Reflexive Bilinear Form is Symmetric or Alternating

Theorem
Let $\mathbb K$ be a field.

Let $V$ be a vector space over $\mathbb K$.

Let $f$ be a bilinear form on $V$.

Let $f$ be reflexive.

Then $f$ is symmetric or alternating.

Proof
Let $x, y, z \in V$.

Because $f$ is bilinear:
 * $\map f {x, \map f {x, y} z - \map f {x, z} y} = \map f {x, y} \, \map f {x, z} - \map f {x, z} \, \map f {x, y} = 0$

Because $f$ is reflexive:
 * $\map f {\map f {x, y} z - \map f {x, z} y, x} = 0$

Because $f$ is bilinear:
 * $(1): \quad \map f {x, y} \, \map f {z, x} = \map f {x, z} \, \map f {y, x}$

Letting $z = x$, we obtain:
 * $\map f {x, x} \, \map f {x, y} = \map f {x, x} \, \map f {y, x}$

Thus:
 * $(2): \quad$ if $\map f {x, x} \ne 0$, then $\map f {x, y} = \map f {y, x}$ for all $y \in V$.

Suppose $f$ is not symmetric.

We show that $f(u,u) = 0$ for all $u\in V$.

Let $v,w\in V$ with $f(v,w) \neq f(w,v)$.

By $(2)$, $f(v,v) = f(w,w) = 0$.

Suppose $u\in V$ with $f(u,u)\neq0$.

By $(2)$, $f(u,v) = f(v,u)$ and $f(u,w) = f(w,u)$.

By $(1)$, we have:
 * $\begin{cases}f(v,w) f(u,v) = f(v,u) f(w,v)\\ f(w,v) f(u,w) = f(w,u) f(v,w)\end{cases}$.

Because $f(v,w)\neq f(w,v)$:
 * $\begin{cases}f(u,v) = f(v,u) = 0 \\ f(u,w) = f(w,u) = 0\end{cases}$.

Because $f$ is bilinear, $f(u+v, w) = f(v,w) \neq f(v,w) = f(u+v, w)$.

By $(2)$, $f(u+v, u+v) = 0$.

Because $f$ is bilinear and $f(u,v) = f(v,u) = f(v,v) = 0$, $f(u,u) = 0$.

This is a contradiction with the assumption $\map f {u, u} \ne 0$.

Thus $f$ is alternating.

Also see

 * Bilinear Form is Reflexive iff Symmetric or Alternating