TITLE: Analogues between the adjoint and conjugation
QUESTION [5 upvotes]: I have been told that the adjoint of an operator behaves much like complex conjugation, and so self-adjoint operators are like real numbers. Can someone explain this analogy more in depth, or give a reference?
For example, a corollary on p. 129 of Axler's "Linear Algebra Done Right" says that the proposition
$T$ is self-adjoint iff $(Tv, v) \in \mathbb{R}$ for every $v \in V$.
is an example of how self-adjoint operators behave like real numbers. I don't see what he's talking about here, and am wondering if there are other propositions that illustrate parallels between self-adjoint operators and real numbers.

REPLY [4 votes]: Let $T$ be a normal operator on a finite-dimensional Hilbert space $H$. By the spectral theorem, $T$ admits an orthonormal basis of eigenvectors. If $\lambda_1, ... \lambda_n$ denote its eigenvalues, then $T^{\ast}$ has eigenvalues $\overline{\lambda_1}, ... \overline{\lambda_n}$. In particular, as  Jonas mentions in the comments, the adjoint of scalar multiplication by $\lambda$ is scalar multiplication by $\bar{\lambda}$, and if $T$ is self-adjoint, all of its eigenvalues are real. 
Moreover, any normal operator $T$ can be written as the sum of its "real part" and "imaginary part"
$$T = \frac{T + T^{\ast}}{2} + \frac{T - T^{\ast}}{2}.$$
The first term is self-adjoint and has eigenvalues $\text{Re}(\lambda_1), ... \text{Re}(\lambda_n)$. The second term is skew-adjoint (analogous to purely imaginary) and has eigenvalues $\text{Im}(\lambda_1), ... \text{Im}(\lambda_n)$. Both have the same eigenvectors as $T$.
There are deeper levels on which to understand this analogy, but I'm not sure how useful they would be at this point.