Bilinear form with symmetric "perpendicular" relation is either symmetric or skew-symmetric

Solution 1:

Here's a somewhat gruesome but coordinate-free proof; please let me know if anything's the least bit unclear. Since the claim is trivial for $b = 0$, we can consider only $b\neq 0$.

Define $B : V \to V^\ast$ by $$ B : x \mapsto b(x,\cdot); $$ then $B^T : V = (V^\ast)^\ast \to V^\ast$ is given by $$ B^T : x \mapsto b(\cdot,x), $$ for if $y \in V = (V^\ast)^\ast$, then $$ B^T(y) : x \mapsto y(B(x)) = y(b(x,\cdot)) = b(x,y). $$ Our goal is to show that $B^T = \pm B$, implying immediately that $b$ is symmetric (if $B^T = B$) or antisymmetric (if $B^T = -B$).

First, by the fundamental theorem of linear algebra, $$ \ker B = (\operatorname{im} B^T)^0, \quad \quad \ker B^T = (\operatorname{im} B)^0; $$ by your assumption on $b$, $B(x) = 0$ iff $B^T(x) = 0$, so that $\ker B = \ker B^T$, and hence, by taking annihilators, $\operatorname{im} B = (\ker B^T)^0 = (\ker B)^0 = \operatorname{im} B^T$. As a result, if $$ R := \operatorname{im} B = \operatorname{im} B^T, \quad N := \ker B = \ker B^T, $$ then $B$ and $B^T$ descend to invertible linear transformations $$ \tilde{B}, \; \tilde{B^T} : V/N \to R \subset V^\ast, $$ and hence, $B^T = CB$ for $C := \tilde{B^T}\tilde{B}^{-1} \in L(R)$; it therefore suffices to show that $C = \pm I_R$.

Now, I want to show that $C = cI_R$ for some $c \in F$, so that $B^T = c B$. I'll do this by showing that every $r \in R$ is an eigenvector of $C$. Let $r \in R$ be non-zero, so that $r = \tilde{B}([x]) = b(x,\cdot)$ for some non-zero $[x] \in V/N$. But then, by your assumption on $b$, $$ \ker Cr = \ker C\tilde{B}([x]) = \ker \tilde{B^T}([x]) = \ker b(\cdot, x) = \ker b(x,\cdot) = \ker \tilde{B}([x]) = \ker r, $$ so, as you observed, there exists some constant $\lambda \in F$ such that $Cr = \lambda r$, i.e., $r$ is an eigenvector for $C$ with eigenvalue $\lambda$.

Finally, I want to show that $c = \pm 1$. However, this is trivial, since $B^T \neq 0$ (as $b \neq 0$) and $$ B^T = cB = c(B^T)^T = c(cB)^T = c^2 B^T, $$ so that $c^2 = 1$, as required.