I recently read a little about super vector spaces and naturally I have a question and I don't know if I'm correct or if there's something in the definition that I don't quite understand.
My understanding is that a super vector space is a $\mathbb{Z}_2$-graded vector space, that is a vector space $V$ such that $V=V_0\oplus V_1$, $0,1\in \mathbb{Z}_2$ and $\forall x \in V_i$ we denote the parity of $x$ by $\left|x\right|=i$ (so $x\in V_0$ has parity $0$ and $x\in V_1$ has parity $1$).
Also, given a finite dimensional inner product space $(V,\langle\cdot,\cdot \rangle)$, and a subspace $F$ we can construct the subspace $$F^{\perp}=\left\lbrace u \in V, \langle u,v \rangle =0, v\in F\right\rbrace,$$ the subspace orthogonal to $F$. Furthermore, one can readily show that $V=F\oplus F^\perp$. (Not sure if the same can be said about infinite dimensional?). So call $F=V_0$ and $F^\perp=V_1$ then we can make $(V,\langle\cdot,\cdot\rangle)$ into a super vector space by fixing a subspace $F$, and viewing $V$ as the direct sum of $F$ and $F^\perp$ and saying that if $x\in F, \left|x\right|=0$ and if $x\in F^\perp$ then $\left|x\right|=1$.
So is it correct to say that any finite dimensional inner product space can be made into a super vector space ?