# Orthogonal Complements in Vector Spaces

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). – Alex Becker Jan 19 '12 at 9:20
Dear Radk: Are you sure you don't you want any sesqui-linearity condition? – Pierre-Yves Gaillard Jan 19 '12 at 9:42
@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. – Radk Jan 19 '12 at 9:44
Dear Radk: How do you define $\dim(W')$? – Pierre-Yves Gaillard Jan 19 '12 at 9:57
Dear Radk: How do you know that $W'$ is a sub-$F$-vector space of $V$? – Pierre-Yves Gaillard Jan 19 '12 at 10:15

Let $b$ be a $\mathbb C$-bilinear form on $\mathbb C^2$. Write $b=s+a$ with $s$ symmetric and $a$ anti-symmetric.

There is an $s$-isotropic line $L$ in $\mathbb C^2$. As $L$ is also $a$-isotropic, it is $b$-isotropic.

Alternative wording: There is a line $L$ in $\mathbb C^2$ satisfying $s(L,L)=0$. As $L$ satisfies also $a(L,L)=0$, it satisfies $b(L,L)=0$.

In particular, we have $L\subset L'$, so $L\cap L'\neq0$.

EDIT. To make the answer self-contained, here is a proof of the fact that a symmetric bilinear form $s$ on $\mathbb C^2$ admits an isotropic line:

Assuming this is false, let $L$ be a line in $\mathbb C^2$, and let $L'$ be its orthogonal. We have $\mathbb C^2=L\oplus L'$. There are vectors $v\in L$ and $v'\in L'$ satisfying $s(v,v)=1$ and $s(v',v')=-1$, and $v+v'$ is a nonzero isotropic vector, contradiction.

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