Why are alternating matrices only used when $n$ is even? Consider $\textrm{GL}_n(k)$, where $k$ is a field.  If $Q \in \textrm{GL}_n(k)$, then $Q$ induces a $k$-bilinear form $B_Q: k^n \times k^n \rightarrow k$ by the formula $$B_Q(v,w) = v^t Q w$$  $Q$ is called alternating if $B_Q(v,v) = 0$ for all $v$.  If $Q = (a_{ij})$ is alternating, then one can see that $a_{ij} + a_{ji} = 0$ for all $i,j$, or in other words $Q = -Q^t$.  Is it true that if $Q$ is invertible and alternating, then $n$ must be even?  It seems like many references on algebraic groups are assuming this.

I can see how if the characteristic of $k$ is not $2$, then $n$ must be even: if $n$ isn't even, then $\textrm{Det } Q = (-1)^n \textrm{Det } Q^t = -\textrm{Det } Q$, which implies $\textrm{Det } Q = 0$, absurd.  But what about when $k$ does have characteristic $2$?
When the field has characteristic two, an alternating matrix is symmetric with zero diagonal. Suppose $A$ is such a matrix and $x^TAy=1$. Consider the matrix
\[
  B = A - Ayx^TA - Axy^TA.
\]
Then $B$ is symmetric with zero diagonal and, since $\mathrm{ker}(B)$ contains $x$ and $y$, it follows that $\mathrm{rk}(B)=\mathrm{rk}(A)-2$.
(I'm using the fact that $z^TAz=0$ for all $z$.) By induction it follows that the rank of $A$ is even.
For a field $k$ of arbitrary characteristic, a bilinear form $B:V \otimes V \to k$ on a vector space $V/k$ is considered alternating if $B(v, v) = 0$ for all $v\in V$. Away from characteristic $2$, this definition reduces to the one you mentioned:
\begin{align*}
B(v, w) + B(w, v) = B(v + w, v + w).
\end{align*}
In a vague and fuzzy way, alternating forms are annoying over characteristic $2$. For example, we can usually decompose an arbitrary form $B$ into symmetric and antisymmetric forms:
\begin{align*}
B^+(x, y) &amp;= \frac{1}{2}(B(x, y) + B(y, x)) &amp;
B^-(x, y) &amp;= \frac{1}{2}(B(x, y) - B(y, x)).
\end{align*}
(These are the projections of $B$ onto the $+1$ and $-1$ eigenspaces of the operator $(TB)(x, y) = B(y, x)$ on the space of bilinear forms.) In characteristic $2$, this doesn't quite work.