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Given a symmetric square matrix, how can I apply a permutation to the rows and columns (i.e. the same permutation to both the rows and the columns) such a way that the new structure of the matrix should follow the new row and column order?

Consider an order of rows/columns, according to which the elements of a set are arranged in a matrix:

order = Range[5];
matrix = Table[Subscript[a, i, j], {i, 5}, {j, 5}];
TableForm[matrix, TableHeadings -> {order, order}]

Mathematica graphics

Now take a permutation of the row/column order:

newOrder = RandomSample[order]
   ==> {1, 4, 5, 2, 3}

FindPermutation finds the appropriate permutation cycle for the new order:

permutation = FindPermutation[order, newOrder]
Permute[order, permutation]
   ==> Cycles[{{2, 4}, {3, 5}}]
   ==> {1, 4, 5, 2, 3}

Now the question is: How to easily and elegantly apply the above permutation (preferably in its Cycles form) to the matrix to yield the following one:

Mathematica graphics

Some notes:

  • The matrix is always square and symmetric.
  • Column and head orders are always identical.
  • Bear in mind that order, and consequently matrix, can be big (e.g. 4^8 for order)
  • Since matrix can be big, I'm looking for a method, that applies the permutation 'locally', that is without reconstructing the whole table every time, or without applying replacements/functions enumerating each element, thus no dispatch table should be used. This is rather important, as it could be that the new order only swaps two columns/rows, and nothing else.
share|improve this question
Related… – Leonid Shifrin Feb 25 '12 at 17:16
up vote 15 down vote accepted

How about:

ord = {1, 4, 5, 2, 3}
matrix[[ord, ord]]

(You can convert any permutation (including Cycles) to an index list using PermutationList.)

share|improve this answer
Amazingly simple. Shame on me. – István Zachar Feb 25 '12 at 16:57
Glad I could help. – user21 Feb 25 '12 at 16:58
@Istvan the solution is simple, but IMHO not trivial. It's a good question. – Mr.Wizard Feb 25 '12 at 23:16

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