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,
$\langle u,v+w\rangle =\langle u,v \rangle + \langle u,w \rangle$;
$\langle u+v,w\rangle =\langle u,w \rangle + \langle v,w \rangle$;
$\langle u,v \rangle =0 $ for all $u\in V$ iff $v=0$;
$\langle u,v \rangle =0 $ for all $v\in V$ iff $u=0$;
$\langle v,aw \rangle = \langle av,w \rangle = a\langle v,w\rangle$,
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)?