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Why do we define inner product only over real or complex vector spaces? Why can't we define it over an arbitrary field?

In fact $\langle x,\alpha y+\beta z\rangle=\alpha\langle x,y\rangle+\beta\langle x,z\rangle$ makes sense since $F$ is a field and the $\alpha\langle x,y\rangle$ can be thought of as the muliplication (in $F$) of $\alpha$ and $\langle x,y\rangle.$

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What do you mean by arbitrarily? Over an arbitrary field? –  TZakrevskiy Jul 4 '13 at 9:52
    
@TZakrevskiy: yes ... –  Sriti Mallick Jul 4 '13 at 9:55
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Note that one of the requirements for a non-degenerate symmetric bilinear (or sesquilinear if over $\mathbb C$) form $b$ to be an inner product is that $b(v,v) \geq 0$ for all $v$. In particular this means that you can only have such forms on vector spaces over fields that contain an ordered subfield. This immediately rules out vector spaces over e.g. finite fields.

Of course of you have tuples $(v_1,\dotsc,v_n)$ and $(w_1,\dotsc,w_n)$ of elements of a field $F$ it makes perfectly good sense to consider $\sum_{i=1}^n v_iw_i$ and indeed that can be useful in many circumstances even when the field isn't the reals or complexes. So in some sense it's just a matter of definition.

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