Let V be a vector space over a field h and let n be a positive integer. Let f:V -> h^n be a linear map given by(adsbygoogle = window.adsbygoogle || []).push({});

f(v) = (f1(v), f2(v), ..., fn(v)). Call two vectors (g1, ..., gn) and (h1, ..., hn) in h^n "orthogonal" if

g1 h1 + ... + gn hn = 0

Suppose the only vector orthogonal to every vector f(v) is the 0 vector. Is f surjective?

Maybe I'm just missing something, but the only way I can see to get the result is to consider the orthogonal complement of the image of f, but that requires the codomain to be an inner product space, which h^n isn't in general. (In fact, in this application, h can be either a field of prime order or the rationals.)

This question was inspired by Andrea Ferretti's http://mathoverflow.net/questions/13322/slick-proof-a-vector-space-has-the-same-dimension-as-its-dual-if-and-only-if-it" [Broken] that the dimension of an infinite dimensional vector space is less than that of its dual. I'm not convinced the end of the proof works out. Considering orthogonality in basically the situation I've outlined is suggested at the end of the comments by KConrad.

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# Orthogonality in h^n for a field h

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