I'm stuck on this problem about differentiating an expression with matrices.
The expression it's the following: $F(u)= (D_1 * \vec{u})*(I\vec{u})$, where $I$ is the identity, $\vec{u}$ is a vector of length $m \in \mathbb{N}$ and $\epsilon \in \mathbb{R_{>0}}$
To find the jacobian matrix $JF(u)$, applicated to a vector $v$, I have to solve this limit:
$\lim_{\epsilon \rightarrow 0} \frac{F(\vec{u} +\epsilon \vec{v}) - F(\vec{u})}{\epsilon}$.
So I have to solve $\frac{(D(u +\epsilon v)*(I(u+\epsilon v)) - (Du)(Iu))}{\epsilon}$, but I can't go on because I don't know how to do those products.
Any hint, or reference?
Since we're taking derivatives, $D_1$ is a confusing name for the matrix, so I will use $M$ instead.
The gradient of a matrix-valued function like $F=Muu^T$ with respect to a vector will be a 3rd order tensor. So let's use index notation and calculate the gradient via the differential $$\eqalign{ F_{ik} &= M_{ij}u_ju_k \cr\cr dF_{ik} &= M_{ij}\,du_j\,u_k + M_{ij}u_j\,du_k \cr\cr G_{ikp}=\frac{\partial F_{ik}}{\partial u_p} &= M_{ij}\,\frac{\partial u_j}{\partial u_p}\,u_k + M_{ij}u_j\,\frac{\partial u_k}{\partial u_p} \cr &= M_{ij}\,\delta_{jp}\,u_k + M_{ij}u_j\,\delta_{kp} \cr &= M_{ip}u_k + M_{ij}u_j\,\delta_{kp} \cr }$$ Finally, you wanted to take the derivative in a particular direction $(v)$, which means $$\eqalign{ dF_{ik} &= G_{ikp} (v_pd\epsilon) \cr &= (M_{ip}v_pu_k + M_{ij}u_j\,\delta_{kp}v_p)\,d\epsilon \cr\cr \frac{dF}{d\epsilon} &= M(vu^T + uv^T) \cr }$$