# Orthogonality in h^n for a field h

1. Aug 31, 2011

### Josh Swanson

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
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.

Last edited by a moderator: May 5, 2017
2. Aug 31, 2011

### micromass

Staff Emeritus
Let f(V) be our image. It is proven in www.maths.bris.ac.uk/~maxmr/la2/notes_5.pdf[/URL] that it still holds for every subspace W of $h^n$ that

$$\dim{W}+\dim{W^\bot}=n$$

In particular (since $f(V)^\bot=\{0\}$)

$$\dim{f(V)}=n$$

so our map is surjective.

EDIT: I should probably say why our "inner product" is non-degenerate, since I can imagine that this is not always the case. It's because $f(V)^\bot=\{0\}$, from which it follows that $(h^n)^\bot=\{0\}$. So the radical is {0}, which is equivalent to non-degeneracy.

Last edited by a moderator: Apr 26, 2017
3. Sep 1, 2011

### Josh Swanson

Ah, wonderful, thanks for the link! I was hoping the result was still true outside of inner product spaces, and so it is. That completes the proof then.

P.S. Non-degeneracy can also be deduced easily without recourse to the existence of $f$ from

$$(c_1, ..., c_i, ..., c_n) . (0, ..., 0, 1, 0, ..., 0) = c_i$$
so if c is orthogonal to $h^n$ then
$$c . e_i = c_i = 0$$
so $c = 0$.