Equivalence between the product of a skew symmetric matrix and the product of a bivector and a vector I stumbled upon the following statement on the wikipedia page (https://en.wikipedia.org/wiki/Cross_product#cite_note-lounesto2001-14) about the cross product:

The vector cross product also can be expressed as the product of a skew-symmetric matrix and a vector:
  $$
a \times b = [a]_{\times} b = \left[\begin{array}{ccc}
0 & -a_3 & a_2 \\ 
a_3 & 0 & -a_1 \\ 
-a_2 & a_1 & 0
\end{array}  \right] \left[\begin{array}{ccc}
b_1\\ 
b_2\\ 
b_3
\end{array}  \right].
$$
  This result can be generalized to higher dimensions using geometric algebra. In particular in any dimension bivectors can be identified with skew-symmetric matrices, so the product between a skew-symmetric matrix and vector is equivalent to the grade-1 part of the product of a bivector and vector.[13] In three dimensions bivectors are dual to vectors so the product is equivalent to the cross product, with the bivector instead of its vector dual. In higher dimensions the product can still be calculated but bivectors have more degrees of freedom and are not equivalent to vectors.[13]

I checked the cited source 13, but I couldn't find anything helpful. How are these two things equivalent and in what explicit way? I am especially interested in the four-dimensional case.
 A: Consider a bivector $A=\langle A\rangle_2$ with components $a_{ij}=\langle e_je_iA\rangle_0=-a_{ji}$ with respect to an orthonormal basis $\{e_i\}$:
$$A=\sum_{i<j}a_{ij}\,e_i\wedge e_j=\sum_{i<j}a_{ij}\,e_ie_j=\frac12\sum_{i,j}a_{ij}\,e_ie_j.$$
(The geometric product $e_ie_j$ is the same as the wedge product for $i\neq j$, and for $i=j$ we have $e_ie_i=1$ but $a_{ii}=0$. The sum on the right is $\sum_{i,j}=\sum_{i<j}+\sum_{i=j}+\sum_{i>j}$.)
In particular in $4$ dimensions,
$$A=a_{12}e_1e_2+a_{13}e_1e_3+a_{23}e_2e_3+a_{14}e_1e_4+a_{24}e_2e_4+a_{34}e_3e_4.$$
This acts on vectors $b=\langle b\rangle_1=\sum_k b_ke_k$ by
$$b\mapsto \langle Ab\rangle_1$$
$$=\left\langle\sum_{i<j}a_{ij}\,e_ie_j\sum_k b_k\,e_k\right\rangle_1$$
$$=\sum_{i<j}\sum_k a_{ij}b_k\langle e_ie_je_k\rangle_1$$
(using linearity of grade projection). Now look at the products of basis vectors. If $k=j$, then we have $e_ie_je_j=e_i=\langle e_i\rangle_1$; similarly if $k=i$, then $e_ie_je_i=-e_je_ie_i=-e_j=\langle-e_j\rangle_1$. Finally if $i,j,k$ are all different, then $e_ie_je_k=\langle e_ie_je_k\rangle_3=e_i\wedge e_j\wedge e_k$. These cases can be combined into
$$\langle e_ie_je_k\rangle_1=e_i(e_j\cdot e_k)-(e_i\cdot e_k)e_j=e_i\delta_{jk}-\delta_{ik}e_j,$$
which is a special case of the identity $\langle(a\wedge b)c\rangle_1=a(b\cdot c)-(a\cdot c)b$. So our bivector product simplifies to
$$\langle Ab\rangle_1=\sum_{i<j}\sum_k a_{ij}b_k(e_i\delta_{jk}-\delta_{ik}e_j)$$
$$=\sum_{i<j}a_{ij}b_je_i-\sum_{i<j}a_{ij}b_ie_j$$
$$=\sum_{i<j}a_{ij}b_je_i+\sum_{j>i}a_{ji}b_ie_j$$
$$=\sum_{i<j}a_{ij}b_je_i+\sum_{i>j}a_{ij}b_je_i+\sum_i a_{ii}b_ie_i\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$$
$$=\sum_{i,j}a_{ij}b_je_i.$$
This is exactly the same as the product of the matrix $[a_{ij}]$ with the vector $b$.
Clearly, given either a bivector or an antisymmetric matrix with components $a_{ij}$, we can construct the other with these same components, and they give the same product with any vector.

The Hodge dual in $n$ dimensions is multiplication with the unit $n$-blade $(\pm)e_1e_2e_3\cdots e_n$. This acts on bivectors in $3$ dimensions as
$$e_3e_2e_1(a_{12}e_1e_2+a_{13}e_1e_3+a_{23}e_2e_3)=a_{23}e_1-a_{13}e_2+a_{12}e_3$$
(resulting in a vector), and in $4$ dimensions as
$$-e_1e_2e_3e_4(a_{12}e_1e_2+a_{13}e_1e_3+a_{23}e_2e_3+a_{14}e_1e_4+a_{24}e_2e_4+a_{34}e_3e_4)$$
$$=a_{34}e_1e_2-a_{24}e_1e_3+a_{14}e_2e_3+a_{23}e_1e_4-a_{13}e_2e_4+a_{12}e_3e_4$$
(resulting in another bivector).
