Bilinear form with symmetric "perpendicular" relation is either symmetric or skew-symmetric Let $b$ be a bilinear form on a finite-dimension vector space $V$ (over a field with char $\neq$ 2) such that for each $x,y\in V$ one has $b(x,y)=0\Leftrightarrow b(y,x)=0$. Prove that $b$ is symmetric or skew-symmetric.
The condition is equal to this: for every vector $x$ co-vectors $b(\cdot,x)$ and $b(x,\cdot)$ have the same kernels, so for some non-zero  constant $c:b(\cdot,x)=cb(x,\cdot)$. But I didn't manage to prove that $c$ is equal to $1$ or $-1$ for every $x$.
 A: 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.
A: We may assume WLOG that $b$ is non-degenerate on $V$.  (Otherwise restrict to the maximal sized subspace where it is non-degenerate and run the argument there; in the special case of total isotropy, i.e. $b$ being identically zero, then there is nothing to do and the form is trivially both symmetric and skew symmetric. )
lemma:
If $V= W\oplus W^\perp$
where the OP criterion $b(x,y)=0\Leftrightarrow b(y,x)=0$ holds, and $b$ is symmetric on $W$ & skew symmetric on $W^\perp$, then one of these subspaces is zero dimensional.
proof: Supposing each subspace is non-trivial, then we may select some $w\in W$ such that $b(w,w) = \alpha \neq 0$ and $v,v'\in W^\perp$ such that $b(v,v')=-\alpha$.
(we can always find $b(x,v')=-1$ for any non-zero $v'\in W^\perp$ so let $v:=\alpha\cdot x)$.  Then
$0=b(w+v,w+v')=b(w,w)+b(v,v')+b(w,v')+b(v,w)=b(w,w)+b(v,v')$
$\neq 2 \alpha = b(w,w)+b(v',v)= b(w,w)+b(v',v)+b(w,v')+b(v,w)=b(w+v',w+v)$
violating the OP criterion.  Notice this relies on the fact that $2\neq 0$.
main proof:
The proof proceeds by induction on $\dim V$.
Base Case:
If $\dim V=1$ then there is nothing to do as the form is necessarily symmetric.
Inductive Case:
if $b(u,u)=0$ for all $u\in V$ then $b$ is skew symmetric so we are done.  If not, select some $u$ such that
$b(u,u)=\alpha\neq 0$.  Now define $U$ to be the 1 dimensional subspace generated by $u$.
$V = U\oplus U^\perp$
where $b$ is non-degenerate on $U^\perp$ (otherwise there is a non-zero null vector $\in V$), the OP criterion holds for all $x,y\in U^\perp$ and so by induction hypothesis $b$ is either symmetric or skew symmetric on $U^\perp$.  Since $b$ is symmetric on $U$ and neither subspace is zero dimensional, then the lemma tells us that $b$ must be symmetric on $U^\perp$ as well, i.e. $b$ is symmetric on $V$.
A: We may assume that $V=\mathbb F^n$ and $b(x,y)=x^TBy$ for some matrix $B$. Since every symmetric bilinear form is diagonalisable by congruence, we may also assume that
$$
B=H+K=\pmatrix{D_{r\times r}\\ &0}+K
$$
for some nonzero $r\times r$ nonsingular diagonal matrix $D$ and for some skew-symmetric matrix $K$.
Now suppose that $b$ is neither symmetric nor skew-symmetric. Then $H,K\ne0$. So, we can always pick two vectors $x$ and $w$ such that $x^THx\ne0$ and $x^TKw\ne0$:

*

*if some of the first $r$ rows/columns of $K$ are nonzero, i.e., if $k_{ij}\ne0$ for some $i\le r$, pick $x=e_i$ and $w=e_j$;

*if the first $r$ rows and columns of $K$ are zero, then $k_{ij}\ne0$ for some $i,j>r$ (because $K\ne0$) and we may pick $x=e_1+e_i$ and $w=e_1+e_j$.

Now let $y=tx+w$. Then
$$
\begin{aligned}
x^TBy&=tx^THx+x^THw+x^TKw,\\
y^TBx&=tx^THx+x^THw-x^TKw.
\end{aligned}
$$
As $x^THx\ne0$, there exists $t\in\mathbb F$ such that $x^TBy=0$. Hence $b(x,y)=0\ne -2x^TKw=y^TBx-x^TBy=y^TBx=b(y,x)$.

