Permute any two entries in a $n\times n$ matrix.

Question

Let $A$ be an $8 \times 8$ matrix with integer coefficients. I want to permute two entries in $A$, any two entries as needed: In general, for any two entries $a_j,b_k$ in the matrix is it possible to do this with some matrix $B$ dependent on $a_j,b_k$?

I know about permutation matrices, but they only permute entire rows and columns not individual entries.

Edit 1:

For example: Say I want to permute $x_{12}$ with $x_{33}$, I want a matrix $B$ suh that :

$$ \begin{bmatrix} x_{11} & x_{12} & x_{13} & \dots & x_{1n} \\ x_{21} & x_{22} & x_{23} & \dots & x_{2n} \\ x_{31} & x_{32} & x_{33} & \dots & x_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n1} & x_{n2} & x_{n3} & \dots & x_{nn} \end{bmatrix}B= \begin{bmatrix} x_{11} & x_{33} & x_{13} & \dots & x_{1n} \\ x_{21} & x_{22} & x_{23} & \dots & x_{2n} \\ x_{31} & x_{32} & x_{12} & \dots & x_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n1} & x_{n2} & x_{n3} & \dots & x_{nn} \end{bmatrix}$$

Thanks.

P.s[Moderators]: Edit tags as appropriate, I added as many as made sense to me.

Answer 1

I doubt you can do it with "a matrix" $B$.

Here's an ugly algorithm using matrix arithmetic. You can multiply on the left and on the right by matrices with one $1$ and all the rest $0$ to build a matrix with one entry untouched and all the other entries $0$. Do this for all $64$ entries in the original matrix, permute rows and columns for the pair of entries you want to swap, and add everything back up.

I doubt this is what you're after.

Answer 2

The dimension of $B$ is not a restriction...it could be as big as it needs to be, so long as I know how to create them.

Consider a "flatten" operator $\iota$ that maps $A$ to a single row of length $n^2$ by concatenating successive rows of $A$. Then let $B$ be the $n^2\times n^2$ permutation matrix that swaps entries $(i-1)n+j$ and $(k-1)n+m$ for any fixed $i,j,k,m \in \{1,\ldots,n\}$.

Then $\iota(A)B = \iota(\tilde{A})$ where $\tilde{A}$ is the result of swapping the $i,j$ and $k,m$ entries of $A$.

So if you allow for "flattening" of $A$ and unflattening the result of multiplying the long row by $B$ to $\tilde{A}$, then we can swap two specific entries of matrix $A$ through multiplication by (much larger) matrix $B$. Some programming languages (such as Matlab/Octave/MLab) support this through a "reshape" function.

Answer 3

Of course there is no such $B$ (EDIT: assuming you want a fixed $B$ that works for every $A$). The $j$'th row of $AB$ is (the $j$'th row of $A$) $B$. It doesn't depend on any other row. You can do any linear transformation of that row (but you will be doing the same linear transformation to each of the other rows).