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Let's start with the $2$ by $ 2$ case:

We're given a matrix A

$$\begin{pmatrix} a & b \\ c & d. \end{pmatrix}$$

What class of matrices "rotates" or "permutes" the entries upon left-multiplication, such that we obtain, for example,

$$BA = \\ B\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} b & c \\ d & a \end{pmatrix}? $$

As another example, let's consider a $3$ by $3$ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{pmatrix}. What matrix $B$ would permute these entries, such as, for instance, into

\begin{pmatrix} d & e & a \\ c & h & i \\ f & g & b \\ \end{pmatrix}.

Does there exist a general class of matrices that permutes the entries of an $n$ by $n$ matrix to any desired result?

  • You may be interested in permutation matrices. https://en.wikipedia.org/wiki/Permutation_matrix – CyclotomicField Oct 22 '22 at 23:39
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    If you intend $B$ to be a $2 \times 2$ matrix and $BA$ to mean matrix multiplication, then there is no such matrix $B$ as this is nonlinear. Interpreting $A$ as a vector of length $4$, there is such a matrix $B$ that is $4 \times 4$. This $B$ is one of the permutation matrices as linked to in the other comment. – davidlowryduda Oct 22 '22 at 23:50
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    @davidlowryduda: what do you mean by "nonlinear"? Every permutation of the entries is linear. – Qiaochu Yuan Oct 23 '22 at 00:04
  • @davidlowryduda Well, given that the identity matrix permutes entries to their original positions, there is at least one matrix with at least one existent permutation. I'm not ruling out the possibility that such a matrix is specific to each 2 x 2 or 3 x 3 case, being in terms of the original entries $a, b, c, d, e,f,g...$ – askquestions4 Oct 23 '22 at 00:07
  • @askquestions4 the issue is that you can only do row permutations or column permutations depending on which side of the matrix you multiply on. If you allow a $B$ and a $C$ such that $BAC$ is the permuted matrix then it's possible with permutation matrices. – CyclotomicField Oct 23 '22 at 00:12

3 Answers3

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It's very easy to see that such matrices don't always exists.

For example, there's no matrix $B$ such that $$B\begin{pmatrix}1&0\\0&0\end{pmatrix} = \begin{pmatrix}0&0\\0&1\end{pmatrix}$$

To see why, just write the entries of $B$ and calculate the product on the LHS.

jjagmath
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Permutations of $n^2$ elements can best be described by elements of the symmetric group of order $n^2$, written $S_{n^2}$. You are asking now how these permutations act on matrices. We could also ask more generally how these permutations act on elements of any vectorspace (matrices are, after all, also elements of the vector space $\mathbb{R}^{n^2}$).

It turns out that there is a whole theory behind this, called representation theory. A representation is a function $\rho$ (more precisely a group homomorphism) from a group $G$ to the invertible linear operators over a vector space $V$, i.e.: $\rho: G \rightarrow \operatorname{GL}(V)$. The objects in $\operatorname{GL}(V)$ can be described (in the finite dimensional case) as matrices. So $\rho$ associates a matrix to every group element.

In your case the group is $S_{n^2}$ and we wish to act with it on the vector space of $n \times n$ matrices, this corresponds to a representation $\rho: S_{n^2} \rightarrow \operatorname{GL}(\mathbb{R}^{n^2})$. If we let $\sigma \in S_{n^2}$ be a permutation we would expect something like this to happen:

$$ \rho(\sigma) \begin{pmatrix} a_{1} & a_{2} & \dots \\ a_{n+1} & a_{n+2} \dots \\ & \dots & \\ a_{(n-1)n} & a_{(n-1)n +1} & \dots \end{pmatrix}= \begin{pmatrix} a_{\sigma^{-1}(1)} & a_{\sigma^{-1}(1)} & \dots \\ a_{\sigma^{-1}(n+1)} & a_{\sigma^{-1}(n+2)} \dots \\ & \dots & \\ a_{\sigma^{-1}((n-1)n)} & a_{\sigma^{-1}((n-1)n +1)} & \dots \end{pmatrix} $$

(We have to put $\sigma^{-1}$ there for $\rho$ to be a homomorphism, but this doesn‘t matter).

The matrix you were looking for is therefore exactly $\rho(\sigma)$, $\rho$ is in this case called the standart representation of $S_{n^2}$, and a well understood function. You can now go further and ask interesting questions like „Are there subspaces in the vector space $\mathbb{R}^{n^2}$, that are left invariant by all possible permutations?“, this is exactly what representation theory deals with.

I now this is a very theoretical answer that doesn’t tell you how the $\rho(\sigma)$ look like, but I hope that this could still show you how much fascinating theory there is behind your question:)

Edit: The $\rho(\sigma)$ are exactly the permutation matrices linked to in the comment under your post. Notice though that we consider the matrices upon which is acted here as elements of a vector space and represent them therefore as vectors, so the permutation matrices/ the $\rho(\sigma)$ don’t act on the matrices by matrix multiplication but by „matrix vector multiplication“, i.e. you have to write the matrix that $\rho(\sigma)$ acts on as a column vector.

Henry T.
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    So...what do these matrices actually look like? Can you provide a couple examples? Thank you for your details though otherwise. – askquestions4 Oct 22 '22 at 23:36
  • Made an edit regarding to what they look like – Henry T. Oct 22 '22 at 23:58
  • I find it hard to follow you... – Jean Marie Oct 23 '22 at 20:52
  • @Jean Marie In which step, if I might ask:)? – Henry T. Oct 23 '22 at 23:35
  • Re-reading your text, I realize that my above comment isn't fair. Your answer is understandable. I should have made my comment about the effectiveness of this approach : how this representation theory can help in the case proposed by the asker for example to prove that it is only permutation matrices that can "act" as "scramblers" by left-multiplication ? – Jean Marie Oct 24 '22 at 04:33
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Concerning your $2 \times 2$ example : it is impossible that a (fixed) matrix $B$ exist such that for any $A$, we have this "scrambled" result:

$$ B\underbrace{\begin{pmatrix} a & b \\ c & d \end{pmatrix}}_A = \begin{pmatrix} b & c \\ d & a \end{pmatrix}$$

for a simple reason: if we take the determinants on both sides, we should have:

$$\det(B)(ad-bc)=ab-cd \ \iff \ \det(B)=\frac{ab-cd}{ad-bc}$$

with a fixed value on the left and a variable value on the right, which is impossible.

Same comment for your second $3 \times 3$ example.

I think (it remains to be proven) that the only "possible scrambling operators" $B$ of the entries of a matrix by left-multiplication are those where $B$ is permutation matrix (a single entry $1$ on each row and each column) ; in such a case, one obtains the permutation of the lines of the matrix $A$ like here :

$$\underbrace{\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}}_B\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}=\begin{pmatrix} a & b & c \\ g & h & i \\ d & e & f \end{pmatrix},$$

where the fact that $B(e_1)=e_1, B(e_2)=e_3, B(e_3)=e_2$ make $B$ "act" on the rows of matrix $A$ in the following way: a "stand-still" for the first row and and a transposition of the 2nd and third row.

Remark: post(=right)multiplication by this matrix $B$ gives a similar action on columns of $A$.

Jean Marie
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