= Alternating bilinear form
{title2=$B(v,v)=0$}
A <bilinear form> is alternating if $B(v,v)=0$ for every <vector> $v$. Expanding $B(v+w,v+w)=0$ gives $B(v,w)+B(w,v)=0$, hence $B(v,w)=-B(w,v)$. If the <characteristic> is not two, the converse follows from $B(v,v)=-B(v,v)$ by dividing by two. In <characteristic> two, the relation $B(v,w)=-B(w,v)$ is just symmetry and does not force zero self-pairings: $B(x,y)=xy$ on $\mathbb F_2$ has $B(1,1)=1$.https://webspace.maths.qmul.ac.uk/p.j.cameron/notes/linalg.pdf[Peter J. Cameron, Notes on Linear Algebra, §§9.1–9.2: alternating forms in arbitrary characteristic]{ref}
The <matrix>
$$
J=\begin{pmatrix}0&1\\-1&0\end{pmatrix},\qquad B(v,w)=v^TJw=v_1w_2-v_2w_1=\det\begin{pmatrix}v_1&w_1\\v_2&w_2\end{pmatrix}
$$
gives an <alternating bilinear form> over every <field>. Over the <real numbers>, its value is the signed <area> of the <parallelogram> with sides $v,w$: $B((2,1),(1,3))=6-1=5$, and swapping the sides gives $-5$. Every self-pairing is zero, yet $\det J=1$, so this <bilinear form> is <nondegenerate>. Directly, pairing $(x,y)$ with $(1,0)$ gives $-y$, and pairing it with $(0,1)$ gives $x$; both vanish only when $(x,y)=0$.
In <characteristic> different from two, a <skew-symmetric matrix> of <odd integer> size has zero <determinant>, since $\det A=\det(-A^T)=(-1)^n\det A$. Thus an <alternating bilinear form> on a finite odd-dimensional <vector space> is a <degenerate bilinear form>. The stronger statement that the <rank> of an <alternating bilinear form> is even holds in all <characteristics>: splitting off a two-dimensional block with <matrix> $J$ repeatedly proves it. A <nondegenerate> <alternating bilinear form> defines a <symplectic vector space>, the linear setting for signed <area> rather than squared length.
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