$\langle v, v \rangle \geq 0$ meaning in inner product space. In Linear Algebra Done Right: https://zhangyk8.github.io/teaching/file_spring2018/linear_algebra_done_right.pdf
Inner product space is defined as a vector space V together with an inner product $V \times V \to \mathbb{F}$ where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$.
The inner product must satisfy $\langle v, v \rangle \geq 0$.
How would it be defined if $\langle v, v \rangle \in \mathbb{C}$?
In the definition of norm, they defined norm as $||v|| = \sqrt{\langle v, v \rangle}$. From another source, norm is defined as a nonnegative real value function, that implies $\langle v, v \rangle$ must be nonnegative real.
Additionally, they defined a positive operator as a linear map from $V$ to $V$ that is self-adjoint and $\langle Tv, v \rangle \geq 0 \text{ } \forall v \in V$
How would it be defined if $\langle Tv, v \rangle \in \mathbb{C}$?
Thank you!
 A: This is sometimes an annoying convention, but for complex numbers when we write (for instance) $z\ge 0$, we really mean that $z$ is actually a positive real number. In this case, note that the standard Hermitian inner product on $\Bbb{C}^n$ given by $\langle z,w\rangle =\sum_{i=1}^n \overline{z}^iw^i$ has
$$
\langle z,z\rangle =\sum_{i=1}^n \overline{z}^iz^i=\sum_{i=1}^n \lvert z\rvert^2\in \Bbb{R}.
$$
A: By definition it means that for every $v\in V$, $\langle v,v\rangle\in\Bbb{R}$ and $\langle v,v\rangle\geq 0$; or simply, because the set $[0,\infty)$ is a subset of $\Bbb{C}$, we can just say $\langle v,v\rangle\in [0,\infty)$. It is common to avoid saying all of this, and just write $\langle v,v\rangle\geq 0$.
Here's another long-winded way of saying it. The inner product on a complex vector space is a function $\langle\cdot,\cdot\rangle:V\times V\to \Bbb{C}$. From this function we can define the "quadratic function" $Q:V\to\Bbb{C}$ defined as $Q(v):=\langle v,v\rangle$. Now although $Q$ has the complex numbers $\Bbb{C}$ as the target space, what the axiom is requiring is that the image of the function $Q$ is actually a subset of $[0,\infty)\subset \Bbb{C}$.
It's the same story with $\langle T(v),v\rangle$.
