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Say I would like to swap the two rows, $i$th and $j$th in the matrix $A$. Then the appropriate permutation matrix is

$$ P_{ij}^T = \left[e_1 \cdots e_{i-1} e_j e_{i+1} \cdots e_{j-1} e_i e_{j+1} \cdots e_n \right] $$

where WLOG $i < j$ assumed,

which can be premultiplied to $A$ such that

$$ \tilde{A}_{ij} = P_{ij} A $$

where $\tilde{A}_{ij}$ denotes the $i,j$-column swapped version of $A$.


Question

The above reasoning is quite straightforward to me, but I would like to know when the permutation matrix is unique as $P_{ij}$.

The permutation matrix is not unique, because When $A=0$ any row permutation of $A$ can be denoted as follows:

$$ PA = A $$

where $P$ can be, any matrix with the appropriate size.

So I'm wondering if there is any necessary/sufficient condition for the $P_{ij}$ to be the unique row exchanging matrix. Any help will be appreciated.


Edited

From a clever comment, I realized that

$$ P_{ij} = \tilde{A_{ij}} A^{-1} $$

if $A^{-1}$ exists. However, I would like to know if this uniqueness is preserved even when $A^{-1}$ does not exist.

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    $\begingroup$ Do you allow any matrix for $P$ or at least require it to be actually a permutation matrix(i.e., obtained form the identity by permuting rows)? Clearly, when $A$ is invertible, we must have $P=\tilde AA^{-1}$ $\endgroup$ – Hagen von Eitzen Apr 14 at 3:40
  • $\begingroup$ @HagenvonEitzen What problems might arise when we allow $P$ to be any matrix, rather than a permutation matrix derived from $I$? $\endgroup$ – Moreblue Apr 14 at 3:50
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    $\begingroup$ Let $A$ be invertible, but with one row replaced by zeroes. Then there is only one permutation matrix $P$, but many arbitrary matrices $P$. So the uniqueness condition with permutation matrices is weaker than with arbitrary matrices (in fact it then boils down to pairwise distinct rows in$A$ instead of invertibility) $\endgroup$ – Hagen von Eitzen Apr 14 at 3:54

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