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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!

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    $\begingroup$ In the context of complex numbers, $z \geq 0$ generally means that $z$ is real-valued and nonnegative. That's the intended meaning for $\langle v, v \rangle \geq 0$. $\endgroup$
    – user169852
    Feb 18 at 6:15
  • $\begingroup$ Thanks, personally I think it is not satisfying since they never defined $\geq$ in $\mathbb{C}$ $\endgroup$
    – khanh
    Feb 18 at 6:30
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    $\begingroup$ When the book says ${\mathbb F} = {\mathbb C}$, it means that $\langle u, v\rangle \in {\mathbb C}$ in general, but still $\langle v, v\rangle \in {\mathbb R}$. $\endgroup$ Feb 18 at 6:33
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    $\begingroup$ @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." $\endgroup$
    – user169852
    Feb 18 at 6:45
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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 $$ \langle z,z\rangle =\sum_{i=1}^n \overline{z}^iz^i=\sum_{i=1}^n \lvert z\rvert^2\in \Bbb{R}. $$

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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 $\langle T(v),v\rangle$.

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