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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.

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    @HuiYu If $a$ is a normal operator then the converse holds.2012-06-15

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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.