Basis for span of tensor power of permutation matrices? Set 
$$
P^{(r)}=\{\sigma ^{\otimes r}\,|\,\sigma\,\, \textrm{ is an $n\times n$ permutation matrix}\}
$$
We know that for $r=1$ $\mathbb{F}P^{(r)}$ is the space of all matrices over $\mathbb{F}$ with constant column and row sum. We can also show that this space is of dimension $(n-1)^2+1$ over any field, and compute an explixit basis for it. Namely
$$
\{E_{11}+E_{ij}-E_{1j}-E_{i1}\,|\,1<i,\,j\leq n\}\cup\{\mathbb{I}\}
$$
where $E_{ij}$ is the matrix with $ij$ entry $1$ and $0$ elsewhere and $\mathbb{I}$ is the identity matrix.
Is there a similar result for $P^{(r)}$ defined as above, that is, an explicit closed form for its basis for any $r\in \mathbb{N}$? And is there a closed form for the dimension of $\mathbb{F}P^{(r)}$?
I suppose one could use representation theoretic arguments by decomposing $\mathbb{F}P^{(r)}$ whenever it is semisimple, but I'm just wondering if things were as simple as the case where $r=1$.
 A: Write $P(w)$ for the permutation matrix corresponding to a given permutation $w$. Then the set of all $P(w)^{\otimes r}$ such that $w$ has an increasing subsequence of length $n-r$ is a basis for the linear span of all the $P^{\otimes r}$, $P$ a permutation matrix.
A proof is given in
https://arxiv.org/abs/2109.00107.
Here $w$ is being regarded as a sequence $w_1w_2 \cdots w_n$, where $w_i = w(i)$ for all $i=1, \dots, n$. For example, the sequence $w=213465$ defines the permutation interchanging $1 \leftrightarrow 2$ and $5 \leftrightarrow 6$, which would be written as $w=\binom{123456}{213465}$ in the two-line notation for permutations.
Notice that when $r=1$, the basis has $1+(n-1)^2$ elements. They are the permutation matrices corresponding to the set of consecutive cycles. For example, the cycles $$2 \mapsto 3 \mapsto 4 \mapsto 2, \quad 4 \mapsto 3 \mapsto 2 \mapsto 4$$ (that is, $1342$, $1423$ as sequences) are consecutive $3$-cycles, but the $3$-cycle $$1 \mapsto 3 \mapsto 4 \mapsto 1$$ (that is, the sequence $3241$) is non-consecutive.
