TITLE: What is the underlying structure that makes this analogy so good?
QUESTION [5 upvotes]: In "Linear Algebra Done Right", the author draws (in my opinion) a fantastic parallel between $\mathbb{C}$ and $\mathcal{L}(V)$ (where $V$ is an $\mathbb{F}$-inner product space).  In this analogy, he establishes:

A complex number $z$ corresponds to an operator $T$,
The conjugate $\overline{z}$ corresponds to the adjoint $T^*$,
The complex number $z$ is real corresponds to $T$ being self-adjoint,
The complex number $z$ is non-negative corresponds to $T$ being positive-semidefinite,
The complex number $z$ satisfies $|z| = 1$ corresponds to $T$ being an isometry ($TT^* = I$),

among others.  I'm curious though if there is some underlying structure linking $\mathbb{C}$ and $\mathcal{L}(V)$ that makes the parallel so great (I know they are both vector spaces), or is this just a coincidental observation by the author?

REPLY [2 votes]: It is not a coincidence. $\mathbb C$ can be seen as $\mathcal L(V)$ in the case where $V$ is of dimension one. 
If you want, you can see the notions of adjoint, positive, etc., for operators as the right generalizations of the corresponding notions of $\mathbb C$.