In my book "Linear Algebra as an Introduction to Abstract Mathematics", the inner product on vector space $V$ over field $\mathbb F$ is defined as the following map:
$\langle \cdot,\cdot \rangle : V \times V \to \mathbb F$ where $(u,v) \mapsto \langle u, v \rangle$
Now, given this definition, and the post here (Inner product for vector space over arbitrary field) and here (Why are inner product spaces only defined on $\Bbb R$ or $\Bbb C$?), it appears as though the $\mathbb F$ used by the author most likely refers to either $\mathbb R$ or $\mathbb C$.
The book provides $4$ properties that the inner product must obey...
- "Linearity in the first slot"
- "Positivity"
- "Positive Definiteness"
- "Conjugate Symmetry"
I am "good to go" with the 1st, 3rd, and 4th property...however, I am sort of stuck understanding how the positivity property is implemented when $\mathbb F = \mathbb C$.
The positivity property reads as follows:
Positivity: $\langle v,v\rangle \geq 0$ for all $v \in V$.
For $\mathbb F = \mathbb R$, sure...this is something that makes sense. But when the field is set to the complex plane, I am unsure of how to verify this property. Specifically, how does the notion of $\geq$ translate in the context of scalars taken from the complex plane? Is there a different conventional form that we adopt for the positivity property when talking about the inner product in the context of $\mathbb F = \mathbb C$?
After digging around for a little bit, I came across this post: Positive definiteness in an inner product over complex numbers , which (from my interpretation) makes the following claim:
$\langle x,x \rangle = \overline{\langle x,x \rangle} \iff \langle x,x \rangle \in \mathbb R$. If such a claim is true, then I would agree that even in the context of $V$ over $\mathbb C$, the symbol $\geq$ now makes sense.
I went to prove this biconditional...however, this proof required me to assert that the only valid fields to define an inner product for are either $\mathbb F = \mathbb R$ or $\mathbb F = \mathbb C$. And quite frankly, I am unsure if this really is true (because I have seen several other posts...not listed...about how inner products can also be defined on finite fields).
The proof reads as follows:
Prove: if $\langle x,x \rangle \in \mathbb R$, then $\langle x,x \rangle = \overline{\langle x,x \rangle}$
Suppose $a = \langle x,x \rangle \in \mathbb R$. Viewing $a$ in the context of complex numbers, we can say $a = (\lambda, 0)$. By definition of a complex conjugate, $\bar a = (\lambda, -0) = (\lambda,0)$. Therefore, $a = \bar a$ ... $\langle x,x \rangle = \overline{ \langle x,x \rangle }$. Alternatively, you could just say "by Conjugate Symmetry property $\square$ ".
Prove: if $\langle x,x \rangle = \overline { \langle x,x \rangle}$ then $\langle x,x \rangle \in \mathbb R$.
Suppose, by contradiction, $\langle x,x \rangle \notin \mathbb R$. Then $\langle x,x \rangle \in \mathbb C$ (because this is the only other option).
Let $z = \langle x,x \rangle = (\lambda, \eta)$. By definition of complex conjugate, $\bar z = (\lambda, - \eta)$. $z = \bar z$ only if $\eta = -\eta$, but that can only happen if $\eta = 0$, which would mean $z \in \mathbb R$...a contradiction.
Thus, $z \notin \mathbb C$, which means $z = \langle x,x \rangle \in \mathbb R$.
So we've proved the biconditional.
IMPORTANTLY, this proof required that the only fields I could choose from were either $\mathbb R$ or $\mathbb C$. I have some concerns with this...given the posts I have seen about finite fields also being possible.
Any thoughts on the matter? Thank you!
