Assumed underlying field for inequality theorems involving inner products of vectors

Question

The section on inner products in Lang's Linear Algebra has proofs on several inequalities involving norms and inner products, like

$| \langle v, w\rangle| \leq ||v||.||w||$,

$||v+w|| \leq ||v|| + ||w||$.

Is there any sense to these inequalities if the underlying field is not $\mathbb{R}$? For example, in the case of complex fields, what happens if either side of any of the inequalities is complex?

There's a theorem in the proof of which the following step is used:

$||v - a_1v_1 - \ldots - a_nv_n||^2 = ||v - c_1v_1 - \ldots - c_nv_n||^2 + ||(c_1-a_1)v_1 + \ldots + (c_n-a_n)v_n||^2$

$\implies ||v - a_1v_1 - \ldots - a_nv_n||^2 \geq ||v - c_1v_1 - \ldots - c_nv_n||^2$,

which assumes that the term on the far right is positive. But if the underlying field isn't $\mathbb{R}$ (let's say it's $\mathbb{C}$), then there 's no guarantee of that being true. So for such inequality theorems, should I be additionally assuming that the underlying field is $\mathbb{R}$?

Answer 1

In order for an inner product space (or normed linear space) to make any sense, the ground field must be $\mathbb{R}$ or $\mathbb{C}$. Note that, in either case, the norm is always real-valued, as is the modulus of the inner product. So, the Cauchy-Schwarz and Triangle inequalities do indeed make sense over $\mathbb{C}$ (and indeed hold true).

Answer 2

No. Norm is always positive, by definition. So are it's powers. What changes, though, when you replace $\mathbb{R}$ by $\mathbb{C}$ or any other field $F$, is the rule $\| a x \| = |a| \, \| x \|$ for all $a \in F$ and $x \in L$. And it's not clear (for arbitrary field) what $| \cdot |$ is. So, in order to define norm over linear space $L$ over field $F$, you have to define absolute value $|\cdot|$ in F, which satisfies standard properties.

P.S. There is an interesting question that bugs me: is it true, that for each field there exists a non-trivial absolute value function?