# Can matrices do cross product with vectors?

this is a practical problem raised from my chemistry research, apologize if somewhere I used incorrect words.

I'm dealing with the derivatives of a vector with respect to another vector. After some searching, I found the rule for differentiating a column vector with respect to a row vector is (all these vectors are 3 dimensional in Cartesian):

$$\mathbf{e}=\begin{bmatrix}e_1\\e_2\\e_3\end{bmatrix}$$ $$\mathbf{P}^T=\begin{bmatrix}P_1 & P_2 & P_3 \end{bmatrix}$$ $$\frac{\partial \mathbf{e}}{\partial \mathbf{P}^T}= \begin{bmatrix} \frac{\partial e_1}{\partial P_1} & \frac{\partial e_1}{\partial P_2} & \frac{\partial e_1}{\partial P_3} \\ \frac{\partial e_2}{\partial P_1} & \frac{\partial e_2}{\partial P_2} & \frac{\partial e_2}{\partial P_3} \\ \frac{\partial e_3}{\partial P_1} & \frac{\partial e_3}{\partial P_2} & \frac{\partial e_3}{\partial P_3} \\ \end{bmatrix}$$

Here my $\mathbf{e}$ is determined from a cross product, that is, $\mathbf{e}=\mathbf{n}\times\mathbf{m}$. So I tried to do this: $$\frac{\partial \mathbf{n}\times \mathbf{m}}{\partial \mathbf{P}^T}$$

From wiki, it says the product rule can be applied, so I arrived here: $$\frac{\partial \mathbf{n}\times \mathbf{m}}{\partial \mathbf{P}^T}=\frac{\partial \mathbf{n}}{\partial \mathbf{P}^T}\times\mathbf{m}+\mathbf{n}\times\frac{\partial \mathbf{m}}{\partial \mathbf{P}^T}$$ Now I'm confused because $\frac{\partial \mathbf{n}}{\partial \mathbf{P}^T}$ and $\frac{\partial \mathbf{m}}{\partial \mathbf{P}^T}$ are matrices, and they do cross product with vectors.

I didn't found anywhere saying the cross product of matrix, so I'm concerned perhaps there is something wrong in the derivation.

Hope you can give some help. Thanks in advance!

• @RodrigodeAzevedo I agree with your comment, but I think there's no such situation in my question... – Ruixing Wang Oct 26 '16 at 14:01
• Sorry. You're right. – Rodrigo de Azevedo Oct 26 '16 at 14:11

The writing is a bit awkward. If you take the derivative with respect to $P_i$ then $$\frac{\partial n\times m}{\partial P_i} = \frac{\partial n}{\partial P_i}\times m + n\times \frac{\partial m}{\partial P_i}$$ makes more sense. If you want to give a meaning to $\frac{\partial n}{\partial P}\times m$ then it should be that the cross product acts column wise, i.e. that you take each column vector in $\frac{\partial n}{\partial P}$ and replace it by its cross product with m (which is equivalent to the above formula).
• Thanks. I also considered the matrix form of the cross product ($n\times m=[n]_{\times}m$), and the result is equivalent. – Ruixing Wang Oct 27 '16 at 2:25