Given A a singular matrix, and B a symmetric matrix, I find that solving $(A\circ B)x=v$ by $$ x= (A\circ B)^+v $$ gives the correct solution to my physical system ($\circ$ denotes Hadamard product). The physical system
1) actually has one unique solution.
2) $(A\circ B)x=v$ is resulted from making the diagonal of $AXB$ equal the diagonal of $V$, where $X$ is a diagonal matrix belonging to $x$, and $V$ is a diagonal matrix belonging to $v$.
3) $A\circ B$ is singular.
example: $$ A= \begin{bmatrix} -1 & -1 & -1\\ -1 & -1 & -1\\ -1 & -1 & -1\\ \end{bmatrix} $$ $$ B= \begin{bmatrix} 1 & 0.28 & 0.64\\ 0.28 & 1 & 0.64\\ 0.64& 0.64 & 1\\ \end{bmatrix} $$ But why pseudoinverse gives the correct solution ? I looked through the wiki page on Moore-Penrose inverse in vain.
Consider the matrix equation $AX=B $ and let $A^-$ be an inner inverse of $A $ (it could be the Moore-Penrose inverse or any other inner inverse, but importantly we know that such an inverse exists; moreover it is the invese when the matrix is invertible). Then it is easy to see that a solution to our equation exists if and only if $AA^-B=B $. Assuming that this holds, the general solution is then given by \begin{equation} X=A^-B+(I-A^-A)Y,\end{equation} where $Y $ is a generic matrix of appropriate siza. Again, the dolution is unique when $A $ is invertible.
For example, let \begin{equation}A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\end{equation} and choose \begin{equation}A^-=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\end{equation} and $X=(x_1,x_2)$ and $B=(2,0)$. Then \begin{equation}X=\begin{pmatrix} 2 \\ 0 \end{pmatrix}+\{\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}-\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix},\end{equation} with $y_1,y_2$ arbitrary. Hence the general solution is $X=(2,s)$ with $s$ arbitrary.