Problem with the Definition of Inner Product Space.

Let $$V$$ be a vector space over a field $$F=\mathbb{R}/\mathbb{C}.$$ Then an inner product on V is defined to be a function $$(,):V\times V \rightarrow \mathbb{F}$$ satisfying the following property:

1. Linearity in the first coordinate.

2. Conjugate-Symmetry.

3. Positivity: For all $$x \in V$$, if $$x$$ is nonzero then $$(x,x)>0$$.

My Questions:

1. If I take $$\mathbb{F}$$ to be $$\mathbb{C}$$. Since it is not an ordered field, the idea of positivity does not make sense. But I find this definition everywhere. Am I missing something out?

2. Every Inner Product defined on $$V$$ (an $$\mathbb{R}$$-vector space) is bilinear. Does every bilinear map $$f: V\times V\rightarrow \mathbb{R}$$ give rise to an inner product on $$V$$?

• Because of Property 2 one gets that $(x,x)\in \mathbb{R}$.
– mfl
Dec 21 '20 at 20:12

1. The third condition should be: for every $$x\in V$$, $$(x,x)\in[0,\infty)$$. In particular, $$(x,x)$$ is always a real number.
2. No. If, say $$f(x,x)$$ is always equal to $$0$$; then $$f$$ is not an inner product. And, if $$V=\Bbb R$$ and $$f(x,y)=-xy$$, this $$f$$ is also not an inner product.
• @Saikat Done.${}$ Dec 21 '20 at 20:47
• By $f(x,x)=0$ always, you mean a bilinear map with this property. Correct? That is if we take $f(x,y)=0$ always. Then this works well. Am I correct? Dec 21 '20 at 20:57
• Conjugate-symmetry tells us that $(y,x) = \bar{(x,y)}$ for all $x,y \in V$. So, $(x,x) = \bar{(x,x)}$ for all $x \in V$. This shows that $(x,x)$ must be real. So we don't need to modify the third condition. Dec 21 '20 at 20:59