Remark. This exercise is closely related to Hermann Grassmann’s discovery of exterior algebra. If I understand correctly, in his 1862 work “Die Ausdehnungslehre. Vollständig und in strenger Form begründet” (commonly known as “A2”), Grassmann was exploring ways to define a (scalar-, vector or tensor-valued) product of two vectors. He wanted the product to be bilinear (which, in hindsight, is quite natural; dot product, cross product and outer product of two vectors are bilinear, for instances). He also wanted the product to be specified by a set of “defining equations”, i.e. a minimal set of linear equations that implicitly define some products of basis vectors in terms of the others. E.g. on $\mathbb R^2$, given the equations $e_1\cdot e_1-e_2\cdot e_2=0$ and $e_1\cdot e_2=e_2\cdot e_1=0$, then the value of each $e_i\cdot e_j$ and in turn the value of each $u\cdot v$ are uniquely determined by the value of $e_1\cdot e_1$. (If the product is real-valued and we set $e_1\cdot e_1=1$, we obtain the usual dot product.)
He also wanted the defining equations to be invariant under a change of basis, in the sense that if $\{v_1,v_2,\ldots,v_n\}$ and $\{w_1,w_2,\ldots,w_n\}$ are two bases of the same vector space $V$ and
$$
\begin{cases}
\sum_{i,j}c_{ij}^{(1)}(v_i\cdot v_j)=0,\\
\sum_{i,j}c_{ij}^{(2)}(v_i\cdot v_j)=0,\\
\vdots\\
\sum_{i,j}c_{ij}^{(m)}(v_i\cdot v_j)=0\\
\end{cases}
$$
is a set of defining equations (where the $c_{ij}^{(k)}$s are scalar coefficients), we must also have
$$
\begin{cases}
\sum_{i,j}c_{ij}^{(1)}(w_i\cdot w_j)=0,\\
\sum_{i,j}c_{ij}^{(2)}(w_i\cdot w_j)=0,\\
\vdots\\
\sum_{i,j}c_{ij}^{(m)}(w_i\cdot w_j)=0.\\
\end{cases}
$$
He discovered that for the product of two vectors, there are only four possibilities: (1) the set of defining equations is empty, (2) the product is always zero (i.e., all coefficients in the defining equations are zero), (3) the defining equations are $v_i\cdot v_j=v_j\cdot v_i$ for all $i\ne j$, i.e., the product is symmetric, (4) the defining equations are $v_i\cdot v_j+v_j\cdot v_i=0$ for all $i\ne j$, i.e., the product is skew-symmetric.
For products of three vectors, there are other possibilities, but Grassmann‘s investigation of products of two vectors already allowed him to generalise skew-symmetric products of two vectors to wedge products of two or more vectors. He also noticed that when $\dim V=n$, the wedge product of $n$ vectors is nonzero if and only if those vectors are linearly independent, i.e., iff each vector is exterior to (i.e., lies outside) the linear span of the others. He therefore called the wedge product “exterior product”.
The morals of this history are twofold. First, introductions of determinant in many textbooks are often criticised as unmotivated. In my opinion, such criticisms are very problematic, not because “motivation” is a bad thing, but because in the history of mathematics, many if not most concepts were actually discovered without clear motivations. This, I think, is an important fact that a teacher should help their students understand. In case of determinant, in the theory of equations it was most likely discovered after someone saw the patterns in the solutions of systems of linear equations in two and three variables. I don't believe that pioneers like Seki, Leibniz or Cramer had envisioned the concept of determinant without seeing those patterns first. In exterior algebra, determinant shows up quite naturally, but exterior algebra itself was developed out of Grassmann's desire to develop a product of two vectors, not out of any motivation to develop any alternative definition of determinant. As long as one explains how a mathematical concept was possibly come by, I don't see how the introduction of an unmotivated concept that was discovered historically as a by-product rather than created intentionally could pose any problem.
Second, owing to the importance of exterior algebra and geometric algebra in geometric applications and the elegant definition of determinant in terms of top exterior power, many people seem to think that one should use the calculation of the volume of a parallelepiped as a motivation for developing the concept of determinant. But despite the main theme of Grassmann’s A2 was to develop a formal system of mathematics that can be applied to geometry, we see in the above that the starting point of Grassmann's discovery of exterior algebra is mostly if not purely algebraic. It is one thing (and a very good thing) to teach students the geometric interpretation of determinant, but a completely different thing to use calculation of volume as the motivation (countable noun: “the need or reason for doing something”) for studying determinant. The algebraic properties of determinant allow us to use it to calculate volumes of parallelepipeds, but volumetric considerations do not explain why the determinant function still works as expected over other fields or commutative rings. Depicting the geometric interpretation of determinant as a motivation is not only ahistorical, but also severely limiting.
