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The definition of an inner product in Linear Algebra Done Right by Sheldon Axler assumes that the vector space is over either the real or complex field. PlanetMath makes the same assumption.

Is there a definition of an inner product over, for example, finite fields? I sometimes find finite fields easier to reason about, so it would be nice to have a definition of an inner product for vector spaces over them.

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up vote 3 down vote accepted

You could define an inner product on an ordered field, as you need to satisfy the positive-definiteness axiom. However, without a suitable order on the field, this axiom is meaningless. In order to generalise the conjugation in the definition of a hermitian inner product, you can introduce an involution on a field. As long as you can introduce an order and an involution, you should be able to generalise the definition easily enough.

As an aside, it's worth noting that when you do away with the positive-definiteness axiom, what you have is a symmetric bilinear form, which you can define on most (all? I'm not 100%!) fields.

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You can indeed equip any finite-dimensional vector space over an arbitrary field with a symmetric bilinear form. You could also weaken the positive-definite axiom slightly and only insist that the form is non-degenerate, i.e. that for any given nonzero vector $x$ there exists some $y$ such that $\langle x,y\rangle \neq 0$. For many applications of an inner product, you really only need this nondegeneracy (but you won't have, for example, a good notion of the "angle between two vectors" without an inner product). – Brad Oct 19 '12 at 2:29
Hmmm... I think the requirement that the field should be ordered is a bit too strict and it obviously won't even work for complex vector spaces. A presence of an ordered subfield in the field should be enough as in the case of a complex vector space, where the inner product is defined in such a way that it always returns the value from the field of real numbers (which is an ordered subfield of the field of complex numbers) when both arguments are the same. – Wildcat Dec 1 '13 at 13:26

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