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$\newcommand\mat{\mathbf}$A permutation matrix is a matrix whose columns are a permutation of the columns of the identity matrix $\mat I$. In other words, a permutation matrix is a matrix $\mat P$ with precisely one $1$ per row/column and zeros everywhere else.

A few easy observations about permutation matrices are:

  • $\mat P^{-1} = \mat P^\mathsf{T}$ (orthogonality)
  • $\mat P\mat 1 = \mat P^\mathsf{T}\mat1= \mat 1$ (doubly stochastic), where $\mat 1 = (1,\dots,1)$ is the all-ones vector
  • Eigenvalues are $e^{2i\pi k/n}$ for $k=1,\dots,n$, where $n$ is the least positive integer such that $\mat P^n = \mat I$.

But I don't think these three properties suffice to characterise permutation matrices, and the latter two aren't too nice to work with anyway. Is there a nice set of equations one can work with which completely capture the behaviour of permutation matrices?

Rodrigo de Azevedo
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Luke Collins
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    Along the lines of Birkhoff's theorem, you could say that the permutation matrices are the extreme points of the set of doubly-stochastic matrices. – Ben Grossmann Mar 12 '19 at 18:56
  • @Omnomnomnom Yes I've read about that theorem in pursuit of a characterisation, but how can I encapsulate what it says in the form of an equation? In a geometric sense, I believe one can see the set of doubly stochastic matrices as the convex hull of the polytope made up of the set of permutation matrices. – Luke Collins Mar 12 '19 at 18:57
  • Your geometric characterization is correct. Not quite sure how to describe them "in the form of an equation", as you'd like to – Ben Grossmann Mar 12 '19 at 18:59
  • Mostly because I would like to prove some properties about the adjacency matrices of graphs, and I need to work with permutation matrices. – Luke Collins Mar 12 '19 at 19:03
  • Could you be more specific? What do the adjacency matrices look like (or what do you know about the graphs), and what kinds of properties are you trying to prove? – Ben Grossmann Mar 12 '19 at 19:26
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    Relevant: https://math.stackexchange.com/questions/322514/nonnegative-orthogonal-matrices – Dap Mar 12 '19 at 20:19

1 Answers1

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I found a sufficiently nice characterisation of permutation matrices:

Permutation matrices are the only non-negative orthogonal matrices.

Proof: That permutation matrices are non-negative and orthogonal is clear.

Conversely, let $\mathbf P$ be a non-negative matrix with $\mathbf P^{\mathsf T}\mathbf P=\mathbf I$. Each row and column of $\mathbf P$ must contain at least one non-zero entry, otherwise $\mathbf P^{\mathsf T}\mathbf P \neq \mathbf I$ since the product will have a column of zeros.

Now suppose two entries of $\mathbf P$ in the $i$th row are non-zero, say $p_{ij}$ and $p_{ij'}$ where $j\neq j'$. But then the entry in the $jj'$th position in $\mathbf P^{\mathsf T}\mathbf P$ will not be zero, a contradiction. A similar argument yields the corresponding fact for columns.

Finally, suppose one of the non-zero entries of $\mathbf P$ is not equal to $1$, say in the $ij$th position. Then the entry in the $jj$th position of $\mathbf P^{\mathsf T}\mathbf P$ is $\sum_{k=1}^n p_{kj}^2 = p_{ij}^2 \neq 1$, a contradiction.

Luke Collins
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