TITLE: Skew symmetric matrix of vector
QUESTION [10 upvotes]: 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!

REPLY [3 votes]: $\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.