$\langle v, v \rangle \geq 0$ 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!

• In the context of complex numbers, $z \geq 0$ generally means that $z$ is real-valued and nonnegative. That's the intended meaning for $\langle v, v \rangle \geq 0$.
– user169852
Feb 18 at 6:15
• Thanks, personally I think it is not satisfying since they never defined $\geq$ in $\mathbb{C}$ Feb 18 at 6:30
• When the book says ${\mathbb F} = {\mathbb C}$, it means that $\langle u, v\rangle \in {\mathbb C}$ in general, but still $\langle v, v\rangle \in {\mathbb R}$. Feb 18 at 6:33
• @khanh The author does define it. See the second-to-last bullet point on page 165 of your PDF. "If $\lambda$ is a complex number, then the notation $\lambda \geq 0$ means that $\lambda$ is real and nonnegative."
– user169852
Feb 18 at 6:45

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 $$\langle z,z\rangle =\sum_{i=1}^n \overline{z}^iz^i=\sum_{i=1}^n \lvert z\rvert^2\in \Bbb{R}.$$
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 $$\langle T(v),v\rangle$$.