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Suppose I have a square matrix, $\textbf{A}$. I want to know whether I can represent it, it up to a permutation of the columns, using a symmetric or triangular matrix instead, since these only have ${n}\choose{2}$ entries instead of $n^2$.

In other words, what properties would $\textbf{A}$ need to have for a solution to exist for the following equation?

$$\textbf{A P} = \textbf{B},$$

where $\textbf{P}$ is a permutation matrix and $\textbf{B}$ is either triangular or symmetric (whichever is easier to write an answer about).

I don't need to actually find $\textbf{B}$ or compare individual pairs of $\textbf{A}$ and $\textbf{B}$, so this may be simpler than some previous questions such as this one or this one.

David J. Harris
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1 Answers1

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Look for a column whose last $n-1$ entries are zeroes; if there isn't one, then no permutation exists. If there is, remove it, and the first row, and apply the same approach to the remaining $(n-1) \times (n-1)$ matrix. If you reach the empty matrix, then there's a permutation that does what you want. If not, then there is not.

That handles the upper-triangular case.

For the diagonal case, if the ROWS you remove all all zeroes, then you can diagonalize. (I"m not sure that this is a complete answer to the second question, however.)

John Hughes
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  • I think I may have intended to ask a different question than the one I actually did, but your answer was very helpful for pointing that out. Thanks! – David J. Harris Oct 02 '17 at 18:12
  • Glad to have been of service; I tried to imagine what question you WERE trying to ask, but couldn't come up with anything but what you did ask, so I answered that. :) – John Hughes Oct 02 '17 at 18:35