Derivative of squared frobenius norm of hadamard product of outer product of vector with itself and matrix w.r.t. vector I know the title is a mouth full, and there have been many similar (and probably more complicated) questions/answers on this site, but I'm stuck on this specific problem.
I am working with the following function:
$$f(x) = \frac{1}{2}||xx^{T} \circ Y||_{2}^{2}$$
where $x \in \mathbb{R}^{n}$, $Y \in \mathbb{R}^{n \times n}$ and $\circ$ denotes the hadamard (element wise) product. I would like to compute the following gradient:
$$\frac{\partial f(x)}{\partial x} = \frac{\partial }{\partial x} \frac{1}{2}||xx^{T} \circ Y||_{2}^{2}$$
The furthest I've gotten is w.r.t. $xx^{T}$, following this:
$$\frac{\partial f(x)}{\partial xx^{T}} = xx^{T} \circ Y$$
but I do not know how to actually compute this w.r.t. $x$. I am pretty inexperienced with this kind of thing, so any input would be really helpful.
Thank you!
 A: For ease of typing, define the matrices
$$\eqalign{
W &= xx^T\circ Y \quad\implies\quad dW = \left(dx\,x^T+x\,dx^T\right)\circ Y \\
Z &= Y\circ Y \\
}$$
and use a colon as a product notation for the trace, i.e.
$$\eqalign{
A:B &= {\rm Tr}(AB^T) \;=\; \sum_{i=1}^m \sum_{j=1}^n A_{ij} B_{ij} \\
A:A &= \big\|A\big\|^2_F \\
}$$
Write the function in terms of these new variables. Then calculate the differential and gradient.
$$\eqalign{
f &= \tfrac 12\,W:W \\
df &= W:dW \\
 &= (xx^T\circ Y):(dx\,x^T+x\,dx^T)\circ Y \\
 &= (xx^T\circ Y\circ Y):(dx\,x^T+x\,dx^T) \\
 &= xx^T\circ\left(Z+Z^T\right):dx\,x^T \\
 &= \left(xx^T\circ\left(Z+Z^T\right)\right)x:dx \\
\frac{\partial f}{\partial x}
 &= \left(xx^T\circ\left(Z+Z^T\right)\right)x \\
}$$
If $\;Y^T=Y\,$ then the gradient can be simplified to
$$\eqalign{
\frac{\partial f}{\partial x}
 &= 2\left(xx^T\circ Y\circ Y\right)x \\
}$$
A: The function in question is:
$$f(x) = \frac12 \sum_{i, j} x_i^2 x_j^2 y_{ij}^2.$$
The partial derivatives with respect to the $x_i$ are routine. I have no idea what you mean by $\frac{\partial f}{\partial x},$ but it will be easy enough.
