Let $f:V_1\times\dots\times V_N \to W$ be multilinear. Then $f$ s differentiable and $$df(a_1,\dots,a_n)(h_1+\dots+h_n)=f(h_1,a_2,\dots,a_n)+f(a_1,h_2,a_3,\dots,a_n)+f(a_1,\dots,a_{n-1},h_n)\tag{1}$$
Problem: Calculate the derivative of the cross product $\mathbb R^3\times \mathbb R^3 \to \mathbb R^3$.
Solution: Let $f$ be the cross product. It's easy to show that $f$ is multilinear by a direct calculation thus we can use (1). We calculate the derivative at $(u,v)=((u_1,u_2,u_3),(v_1,v_2,v_3))$. We then have $$df(u,v)(h_1+h_2)=h_1\times v + u\times h_2\tag{2}$$ Let $e_i, i=1,2,3$ be the unit vectors in $\mathbb R^3$. We calculate $$e_1\times v=\begin{pmatrix}0\\ -v_3 \\ v_2\end{pmatrix}$$ $$e_2\times v=\begin{pmatrix}v_3\\ 0 \\ -v_1\end{pmatrix}$$ $$e_3\times v=\begin{pmatrix}-v_2 \\ v_1 \\ 0\end{pmatrix}$$ For $u$ we use the identity $u\times e_i=-e_i\times u$. So we get the following derivative matrix: $$df(u,v)=\begin{pmatrix}0&v_3&-v_2&0&-u_3&u_2 \\ -v_3 & 0 & v_1 & u_3 & 0 & -u_1 \\ v_2 & -v_1 & 0 & -u_2 & u_1 & 0\end{pmatrix}$$
Question: 1. Apprently, the $+$ means writing the result of $f$ as a column, so it's not the $+$ from e.g. $\mathbb R^3$. Should I see that from the definition above without the example? How should I know the dimension of the imge of $df$ without them specifying it somehow?
- What $+$ is used in $f(a_i)(h_1+h_2)$? How does $h_1$ and $h_2$ and $h_1+h_2$ look like?