Recent content by jOc3

This is what I come out with: Let y\in<S>^{\bot}, then <y,w>=0, for all w\in<S>. But S\subseteq<S>. Hence for y\in<S>^{\bot}, then <y,w>=0, for all w\inS. Hence y\inS^{\bot}. Hence <S>^{\bot}\subseteqS^{\bot}. Is it correct? Or should I say "for some w"?

Homework Statement Show that <S>^{\bot}=S^{\bot} Homework Equations The Attempt at a Solution I manage to show S^{\bot}\subseteq<S>^{\bot}. What about the other way round? Any way of proving without using the concept of basis?

It's hard to find the proofs of these theorems. Please help me... Thanks! Theorem 1: Let V be a vector space over GF(q). If dim(V)=k, then V has \frac{1}{k!} \prod^{k-1}_{i=0} (q^{k}-q^{i}) different bases. Theorem 2: Let S be a subset of F^{n}_{q}, then we have dim(<S>)+dim(S^{\bot})=n.

Wow! That really answers all my doubt. Thank you!

This is what I get from wikipedia but I can't figure it out. "...it is necessary to restrict the basefield to R & C in the definition of inner product space. Briefly, the basefield has to contain an ordered subfield (in order for non-negativity to make sense) & therefore has to have...

I come over this in my coding theory but can't understand it. It says finite fields do not fulfil the definition of inner product space like other fields (R and C. Why? How is the proof? Thanks!