Derivative of Hadamard Product of two vectors How can I compute the following derivative?
$$\frac{\partial(K u \circ T u)}{\partial u}$$
$K$ and $T$ are constant matrices, $u$ is an unknown vector. and $\circ$ is Hadamard product.
my solution:
$$\frac{\partial(K_{ij} u_j \circ T_{mn} u_n)}{\partial u_p} = K_{ij}\frac{\partial(u_j)}{\partial u_p}\circ T_{mn} u_n+K_{ij} u_j \circ T_{mn} \frac{\partial u_n}{\partial u_p} = K_{ij}\delta_{jp} \circ T_{mn} u_n+K_{ij} u_j \circ T_{mn} \delta_{np}=K_{ip} (\sum_n T_{mn} u_n)+T_{mp} (\sum_j K_{ij} u_j).$$
therefore it can be written as follow $$\frac{\partial(K u \circ T u)}{\partial u} = K^T(Tu)+T^T(Ku).$$
where $\square^T$ is the transpose of the matrix.

Assuming $K,T\in\mathbb R^{n\times n}$ and $u\in\mathbb R^n\simeq\mathbb R^{n\times 1}$, we have (where $\square_{i,j}$ denotes the $(i,j)$-th entry of the matrix $\square$ and $\square^i=\square_i$ denotes the $i$-th entry of the vector $\square$), with the Einstein summation convention within $(\cdot)$,
$$(Ku\cdot_{\text{Hadamard}}Tu)_{i}=(K_{i,k}u^k)(T_{i,k}u^k),$$
so
$$\frac{\partial(Ku\cdot_{\text{Hadamard}}Tu)_{i}}{\partial u^j} = (K_{i,k}u^k) \partial_{u^j}[(T_{i,k}u^k)]+\partial_{u^j}[(K_{i,k}u^k)](T_{i,k}u^k)=(K_{i,k}u^k)T_{i,j}+(T_{i,k}u^k) K_{i,j}.$$
So far we agree.
I disagree with this resulting in the Jacobian of $Ku\cdot_{\text{Hadamard}}Tu$ being equal to $K^\top(Tu)+T^\top(Ku)$. For one, the Jacobian of $Ku\cdot_{\text{Hadamard}}Tu$ is an element of $\mathbb R^{n\times n}$ while $K^\top(Tu)+T^\top(Ku)\in\mathbb R^n$.
Correct is
$$\operatorname{Jac}Ku\cdot_{\text{Hadamard}}Tu = (\underbrace{Ku, Ku, \dots, Ku}_{n\text{ times}})\cdot_{\text{Hadamard}}T+(\underbrace{Tu, Tu, \dots, Tu}_{n\text{ times}})\cdot_{\text{Hadamard}}K.$$
