Jacob Harding

Suppose \(\mathbf{x} \in \mathbb{R}^{3 \times 1}\) and \(\mathbf{f}(\mathbf{x}), \mathbf{g}(\mathbf{x}): \mathbb{R}^{3 \times 1} \rightarrow \mathbb{R}^{3 \times 1}\). The goal is to find the matrix of first-order partial derivatives of the cross product of \(\mathbf{f}\) and \(\mathbf{g}\):

\[ \frac{\partial}{\partial \mathbf{x}}(\mathbf{f}(\mathbf{x}) \times \mathbf{g}(\mathbf{x}))\]

A matrix of first-order partial derivatives is also called a Jacobian matrix so an equivalent notation can be defined:

\[ \mathbf{J}_{\mathbf{x}}(\mathbf{f}(\mathbf{x}) \times \mathbf{g}(\mathbf{x})) = \frac{\partial}{\partial \mathbf{x}}(\mathbf{f}(\mathbf{x}) \times \mathbf{g}(\mathbf{x}))\]

Instead of taking the derivative with respect to the whole \(\mathbf{x}\), its easier to take it with respect to each component of \(\mathbf{x}\) individually, and build the Jacobian from each of these derivatives. For example, we find \(\frac{\partial \mathbf{f} \times \mathbf{g}}{\partial x_0}\), \(\frac{\partial \mathbf{f} \times \mathbf{g}}{\partial x_1}\), and \(\frac{\partial \mathbf{f} \times \mathbf{g}}{\partial x_2}\) which are all column vectors of partial derivatives. Each of these correspond to a column of the Jacobian matrix so they can be assembled into the complete Jacobian like so:

$$ \mathbf{J}_{\mathbf{x}}(\mathbf{f}(\mathbf{x}) \times \mathbf{g}(\mathbf{x})) = \begin{bmatrix} \frac{\partial \mathbf{f} \times \mathbf{g}}{\partial x_0} & \frac{\partial \mathbf{f} \times \mathbf{g}}{\partial x_1} & \frac{\partial \mathbf{f} \times \mathbf{g}}{\partial x_2} \end{bmatrix} $$

The standard form of the product rule works with the cross product with respect to a single variable like we have with each column vector of partial derivatives so we find:

$$ \frac{\partial \mathbf{f} \times \mathbf{g}}{\partial x_0} = (\frac{\partial \mathbf{f}}{\partial x_0} \times \mathbf{g}) + (\mathbf{f} \times \frac{\partial \mathbf{g}}{\partial x_0}) $$

Substituting each column of partial derivatives with its product rule definition gives the matrix:

$$ \begin{bmatrix} (\frac{\partial \mathbf{f}}{\partial x_0} \times \mathbf{g}) + (\mathbf{f} \times \frac{\partial \mathbf{g}}{\partial x_0}) & (\frac{\partial \mathbf{f}}{\partial x_1} \times \mathbf{g}) + (\mathbf{f} \times \frac{\partial \mathbf{g}}{\partial x_1}) & (\frac{\partial \mathbf{f}}{\partial x_2} \times \mathbf{g}) + (\mathbf{f} \times \frac{\partial \mathbf{g}}{\partial x_2}) \end{bmatrix} $$

This can be split into a two matrix sum by gathering all the cross product terms with \(\mathbf{f}\) into one matrix and all the cross products with \(\mathbf{g}\) into the other:

$$ \begin{bmatrix} (\frac{\partial \mathbf{f}}{\partial x_0} \times \mathbf{g}) & (\frac{\partial \mathbf{f}}{\partial x_1} \times \mathbf{g}) & (\frac{\partial \mathbf{f}}{\partial x_2} \times \mathbf{g}) \end{bmatrix} + \begin{bmatrix} (\mathbf{f} \times \frac{\partial \mathbf{g}}{\partial x_0}) & (\mathbf{f} \times \frac{\partial \mathbf{g}}{\partial x_1}) & (\mathbf{f} \times \frac{\partial \mathbf{g}}{\partial x_2}) \end{bmatrix} $$

Now I’m going to introduce some new notation of a matrix crossed with a vector: \(\mathbf{A} \times \mathbf{v}\). This notation means form a matrix with the same size as \(\mathbf{A}\), where each column is the cross product of the corresponding \(\mathbf{A}\) matrix column with the vector \(\mathbf{v}\). For example:

$$ \mathbf{A} \times \mathbf{v} = \begin{bmatrix} \mathbf{A}_{\text{col0}} \times \mathbf{v} \qquad \mathbf{A}_{\text{col1}} \times \mathbf{v} \qquad \mathbf{A}_{\text{col2}} \times \mathbf{v} \end{bmatrix} $$

With this notation established and the fact that: \(\mathbf{v} \times \mathbf{u} = -\mathbf{u} \times \mathbf{v}\), we can pull out the \(\mathbf{g}\) and \(\mathbf{f}\) terms like so:

$$ \begin{bmatrix} \frac{\partial \mathbf{f}}{\partial x_0} & \frac{\partial \mathbf{f}}{\partial x_1} & \frac{\partial \mathbf{f}}{\partial x_2} \end{bmatrix} \times \mathbf{g} - \begin{bmatrix} \frac{\partial \mathbf{g}}{\partial x_0} & \frac{\partial \mathbf{g}}{\partial x_1} & \frac{\partial \mathbf{g}}{\partial x_2} \end{bmatrix} \times \mathbf{f} $$

Now the two matrices have simply become \(\frac{\partial \mathbf{f}}{\partial \mathbf{x}}\) and \(\frac{\partial \mathbf{g}}{\partial \mathbf{x}}\) so this simplifies further into the final form:

$$ \bbox[5px,border:2px solid blue] { \mathbf{J}_{\mathbf{x}}(\mathbf{f}(\mathbf{x}) \times \mathbf{g}(\mathbf{x})) = (\frac{\partial \mathbf{f}}{\partial \mathbf{x}} \times \mathbf{g}) - (\frac{\partial \mathbf{g}}{\partial \mathbf{x}} \times \mathbf{f}) } $$

Afterword On Cross Product Matrix

Its worth noting what happens when the matrix in a matrix-vector cross product is the identity:

$$ \mathbf{I} \times \mathbf{v} = \begin{bmatrix} 0 & v_3 & -v_2 \\\\ -v_3 & 0 & v_1 \\\\ v_2 & -v_1 & 0 \\\\ \end{bmatrix} = -[\mathbf{v}]_{\times} $$

\([\mathbf{v}]_{\times}\) is a special skew-symmetric matrix which is often called the ‘cross product matrix’. The reason its called that is because when multiplied with a vector \(\mathbf{u}\) the result is the cross product \(\mathbf{v} \times \mathbf{u}\).

Also to make it more clear that \(\mathbf{I} \times \mathbf{v} = -[\mathbf{v}]_{\times}:\)

We have

$$ \mathbf{I} \times \mathbf{v} = \begin{bmatrix} \mathbf{e}_1 \times \mathbf{v} & \mathbf{e}_2 \times \mathbf{v} & \mathbf{e}_3 \times \mathbf{v} \end{bmatrix} $$

Where

$$ \mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad \mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \quad \mathbf{e}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$

And knowing cross product in general is:

$$ \mathbf{a} \times \mathbf{b} = \begin{bmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{bmatrix} $$

Taking these column cross products then gives the form of the matrix \(\mathbf{I} \times \mathbf{v}\) shown at the beginning.