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winston
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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 $B$$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)?

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 $B$ 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)?

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)?

Post Reopened by Federico Poloni, Stefan Kohl, John Pardon, Francois Ziegler, Gerry Myerson
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winston
  • 143
  • 7

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 $B$ 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 matrix ,singular, Laplacian matrix. $C=\left[ \begin{array}{ccc} 1&-1/2&-1/2\\-2/3&1&-1/3\\-4/5&-1/5&1\end{array} \right]$ (also singularsingular, 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)?

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 $B$ 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 matrix ,singular, Laplacian matrix. $C=\left[ \begin{array}{ccc} 1&-1/2&-1/2\\-2/3&1&-1/3\\-4/5&-1/5&1\end{array} \right]$ (also singular). 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?

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 $B$ 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)?

added 334 characters in body
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winston
  • 143
  • 7

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 $B$ 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 matrix ,singular, Laplacian matrix. $C=\left[ \begin{array}{ccc} 1&-1/2&-1/2\\-2/3&1&-1/3\\-4/5&-1/5&1\end{array} \right]$ (also singular). 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?

Let me be more specific: If $A=BC$, where $A$ and $C$ are given Laplacian matrices, how to calculate $B$?

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?

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 $B$ 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 matrix ,singular, Laplacian matrix. $C=\left[ \begin{array}{ccc} 1&-1/2&-1/2\\-2/3&1&-1/3\\-4/5&-1/5&1\end{array} \right]$ (also singular). 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?

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winston
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Post Closed as "Not suitable for this site" by Federico Poloni, R.P., Stefan Kohl, Loïc Teyssier, Marco Golla
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winston
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