What are the conditions, for $n$ matrices $A_1$ and $A_2$ ... $A_n$ etc, such that their Hadamard product and Matrix products are identical? What is the possible family of matrices $A_1$, ... , $A_n$ such that their Hadamard (elementwise) products and matrix products are equivalent, and is it valid as $n$ $\rightarrow$ $\infty$?
I.e.:
$A_1\odot A_2\cdots\odot A_{n-1}\odot A_n = A_1A_2\cdots A_{n-1}A_n$
I realise this is true for diagonal matrices, but was wondering what is the set of matrices that this holds true.
Thanks
 A: I did it for $2\times 2$ matrices, there are already many possibilities.
$A=\begin{pmatrix}a&c\\b&d\end{pmatrix}\quad B=\begin{pmatrix}0&0\\0&0\end{pmatrix}\quad$ trivial case $B=0$
$A=\begin{pmatrix}a&0\\b&d\end{pmatrix}\quad B=\begin{pmatrix}0&0\\0&v\end{pmatrix}\quad$ triangular case
$A=\begin{pmatrix}\frac cu(u-v)&c\\0&d\end{pmatrix}\quad B=\begin{pmatrix}s&u\\0&v\end{pmatrix}$
$A=\begin{pmatrix}a&0\\0&d\end{pmatrix}\quad B=\begin{pmatrix}s&0\\0&v\end{pmatrix}\quad$ diagonal case
$A=\begin{pmatrix}a&c\\0&d\end{pmatrix}\quad B=\begin{pmatrix}s&0\\0&0\end{pmatrix}\quad$ triangular case
$A=\begin{pmatrix}a&0\\b&\frac bt(t-s)\end{pmatrix}\quad B=\begin{pmatrix}s&0\\t&v\end{pmatrix}$
$A=\begin{pmatrix}0&0\\0&0\end{pmatrix}\quad B=\begin{pmatrix}s&u\\t&v\end{pmatrix}\quad$ trivial case $A=0$
There are $2$ cases I did not annotate since I think this won't probably extend to higher dimensions nor multiple matrices, but you may want to have look at the triangular case for higher dimensions.

Edit:
Here is for $3\times 3$ matrices (given the size, I'm forced to make it a link so it will disappear after some time sorry...).
It looks like this is not extending well to bigger dimension, basically only $1$ row or $1$ column matrices and zero elsewhere seems to behave correctly.
https://www.docdroid.net/bQn19yo/toto-pdf
Thus I suspect that for multiple matrices, you will end up with all non-diagonal elements of the product forced to be zeroes, and that the diagonal case will be the only one relevant.
