Calculate $\frac{∂L}{∂A}$ given $\frac{∂L}{∂G}$, $D=(A-\iota\cdot B^T)\odot\iota\cdot C^T$, and $G=D \odot(\iota\cdot E^T)+\iota\cdot F^T$
Solution 1:
$
\def\l{\lambda}\def\o{{\iota}}\def\p{\partial}
\def\L{\left}\def\R{\right}
\def\LR#1{\L(#1\R)}
\def\vecc#1{\operatorname{vec}\LR{#1}}
\def\diag#1{\operatorname{diag}\LR{#1}}
\def\Diag#1{\operatorname{Diag}\LR{#1}}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\c#1{\color{red}{#1}}
\def\gg{\LR{\grad{\l}{G}}}
\def\ga{\LR{\grad{\l}{A}}}
$Let's use a convention wherein an uppercase letter denotes a matrix, a lowercase letter a vector, and a Greek letter a scalar. This means renaming the following problem variables
$$\big\{B,C,E,F\big\}\to \big\{b,c,e,f\big\}$$
because we'll need to use those uppercase letters to denote diagonal matrices whose main diagonals are the lowercase letters, i.e.
$$\eqalign{
B = \Diag{b},\quad C = \Diag{c},\quad E = \Diag{e},\quad I = \Diag{\o} = {\it Identity\;Matrix}
}$$
Diagonal matrices can replace Hadamard products via the following rule
$$\eqalign{
M\odot\LR{b\cdot c^T} &= B\cdot M\cdot C \\
}$$
Therefore
$$\eqalign{
D &= {A\cdot C-\o\cdot b^T\cdot C} \\
G &= {D\cdot E-\o\cdot f^T} \\
}$$
Finally, let's use a colon to denote the Frobenius product
$$\eqalign{
A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij}
\;=\; \trace{A\cdot B^T} \\
A:A &= \big\|A\big\|^2_F \\
}$$
This is also called the double-dot or double contraction product.
When applied to vectors $(n=\tt1)$ it reduces to an ordinary dot product.
The properties of the underlying trace function allow the terms in such a product to be rearranged in many different but equivalent ways, e.g. $$\eqalign{ A:B &= B:A \\ A:B &= A^T:B^T \\ C:\LR{A\cdot B} &= \LR{C\cdot B^T}:A = \LR{A^T\cdot C}:B \\\\ }$$
Use the given gradient to write the differential of the function in terms of $G$, then change the independent variable from $G\to D\to A$, then recover the gradient wrt $A$. $$\eqalign{ d\l &= \gg:dG \\ &= \gg:\LR{dD\cdot E} \\ &= \LR{\gg\cdot E}:{dD} \\ &= \LR{\gg\cdot E}:\LR{dA\cdot C} \\ &= \LR{\gg\cdot E\cdot C}:{dA} \\ \ga &= \gg\cdot E\cdot C \;\;\doteq\; \gg\odot\LR{\o\cdot e^T}\odot\LR{\o\cdot c^T} \\ }$$ The other gradients can be calculated in a similar fashion.