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If $V$ is a finite dimensional vector space over any field $F$, we define an inner product on $V$ as a map $\langle \,, \rangle\colon V\times V\rightarrow F$, satisfying,

  1. $\langle u,v+w\rangle =\langle u,v \rangle + \langle u,w \rangle$;

  2. $\langle u+v,w\rangle =\langle u,w \rangle + \langle v,w \rangle$;

  3. $\langle u,v \rangle =0 $ for all $u\in V$ iff $v=0$;

  4. $\langle u,v \rangle =0 $ for all $v\in V$ iff $u=0$;

  5. $\langle v,aw \rangle = \langle av,w \rangle = a\langle v,w\rangle$,

for all $u,v,w\in W$, $a\in F$.

With respect to this inner product, we define orthogonal complement, $W'$, of a subspace $W$ of $V$ to be the set $\{u\in V\colon \langle u,w \rangle=0 \forall w\in W \}$

It can be shown that $dim(W')+dim(W)=dim(V)$. But, we have taken $F$ to be arbitrary field, it can happen that $W\cap W'\neq 0$ (hence $V\neq W\oplus W')$.

Question: (with above assumptions on $V$, $F$) Does there exist an inner product on $V$ such that $W\cap W'=0$ for all subspaces $W$ of $V$ (where $W'$ is orthogonal complement of $W$ w.r.t. corresponding inner product)?

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    Hmmm... so essentially you are asking if the existence of an inner product implies the existence of one which is nondegenerate when we tighten the nondegeneracy condition 3) to require $\langle u,v \rangle =0$ for all $v\in W$ for some subspace $W$ iff $v=0$, and similarly modify 4).2012-01-19
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    Dear Radk: Are you sure you don't you want any sesqui-linearity condition?2012-01-19
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    @Pierre:- This definition I am referring from the book "Coding Theory- Ling, Zing"; and when read about orthogonal complement in the same, I wondered about this natural question.2012-01-19
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    Dear Radk: How do you define $\dim(W')$?2012-01-19
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    Dear Radk: How do you know that $W'$ is a sub-$F$-vector space of $V$?2012-01-19
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    @Pierre: your question is correct; W' is closed under addition, but how about scalar multiplication? I will try to make question precise. (I think I have to add condition ==a; is it so?)2012-01-19
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    @Pierre: So, I have to add condition $=a=$; then $W'$ will be subspace; and $dim(W')+dim(W)=n$ will follow easily from some theory of system linear equations over fields, and question will be clear. If this is so, I will edit question; shall I do it? (thanks too!)2012-01-19
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    Dear Radk: I agree with your last comment. You're welcome!2012-01-19
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    Dear Radk: Suggestion: study the case $F:=\mathbb C,V:=\mathbb C^2$.2012-01-19
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    This definition is a bit troublesome. What happens when $F = V = \mathbb C$? $\langle i\mathbf 1,i\mathbf 1\rangle = i^2 = -1$, so you don't get a norm induced by your "inner product". The fact that we want inner products to induce norms is the reason that we require them to be sesquilinear forms on real or complex vector spaces. What you have here is merely a bilinear form.2012-01-19

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