Skew-symmetric form as matrix From Humphreys' Introduction to Lie Algebras and Representation Theory:

$C_l$: Let $\dim V=2l$, with basis $(v_1,\ldots,v_{2l})$. Define a nondegenerate skew-symmetric form $f$ on $V$ by the matrix $S=\left| \begin{array}{cc}
0 & I_l \\
-I_l & 0  \end{array} \right|$. 

So, a skew-symmetric form is a function $B:V\times V\rightarrow F$ such that $B(v,w)=-B(w,v)$ for all $v,w\in V$. Why is it defined using the matrix $S$? What is the meaning of the matrix $S$ here?
 A: A quadratic form $B$ can always be written as a matrix $M$, with $B(v,w) = v^TMw$. This follows from the fact that, by bilinearity, the form is uniquely determined by its action on pairs of basis vectors, so when $m_{ij} = B(e_i, e_j)$ the matrix $M$ acts the same way on pairs of vectors as $B$.
For your particular $S$, suppose you write your $v$ and $w$ in block form: $v = \left[\begin{array}{c}v^1\\v^2\end{array}\right]$, $w = \left[\begin{array}{c}w^1\\w^2\end{array}\right].$
Then $$B(v,w) = v^T Sw = [v^1\ v^2]\left[\begin{array}{c}w^2\\-w^1\end{array}\right] = v^1w^2-v^2w^1 = -(w^1v^2-w^2v^1)=-w^TSv = -B(w,v),$$
so the form corresponding to $S$ is skew-symmetric. It is also clearly nondegenerate, since the rows are linearly independent.
A: Let $V$ be an $n$-dimensional vector space over a field $F$, and let $\beta = \{v_1,\dotsc,v_n\}$ be a basis for $V$. Recall that $\beta$ defines an isomorphism of $F$-vector spaces
$$
 V \xrightarrow{\sim} F^n, \quad v = \sum_{k=1}^n a_k v_k \mapsto [v]_\beta := \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix}.
$$
Now, you will likely remember that $\beta$ also defines an isomorphism of $F$-algebras
$$
 L(V) \xrightarrow{\sim} M_n(F), \quad T \mapsto [T]_\beta := \begin{pmatrix} [Tv_1]_\beta \mid \cdots \mid [Tv_n]_\beta\end{pmatrix},
$$
such that
$$
 \forall T \in L(V), \; v \in V, \quad [Tv]_\beta = [T]_\beta [v]_\beta,
$$
and such that the various algebraic properties of linear transformations translate precisely to the analogous algebraic properties of the matrices, e.g., $T$ is invertible if and only if $[T]_\beta$ is.
In a similar way, $\beta$ defines an isomorphism of $F$-vector spaces
$$
 L(V,V;F) \xrightarrow{\sim} M_n(F), \quad f \mapsto [f]_\beta := \begin{pmatrix} f(v_1,v_1) & \cdots & f(v_1,v_n) \\ \vdots & \ddots & \vdots \\ f(v_n,v_1) & \cdots & f(v_n,v_n) \end{pmatrix},
$$
such that
$$
 \forall f \in L(V,V;F), \; v, \; w \in V, \quad f(v,w) = [v]_\beta^T [f]_\beta [w]_\beta,
$$
and such that the various algebraic properties of the bilinear forms translate precisely to the analogous algebraic properties of the matrices, e.g., $f$ is non-degenerate/symmetric/anti-symmetric if and only if $[f]_\beta$ is invertible/symmetric/anti-symmetric (though I seem to recall that anti-symmetric is a somewhat finnicky notion if $F$ has characteristic $2$). In particular, then, given your basis $\beta$ of $V$, any matrix $S \in M_n(F)$ defines a unique bilinear form on $V$ via
$$
 f(v,w) := [v]_\beta^T S [v]_\beta,
$$
or more concretely,
$$
 f\left(\sum_{i=1}^n a_i v_i,\sum_{j=1}^n b_j v_j\right) := \begin{pmatrix} a_1 & \cdots & a_n \end{pmatrix} S \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix},
$$
in which case, indeed, $[f]_\beta = S$.
