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!
 A: 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).
A: $\def\e{\varepsilon}\def\p#1#2{\frac{\partial #1}{\partial #2}}$
The Levi-Civita symbol $(\e_{ijk})$ can be used to write the cross product of two vectors as
$$\eqalign{
a\times b &= -a\cdot\e\cdot b \\
 &= +b\cdot\e\cdot a \;\;=\; -b\times a
}$$
Replacing either vector with a matrix effectively defines the cross product for matrices, e.g.
$$\eqalign{
A\times b &= -A\cdot\e\cdot b \\
a\times B &= -a\cdot\e\cdot B \\
A\times B &= -A\cdot\e\cdot B \\
}$$
Having defined the matrix cross product, the naive product rule
$$\eqalign{
\p{(a\times b)}{p} 
\;\ne\; a\times\left(\p{b}{p}\right) \;+\; \left(\p{a}{p}\right)\times b \\
}$$
is still incorrect $\,\big({\rm NB}:$ It is correct when $p$ is a scalar instead of a vector$\big).$
However, the following rule is compatible with the matrix cross product
$$\eqalign{
\p{(a\times b)}{p} 
\;=\; a\times\left(\p{b}{p}\right) \;-\; b\times\left(\p{a}{p}\right) \\\\
}$$
Another approach is to define a skew matrix associated with each vector
$$\eqalign{
{\cal A} &\doteq {\rm Skew}(a) \;=\; (a\times I) \;=\; -a\cdot\e \\
{\cal B} &\doteq {\rm Skew}(b) \;=\; (b\times I) \;=\; -b\cdot\e \\
}$$
Then the cross product can be replace by a normal matrix-vector product
$$\eqalign{
a\times b \;=\; {\cal A}b \;=\; -{\cal B}a \\
}$$
and the gradient can use standard matrix-matrix products
$$\eqalign{
\p{(a\times b)}{p} 
  &=  {\cal A}\left(\p{b}{p}\right) \;-\; {\cal B}\left(\p{a}{p}\right)\\
}$$
$$\eqalign{
}$$
