If $A=BC$, where $A$ and $C$ are given Laplacian matrices, how to calculate $B$? Let me be more specific: If $A=BC$, where $A$ and $C$ are given Laplacian matrices, how to calculate $B$? The graph corresponding to $A$ is a directed ring, which is strongly connected and $1_n$ and $1_n^T$ are right and left eigenvectors respectively. The graph corresponding to $C$ is a weighted directed ring, which is strongly connected but $1_n^T$ is no longer its left eigenvecor while $1_n$  is still its right eigenvector.
For example, $A=\left[ \begin{array}{ccc} 1&-1&0\\0&1&-1\\-1&0&1\end{array} \right]$, that is , $A$ is a circulant ,singular, Laplacian matrix. $C=\left[ \begin{array}{ccc} 1&-1/2&-1/2\\-2/3&1&-1/3\\-4/5&-1/5&1\end{array} \right]$ (singular, non-symmetric Laplacian matrix). Then how to compute $B$ if $A=BC$. 
Note that $A, B, C$ are all square matrices. I don't want numerical solutions. There may be many solutions to this problem, so is there a formulated way to find one of them (maybe we can restrict $B$ to be Laplacian as well)?
 A: If the system of linear equations is consistent, its solution is $B=AC^+$, where $C^+$ is the pseudoinverse of $C$. Depending on the strictness of your definition of "closed formula", this may already fit the requirements. Otherwise, there are more explicit expressions. I list three alternatives in the following. Divide $A$ and $C$ into blocks
$$
\begin{bmatrix}
A_{11} & A_{12}\\
A_{21} & A_{22}
\end{bmatrix} 
=\begin{bmatrix}
B_{11} & B_{12}\\
B_{21} & B_{22}
\end{bmatrix} 
\begin{bmatrix}
C_{11} & C_{12}\\
C_{21} & C_{22}
\end{bmatrix}.
$$
where the blocks labeled $22$ are $1\times 1$. Note that $C_{11}$ is invertible, because it is a submatrix of a singular irreducible M-matrix.


*

*Remove the last column, which is superfluous since it is the opposite of the sum of the previous ones, to obtain the reduced linear system
$$
\begin{bmatrix}
A_{11}\\
A_{21}
\end{bmatrix} 
=\begin{bmatrix}
B_{11} & B_{12}\\
B_{21} & B_{22}
\end{bmatrix} 
\begin{bmatrix}
C_{11}\\
C_{21}
\end{bmatrix}.  \tag{*}
$$
Now you can use the formula for the pseudoinverse of a matrix with linearly independent columns
$$
B = \begin{bmatrix}
A_{11}\\
A_{21}
\end{bmatrix} 
\begin{bmatrix}
C_{11}\\
C_{21}
\end{bmatrix}^+ = 
\begin{bmatrix}
A_{11}\\
A_{21}
\end{bmatrix}
\left(\begin{bmatrix}
C_{11}\\
C_{21}
\end{bmatrix}^*\begin{bmatrix}
C_{11}\\
C_{21}
\end{bmatrix}\right)^{-1}\begin{bmatrix}
C_{11}\\
C_{21}
\end{bmatrix}^*.
$$

*Once you have removed the last column, you can choose $B_{12}$ and $B_{22}$ arbitrarily and solve for $B_{11}$, $B_{21}$ in (*), which is possible since $C_{11}$ is invertible.

*If $u,v$ are norm-1 vectors such that $Cv=0, u^*C=0$, then another formula for the pseudoinverse is 
$$
C^+ = (C + uv^*)^{-1}-vu^*.
$$
You can prove this using the formula for $C^+$ based on the SVD of $C$ -- basically the idea is that we change the last singular value of $C$ from $0$ to 1, then invert, and undo the change.

