty)$, and $a$ is a projection if and only if $a$ is normal and $\sigma(a)\subseteq\{0,1\}$.)
As Qiaochu mentioned, there are easy counterexamples in the $2$-by-$2$ matrices, e.g. $\begin{bmatrix}\text{real}&\text{nonzero}\\0&\text{real}\end{bmatrix}$.  Nonzero quasinilpotents have spectrum $\{0\}$ but are not self-adjoint, because the spectral radius of a self-adjoint element $a$ is $\|a\|$ (as can be seen from the C*-identity and the spectral radius formula $r(a)=\lim\limits_{n\to\infty}\|a^n\|^{1/n}$).  Going back to matrices, any diagonalizable matrix with real eigenvalues whose eigenvectors are not orthogonal will also be a counterexample.
