Double orthogonal complement of a non-degenerate bilinear form

Question

For a non-degenerate bilinear form on vector space $V$, is the right orthogonal complement of the left orthogonal complement of a subspace $W$ still $W$?

Answer 1

Suppose $\newcommand{\ang}[1] {\left\langle{#1}\right\rangle}\ang{\cdot,\cdot} : V'\times V \to k$ is our nondegenerate bilinear form. Let $W\subseteq V$, $W' = W^\perp = \newcommand{\set}[1]{\left\{{#1}\right\}}\set{v'\in V' : \ang{v',w}=0 \text{ for all $w\in W$}}$. The question then is whether $W'^\perp = W$.

If $V'$ and $V$ are infinite dimensional, the answer is in general no. For example, let $X$ be an infinite dimensional vector space. The pairing $X^*\times X^{**} \to k$ is nondegenerate, since $X^{**}\to X^{**}$ is an isomorphism and $X^{*}\to X^{***}$ is injective, but $X\subseteq X^{**}$ has orthogonal complement 0, since if a dual vector is 0 on all of $X$, it is in fact 0. Then the orthogonal complement of 0 is $X^{**}$, which is larger than $X$ when $X$ is infinite dimensional. See wiki for more details.

On the other hand, if $V'$ and $V$ are finite dimensional, we're ok. In this case, if $\dim V = \dim V' = n$, and $\dim W = i$, then we can prove that $\dim W' = n-i$. Then by a symmetric argument, we'll have that $\dim W'^\perp = i$, so $W'^\perp = W$.

Now to prove $\dim W'=n-i$. By nondegeneracy of the form, $w\mapsto \ang{\cdot,w}$ yields an isomorphism of $W$ with an $i$-dimensional subspace of $V'^*$. Let $w_j$ be a basis for $W$ for $1\le j \le i$. Then if $e_j$ is the standard basis for $k^i$, the map $v\mapsto \sum_{j=1}^i \ang{v,w_j}e_j$ is a surjective map from $V'\to k^i$ whose kernel is $W'$, since the $w_j$ were a basis for $W$, so the linear forms $\ang{\cdot,w_j}$ are a basis for the image of $W$ in $V'^*$. Thus the kernel has dimension $n-i$ by rank-nullity as desired.

Answer 2

Let $(V,g)$ be your space, and denote $W^R = \{x \in V \mid g(x,\cdot)|_W=0 \}$, as well as $W^L = \{x \in V \mid g(\cdot,x)|_W = 0\}$. It is easy to see that $W \subseteq (W^R)^L$. If $\dim V< +\infty$, then equality holds, because $\dim W = \dim (W^R)^L$. This follows from the formula$$\dim W + \dim(W^R) = \dim W + \dim (W^L) =\dim V$$applied twice. To check this, define $\varphi\colon V \to W^*$ by $\varphi(x) = g(\cdot,x)|_W$. Then $\ker \varphi = W^L$, and $\varphi$ is surjective by non-degeneracy of $g$. So $$\dim V = \dim \ker \varphi + \dim {\rm Im}\,\varphi = \dim W^L + \dim W^*= \dim W^L+\dim W.$$Similarly for $W^R$.

---

In case surjectivity is not obvious from thinking of the symmetric case, let's do it again paying attention: fix a basis $\mathcal{B} =(v_1,\ldots,v_n)$ of $V$, and take $v = \sum_{i=1}^n a_iv_i$. Then $g(v,v_j) = \sum_{i=1}^n a_i g(v_i,v_j)$. The matrix $(g(v_i,v_j))_{i,j=1}^n$ is non-singular, so take the inverse $(h_{ij})_{i,j=1}^n$. We obtain $a_i = \sum_{j=1}^n h_{ji}g(v,v_j)$, and hence $$v = \sum_{i=1}^n\left(\sum_{j=1}^nh_{ji}g(v,v_j)\right)v_i.$$With this, take $f \in W^*$, and consider a linear extension (which we'll also denote by $f$) to $V^*$. We have $$\begin{align}f(v) &= f\left(\sum_{i=1}^n\left(\sum_{j=1}^nh_{ji}g(v,v_j)\right)v_i \right) \\ &= \sum_{i=1}^n\sum_{j=1}^nh_{ji}g(v,v_j)f(v_i) \\ &= g\left(v,\sum_{j=1}^n \left(\sum_{i=1}^nh_{ji}f(v_i)\right)v_j\right) \\ &= \varphi\underbrace{\left(\color{blue}{\sum_{j=1}^n \left(\sum_{i=1}^nh_{ji}f(v_i)\right)v_j}\right)}_{\mbox{depends only on }\mathcal{B}}(v) ,\end{align}$$ as desired.