Gradient of an optimization function - Frobenius norm and Hadamard product I am trying to solve a problem for my Optimization class, in which it is asked to calculate the gradient of the following function:
$$g(P)=\frac{1}{2}||1_K\circ(R-Q^0P)||_F^2+\frac{\rho}{2}||Q^0||_F^2+\frac{\rho}{2}||P||_F^2$$
where $\rho, R, Q^0$ and $1_K$ are given constants in this case.
But I am totally stuck particularly on the first term (the one that icludes the Frobenius norm and the Hadamard prouct).
I tried to use the definition of the Frobenius norm $||A||_F=\sqrt{Tr(AA^H)}$ but I do not know how to handle it in this situation.
Thank you very much.
 A: Let's rename the variables using lowercase letters for vectors and uppercase letters for matrices, and omit all of the subscripts and superscripts.
$$\eqalign{
p=P,\quad Q=Q^0,\quad r=R,\quad u={\tt1}_K 
}$$
Let's also define the auxiliary vectors
$$\eqalign{
s &= Qp-r \\
w &= u\circ s \\
}$$
We'll also need the trace/Frobenius product
$$\eqalign{
A:B &= {\rm Tr}(A^TB) = B:A \\
A:A &= \big\|A\big\|^2_F \\
}$$
Conveniently, the Frobenius product commutes with the Hadamard product, i.e.
$$A:(B\circ C) = (A\circ B):C$$
Use the above to rewrite the objective function, then calculate the gradient as follows.
$$\eqalign{
g &= \tfrac 12(w:w) + \tfrac \rho2(p:p) + \tfrac \rho2(Q:Q) \\
\\
dg &= (w:dw) + \rho(p:dp) + 0 \\
 &= w:(u\circ ds) + \rho(p:dp) \\
 &= (u\circ w):ds + \rho(p:dp) \\
 &= (u\circ w):Q\,dp + \rho(p:dp) \\
 &= Q^T(u\circ w):dp + \rho(p:dp) \\
 &= \Big(Q^T(u\circ w)+ \rho p\Big):dp \\
 &= \Big(\rho p + Q^T\big(u\circ u\circ(Qp-r)\big)\Big):dp \\
\\
\frac{\partial g}{\partial p}
 &= \rho p + Q^T\big(u\circ u\circ(Qp-r)\big) \\
}$$
