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I am now playing with permutation matrices, http://mathworld.wolfram.com/PermutationMatrix.html.

Also, there is a similar discussion: Symmetric Permutation Matrix. I want to ask more details than this one.

As we know, a permutation matrix is orthogonal, i.e., $E^T=E^{-1}$. I am interested in when it is symmetric, i.e., $E^T=E^{-1} = E$

Suppose

  1. Start from an identity matrix $I_n$.
  2. $n$ can be even or odd number.
  3. Pick $(i,j)$, where $0<i,j\leq n$ and $i, j$ are integer. Exchange $i$-th and $j$-th columns of $I_n$ (identity matrix) and get $E$. Then $E$ is symmetric. This is because $E_{ii}=E_{jj}=0$ and $E_{ij}=E_{ji}=1$.
  4. Based on 3., if I pick a number of sets $(i,j)$, $(k,l)$, $(q,r), \ldots$, without repeated index in each $(\cdot,\cdot)$, and permute columns of $I_n$ according to these sets, then the resulting permutation matrix $E$ is symmetric.

One key thing here is "without repeated index in each $(\cdot,\cdot)$". This is because if I do $(1,2)$ and $(2,3)$ for $I_3$ for example, I get

$$\begin{bmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix},$$

which is not symmetric. In this case, I repeat $2$ in each suit.

Is the above correct? Or I miss some key assumptions?

Rodrigo de Azevedo
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sleeve chen
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    Yes, in general the permutation is idempotent when is a disjoint product of fix points and cycles of length $2.$ – Phicar Apr 19 '19 at 23:41
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    Yes, it's correct. A permutation matrix describes a permutation $\pi$. You want $E^2 = I$, so $\pi\circ\pi = id$. – amsmath Apr 19 '19 at 23:42

2 Answers2

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You’re correct!

We can think of the action of $E$ on the set of $n$ standard basis vectors as a permutation $\sigma$ on $\{1,\dots,n\}$ and vice versa.

Let $E$ be symmetric, and let $i$ be the only nonzero entry in the first row. This means that $e_{1i}=e_{i1}$ by symmetry. Thus $E$ swaps the first and the $i^{th}$ standard basis vectors, so $(1~i)$ is a cycle in the cycle decomposition of $\sigma$. This argument applies to the rest of the rows to show that $\sigma$ is a product of disjoint transpositions.

Santana Afton
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As you have noted condition for a permutation matrix $E$ to be symmetric is that $E^{-1}=E$, and this condition can be expressed as $E^2=I$.

Interpreting the last condition as repeating the permutation is identity. So this represents a permutation that is its own inverse. That is if $E$ sends a basis vector $v$ to $W$ $E^2=I$ implies $Ew=v$. (possible that $v=w$)

So this corresponds to a permutation where an element is fixed, or if it sends $x$ to $y$ then it has to send $y$ to $x$. Thus this consists of many disjoint swaps (and possibly some fixed points).

In group theory it is a permutation of cycle type corresponding to the partition of $n$ into $2$'s and $1$'s. For example $9=2+2+2+ 1^6 $ (that is 1 repeated six times).

P Vanchinathan
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