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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 $V$?

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    $\begingroup$ Because of Property 2 one gets that $(x,x)\in \mathbb{R}$. $\endgroup$
    – mfl
    Dec 21 '20 at 20:12
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  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.
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  • $\begingroup$ Thanks for the answer. I have added one more question to this. Can you please provide an answer to the added question? $\endgroup$
    – Saikat
    Dec 21 '20 at 20:44
  • $\begingroup$ @Saikat Done.${}$ $\endgroup$ Dec 21 '20 at 20:47
  • $\begingroup$ 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? $\endgroup$
    – Saikat
    Dec 21 '20 at 20:57
  • $\begingroup$ 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. $\endgroup$
    – littleO
    Dec 21 '20 at 20:59
  • $\begingroup$ @Saikat Yes, that is correct. $\endgroup$ Dec 21 '20 at 21:39

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