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This question already has an answer here:

Can we define inner product on every vector space?

I don't know any example of any vector space that do not have any inner product .

Help me

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marked as duplicate by Rhys Steele, Chappers, Gerry Myerson, José Carlos Santos, mechanodroid Jul 25 '18 at 13:16

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  • $\begingroup$ $L^p$ spaces with $p$ different of two are the typical example . $\endgroup$ – Gustave Jul 25 '18 at 13:44
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If you have a finite dimensional vector space just choose a basis and then define an inner product by assuming that basis is orthonormal. Another way to say the same thing: choosing a basis of a finite dimensional vector space over a field $K$ establishes an isomorphism with $K^n$ where there's a natural inner product.

But I think your question misses an important point. We don't find inner products on vector spaces at random. They come to us because they provide useful information that comes essentially from the source of the vector space itself. You've tagged your question "functional analysis". There you regularly encounter inner products that help you do functional analysis - for example, for Fourier analysis.

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    $\begingroup$ My question is- Can we define inner product on any vector space? $\endgroup$ – M. A. SARKAR Jul 25 '18 at 12:38
  • $\begingroup$ See this essential duplicate @RhysSteele found: math.stackexchange.com/questions/247425/… $\endgroup$ – Ethan Bolker Jul 25 '18 at 12:43
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    $\begingroup$ Even in an infinite-dimensional vector space (over $\mathbb R$ or $\mathbb C$), your method works fine. You need the axiom of choice to ensure that there is a basis, but then every basis gives rise to an inner product, in which that basis is orthonormal. $\endgroup$ – Andreas Blass Jul 25 '18 at 13:05

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