Skew symmetric matrix of vector

Imagine a column vector ${\bf A} = (A_1, A_2, A_3)$ and define the matrix

$$ A_\times = \left(\begin{array}{ccc} 0 & -A_3 & A_2 \\ A_3 & 0 & -A_1 \\ -A_2 & A_1 & 0 \end{array}\right) $$

Note that if ${\bf B}$ is another column vector, then

$$ A_\times {\bf B} = {\bf A}\times {\bf B} $$

Moreover

$$ {\rm Transpose}(A_\times) = -A_\times $$

The skew-symmetric tensor product of two vectors with components $A_i$ and $B_i$ is the tensor represented by the matrix with components $S_{ij}=A_iB_j - A_jB_i$. It is skew-symmetric (antisymmetric) because $S_{ij}=-S_{ji}$.

The advantage of this representation is that unlike the vector cross product, which is specific to three dimensions, the skew-symmetric product generalizes the concept to arbitrary dimensions.

Explicitly (in three dimensions),

$$A_iB_j-A_jB_i=\begin{pmatrix}0&A_1B_2-A_2B_1&A_1B_3-A_3B_1\\A_2B_1-A_1B_2&0&A_2B_3-A_3B_2\\A_3B_1-A_1B_3&A_3B_2-A_2B_3&0\end{pmatrix}.$$