Skew-symmetric matrix representation of cross product

We can define $[{\bf a}]_\times = \begin{pmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{pmatrix}\triangleq {\bf a}^\wedge$

${\bf a} \times {\bf b} = {\bf a}^\wedge {\bf b}.$

Note that we have

${\bf a}^\wedge {\bf a}^\wedge {\bf a}^\wedge = - {\bf a}^\wedge$ and

${\bf a}^\wedge {\bf a}^\wedge = {\bf a} {\bf a}^\top -I$

Triple product expansion

We will show the above with the triple product expansion: ${\bf u} \times ({\bf v} \times {\bf w}) = ({\bf u}\cdot {\bf w}) {\bf v} - ({\bf u}\cdot {\bf v}){\bf w}.$

Proof:

(1) $\begin{align*} (\mathbf{u} \times (\mathbf{v} \times \mathbf{w}))_x &= \mathbf{u}_y(\mathbf{v}_x\mathbf{w}_y - \mathbf{v}_y\mathbf{w}_x) - \mathbf{u}_z(\mathbf{v}_z\mathbf{w}_x - \mathbf{v}_x\mathbf{w}_z) \\ &= \mathbf{v}_x(\mathbf{u}_y\mathbf{w}_y + \mathbf{u}_z\mathbf{w}_z) - \mathbf{w}_x(\mathbf{u}_y\mathbf{v}_y + \mathbf{u}_z\mathbf{v}_z) \\ &= \mathbf{v}_x(\mathbf{u}_x\mathbf{w}_x + \mathbf{u}_y\mathbf{w}_y + \mathbf{u}_z\mathbf{w}_z) - \mathbf{w}_x(\mathbf{u}_x\mathbf{v}_x + \mathbf{u}_y\mathbf{v}_y + \mathbf{u}_z\mathbf{v}_z) \\ &= (\mathbf{u}\cdot\mathbf{w})\mathbf{v}_x - (\mathbf{u}\cdot\mathbf{v})\mathbf{w}_x \end{align*}$

Similarly for the $y$ and $z$ components. $\square$

Proof of ${\bf a}^\wedge {\bf a}^\wedge {\bf a}^\wedge = - {\bf a}^\wedge$

Note that for any ${\bf x},$

${\bf a}^\wedge {\bf a}^\wedge {\bf a}^\wedge {\bf x}={\bf a} \times({\bf a}\times ({\bf a} \times {\bf x}))=({\bf a} \cdot ({\bf a}\times {\bf x})) {\bf a} - ({\bf a} \cdot {\bf a}) ({\bf a} \times {\bf x})=-{\bf a} \times {\bf x},$

thus ${\bf a}^\wedge {\bf a}^\wedge {\bf a}^\wedge = - {\bf a}^\wedge.$

$\square$

Proof of ${\bf a}^\wedge {\bf a}^\wedge = {\bf a} {\bf a}^\top -I$

For any ${\bf x},$

(2) $\begin{align*}{\bf a}^\wedge {\bf a}^\wedge {\bf x} &= {\bf a} \times ({\bf a} \times {\bf x}) = ({\bf a}\cdot {\bf x}) {\bf a} - ({\bf a}\cdot {\bf a}) {\bf x} \\ &= {\bf a} {\bf a}^\top {\bf x} - {\bf x} = ({\bf a} {\bf a}^\top - I) {\bf x}. \end{align*}$

Thus, ${\bf a}^\wedge {\bf a}^\wedge = {\bf a} {\bf a}^\top -I.$