Express Hadamard product as a normal matrix product I have $N^2$ equations which I can write as the following Hadamard product. Is there a way I can get rid of the Hadamard product and express this using usual matrix operations?
$\left[ 
\begin{matrix}
0 & a_{21} & \cdots & a_{n1} \\
a_{12} & 0  & \cdots & a_{n2} \\
 \vdots & \vdots & \ddots & \vdots\\
a_{1n} & a_{2n} &\cdots & 0\\
\end{matrix} 
\right]\bigcirc \left[\begin{matrix}
b_1 & b_2 & \cdots & b_n \\
b_1 & b_2 & \cdots & b_n \\
 \vdots & \vdots & \ddots & \vdots\\
b_1 & b_2 & \cdots & b_n\\
\end{matrix} 
\right]=\left[ 
\begin{matrix}
c_{11} & c_{21} & \cdots & c_{n1} \\
c_{12} & c_{22}  & \cdots & a_{n2} \\
 \vdots & \vdots & \ddots & \vdots\\
c_{1n} & c_{2n} &\cdots & c_{nn}\\
\end{matrix} 
\right]$
 A: You can achieve the stated goal by applying the vec operation to both sides of the equation
$$\eqalign{
 A\circ B &= C \cr
 {\rm vec}(A)\circ {\rm vec}(B) &= {\rm vec}(C) \cr
 {\rm Diag}\Big({\rm vec}(A)\Big)  {\rm vec}(B) &= {\rm vec}(C) \cr
 {\cal A}\,b &= c \cr
}$$
but why on earth would you want to do that?  
Since $\,b,c\in {\mathbb R}^{N^2}$ and $\,{\cal A}\in{\mathbb R}^{N^2\times N^2}$ the operation count went from $N^2$ to $N^4$ 
A: As stated in this thread given task is impossible in general. If you try to express it in terms of being
$$
        A \circ B = X \cdot B
$$ 
for your case counter-example reduces to the following:
$$
        A=\begin{pmatrix}
        0 & 1 \\
        1 & 0 \\
        \end{pmatrix},\,
        B=\begin{pmatrix}
        1 & 1 \\
        1 & 1 \\
        \end{pmatrix}
$$
then
$$
        2 = rank (A) = rank(A \circ B) = rank(X \cdot B) \le rank (B) = 1
$$
which is obviously incorrect. The same could be deduced for other cases of general A or B, satisfying your conditions.
I believe this still may be possible if you restrict A and B to meet rank, and maybe other conditions.
A: Define the ${\mathbb R}^n$ vectors
$$b = \pmatrix{b_1\\b_2\\ \vdots\\b_n},\quad u = \pmatrix{1\\1\\ \vdots\\1}$$
and note that
$B=ub^T$
Then we can use the Diag() operator to write
$$\eqalign{
C &= A\circ ub^T \cr
  &= {\rm Diag}(u)\cdot A\cdot{\rm Diag}(b) \cr
  &= I\cdot A\cdot{\rm Diag}(b) \cr
  &= A\cdot{\rm Diag}(b) \cr
}$$
