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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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    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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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.