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Show that if $f$ is a non-singular bilinear form, and $U$ is a subspace of $V$ , then $(U^\perp)^\perp = U$ and $\dim(U) + \dim(U^\perp) = \dim(V)$.

It is clear to me that $U$ is contained in $(U^\perp)^\perp$. But I am struggling to use non-singularity to show the converse.

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Are you assuming finite dimensionality? –  rschwieb Jan 16 '13 at 20:11
    
Oh yes - sorry. Finite dimensionally is assumed. –  user58514 Jan 16 '13 at 20:15

2 Answers 2

For $u \in U$, if $f(u,v) = 0$ for all $v \in V$ then $u = 0$ by non-singularity of $f$. And for $v \in V$, if $f(u,v) = 0$ for all $u \in U$ then $v \in U^\perp$ by definition. Therefore if we restrict the bilinear form $f \colon V \times V \rightarrow F$ to a pairing $U \times V \rightarrow F$, it further descends to a pairing $U \times (V/U^\perp) \rightarrow F$ that is non-singular (i.e., if $\langle u,\overline{v}\rangle = 0$ for all $u$ then $\overline{v} = 0$ while if $\langle u,\overline{v}\rangle = 0$ for all $\overline{v}$ then $u = 0$).

Thus the linear maps $U \rightarrow (V/U^\perp)^*$ and $V/U^\perp \rightarrow U^*$ that we get to the dual spaces by fixing the first coordinate or the second coordinate of the pairing are both embeddings. So by counting dimensions we have $\dim U \leq \dim(V/U^\perp)$ and $\dim(V/U^\perp) \leq \dim U$ (we're in finite dimensions, so dual spaces don't change dimensions). Putting these together gives us $\dim U = \dim V - \dim(U^\perp)$, so $\dim U + \dim(U^\perp) = \dim V$. Replacing $U$ with $U^\perp$ in that last equation then shows us $\dim((U^\perp)^\perp) = \dim V - \dim(U^\perp) = \dim(U)$, so a containment of $U$ inside $(U^\perp)^\perp$ has to be an equality.

By the way, you left out a hypothesis to assure us that $U^\perp$ is well-defined. In general there is a left and right orthogonal subspace: $U^{\perp_L} = \{v \in V : f(v,u) = \text{ for all } u \in U\}$ and $U^{\perp_R} = \{v \in V : f(u,v) = 0 \text{ for all } u \in U\}$, and these need not be the same. If they are not then the notation $U^\perp$ is ambiguous (all the more for $(U^\perp)^\perp$). You need to know that the perpendicularity relation $f(v,w) = 0$ is symmetric in $v$ and $w$ to be sure that $U^\perp$ is well-defined.

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Fix a basis such that $B$ is the matrix for the bilinear form is given by: $xBy^T=\langle x,y\rangle$. Since the form is nondegenerate, $B$ has full rank.

Now form a matrix $M$ whose rows are a basis for $U$, and consider $MB$. This is a matrix with the same rank as $M$, which is just $\dim(U)$. (If $B$ did not have full rank, this could potentially be lower.)

View $MB$ as a transformation from $R^n\rightarrow R^{\dim(U)}$. Notice that the kernel of this transformation is precisely $U^\perp$. By the rank-nullity theorem, the dimension of $\dim(U^\perp)=n-\dim(U)$.

By the same result, $\dim(U^{\perp\perp})=n-\dim(U^\perp)=n-(n-\dim(U))=\dim(U)$.

Since you already established $U\subseteq U^{\perp\perp}$, and the dimensions are equal, equality of $U$ and $U^{\perp\perp}$ must hold.

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