Can You Completely Permute The Elements Of A Matrix By Applying Permutation Matrices?


Answer :

It is not generally possible to do so.

For a concrete example, we know that there can exist no permutation matrices P,QP,Q such that P\pmatrix{1&2\\2&1}Q = \pmatrix{2&1\\2&1} If such a PP and QQ existed, then both matrices would necessarily have the same rank.


Let me add one more argument:

For n2n \ge 2:

Suppose the entries in the n×nn \times n matrix AA are all distinct. Then there are (n2)!(n^2)! distinct permutations of AA.

There are n!n! row-permutations of AA (generated by premultiplication by various permutation matrices), and n!n! col-permutations of AA (generated by post-multiplication by permutation matrices). If we consider all expressions of the form RAC RAC where RR and CC each range independently over all n!n! permutation matrices, we get at most (n!)2(n!)^2 possible results. But for n>1n > 1, we have (n!)2=[n(n1)21][n(n1)21]<[2n(2n1)(n+2)(n+1)][n(n1)21]=(2n)!(n2)!\begin{align} (n!)^2 &= [ n \cdot (n-1) \cdots 2 \cdot 1 ] [ n \cdot (n-1) \cdots 2 \cdot 1 ] \\ &< [ 2n \cdot (2n-1) \cdots (n+2) \cdot (n+1) ] [ n \cdot (n-1) \cdots 2 \cdot 1 ] \\ &= (2n)! \\ &\le (n^2)! \end{align} because 2nn22n \le n^2 for n2n \ge 2, and factorial is an increasing function on the positive integers. So the number of possible results of applying row- and col-permutations to AA is smaller than the number of possible permutations of the elements of AA. Hence there's some permutation of AA that does not appear in our list of all RACRAC matrices.

BTW, just to close this out: for 1×11 \times 1 matrices, the answer is "yes, all permutations can in fact be realized by row and column permutations." I suspect you knew that. :)


Given two elements a1a_1 and a2a_2, the properties "a1a_1 and a2a_2 are on different rows" and "a1a_1 and a2a_2 are on different columns" are preserved by any permutation. Proof:

A column permutation won't affect what row anything is on. A row permutation has to send an entire row to the same row, so if they start on the same row, they end on the same row. Permutations are invertible, so if they can't take two elements on the same row to different rows, they can't take elements on different rows to the same row.

An analogous argument holds for being on the same or different columns.

Thus, a row and column permutation is completely characterized by what it does to a diagonal; to find out where it sends an arbitrary element, just take the row that its row was sent to, and the column its column was sent to.


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