I'm trying to show that if $A$ commutes with all $3\times 3$ permutation matrices, then $A$ has to be of the following form:
$ A =
\begin{pmatrix}
a & b & b \\
b & a & b \\
b & b & a \\
\end{pmatrix}
$ What i've tried so far: Let $A$ be a general $3 \times 3$-Matrix. We are trying to find $A$ such that $AP = PA$. This equation basically means the following:
$AP$ is A with permuted columns. $PA$ is A with permuted rows. This means that $A$ has to be such that permuting columns and rows in the same way leaves the marix looking the same.
For example for the permutations which flip $1$ and $2$:
$
\begin{pmatrix}
d & e & f \\
a & b & c \\
g & h & i \\
\end{pmatrix}$ =
$
\begin{pmatrix}
b & a & c \\
e & d & f \\
h & g & i \\
\end{pmatrix}$. If we do the same for flipping $2$ and $3$ and flipping $1$ and $3$ we get three matrix equations for the 9 entries $a$ to $i$
Resulting in the following equations: $a = i, b = h, g = c, f = d , b = d, a = e, c = f, h = g, b = c, f = h, i = e, g = d$ which means : $ a= e = i$ and $b=c=d=g=h$ which is exactly what I was looking for.
The thing now is. If I look at permutations which not only swap two rows/columns (for example $(123) \implies (312)$ ) I get equations which contradict my above equations.
What am I doing wrong? Is the proposition even true i.e. provable?
Thanks in advance!
Best Answer
Here's the problem:
$$P[3,1,2]=P[2,1,3]P[3,2,1]$$
$$P[3,1,2]A=AP[3,1,2]$$
$$P[2,1,3]P[3,2,1]A=AP[2,1,3]P[3,2,1]$$
When you premultiply you're doing $[3,2,1]$ and then $[2,1,3]$. When you post multiply you're doing $[2,1,3]$ and then $[3,2,1]$, so your calculations only hold for a permutation matrix that can not be written as the product of two other permutation matrices.