During my course in linear algebra, the instructor stated that A cross B is the same as the "skew symmetric matrix" of A times B. So, first of all, can someone clarify or provide sources about skew symmetric matrices? Secondly, I can't really comprehend the idea of how a single column vector crossed with another could be represented by a matrix.
Anyhow, thanks in advance!
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),
$$A_iB_j-A_jB_i=\begin{pmatrix}0&A_1B_2-A_2B_1&A_1B_3-A_3B_1\\A_2B_1-A_1B_2&0&A_2B_3-A_3B_2\\A_3B_1-A_1B_3&A_3B_2-A_2B_3&0\end{pmatrix}.$$
We can substitute vector product $ \mathbf{a} \times \mathbf{b}$ by multiplying the vector $\mathbf{b}$ by a matrix because skew-symmetric matrix corresponding to the first vector $\mathbf{a}$ is defined as
$S(\mathbf{a})=[\mathbf{a} \times \mathbf{i} \ \ \mathbf{a} \times \mathbf{j} \ \ \mathbf{a} \times \mathbf{k} ]$,
where $\mathbf{i},\mathbf{j},\mathbf{k}$ are standard basis vectors forming as columns identity matrix $ {I} = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 &0 \\
0 & 0 &1
\end{bmatrix}$.
This gives formula presented above by caverac (you can notice for example that columns are (easy to check) orthogonal vectors to both $\mathbf{a}$ and appropriate standard basis vectors $\mathbf{i},\mathbf{j},\mathbf{k}$ - also lengths of $S(\mathbf{a})$ columns are coherent with properties of cross product for this case).
In this case we have below formula with the use of multiplication the vector by the matrix interpreted as the sum of products of vector columns of matrix by components of vector (scalars):
$S(\mathbf{a})\mathbf{b}= (\mathbf{a} \times \mathbf{i})b_x + (\mathbf{a} \times \mathbf{j})b_y + (\mathbf{a} \times \mathbf{k}) b_z =\mathbf{a} \times (b_x\mathbf{i} + b_y\mathbf{j} + b_z\mathbf{k} )=\mathbf{a} \times \mathbf{b} $,
$b_x , b_y , b_z$ are coordinates of $\mathbf{b}$ vector.
$\mathbf{a} \times \mathbf{b} = (a_{x}\mathbf{i} + a_{y}\mathbf{j} + a_{z}\mathbf{k}) \times \mathbf{b} $
Cross products can be explained as the rotation of one vector about another vector as the axis.
But rotations can also be described as multiplications of matrices. For example, rotation about the $x$ axis is given by the $\Omega_{x}$ matrix. So now we just need to find the rotation matrix $M_{a}$ corresponding to $\mathbf{a}$.
$\mathbf{a} \times \mathbf{b} = (a_{x}\mathbf{i} + a_{y}\mathbf{j} + a_{z}\mathbf{k}) \times \mathbf{b} = M_{a}\ \mathbf{b} $
I am not very clear here. But hopefully, someone can elaborate better.
For rotations we have
$\Omega_{x}$ matrix for rotation about $x$ axis \begin{bmatrix}0&&0&0\\ 0&&0&-1\\ 0&&1&0\end{bmatrix}
$\Omega_{y}$ matrix for rotation about $y$ axis \begin{bmatrix}0&&0&&1\\ 0&&0&&0\\ -1&&0&&0 \end{bmatrix}
$\Omega_{z}$ matrix for rotation about $z$ axis \begin{bmatrix}0&-1&&0\\ 1&0&&0\\0&0&&0 \end{bmatrix}
All these matrices are skew-symmetric. (as one can see by inspection). One can compose these matrices to form the rotation matrix $M_{a}$. As it turns out the rotation about the vector $\mathbf{a}$ is given by the matrix $M_{a} = a_{x}\Omega_{x} + a_{y}\Omega_{y} + a_{z}\Omega_{z} =$ $\begin{bmatrix}0&-a_{z}&a_{y}\\ a_{z}&0&-a_{x}\\ -a_{y}&a_{x}&0 \end{bmatrix} $ which is also skew symmetric by construction.
Thus, $\mathbf{a} \times \mathbf{b} = (a_{x}\mathbf{i} + a_{y}\mathbf{j} + a_{z}\mathbf{k}) \times \mathbf{b} = \begin{bmatrix}0&-a_{z}&a_{y}\\ a_{z}&0&-a_{x}\\ -a_{y}&a_{x}&0 \end{bmatrix} \begin{bmatrix}b_{x}\\b_{y}\\b_{z}\end{bmatrix}$
Probably this can also be done in 2 D with Pauli matrices.
