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Let $J_n$ be $n \times n$ matrix of ones, so that $J_{ij} = 1$ for all $i,j \in\{1,\ldots,n\}$. I am interested to find the class of matrices which commute with $J_n$, i.e. $$M J_n = J_n M.$$It is not difficult to see that a permutation matrix satisfies this property. By linear superposition, the doubly stochastic matrix also commutes with $J_n$.

Are there any other matrices, apart from doubly stochastic ones, with this property?

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    $\begingroup$ These are the matrices for which the sum of each row and each column is the same. Any $(n-1)\times(n-1)$-matrix has infinitely many extensions to such an $n\times n$-matrix; one for each value of the row- and column-sum. $\endgroup$
    – Servaes
    Mar 4 '19 at 13:01
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The set $V$ of such matrices is easily verified to be a linear subspace of $\operatorname{Mat}(\Bbb{R},n)$. the space of real $n\times n$-matrices. In fact $V$ is the subspace spanned by the doubly stochastic matrices:

First note that $J_n\in V$, and that $\frac{1}{n}J_n$ is doubly stochastic.

Let $M$ be a matrix such that $MJ_n=J_nM$. Then the sum of the entries of each row and each column is the same, say $c$. Let $m$ be the minimum of the entries of $M$. Then all entries of $M-mJ_n$ are nonnegative, and each row and column of $M-mJ_n$ sums to $c-mn$, so in particular $c-mn\geq0$. If $c-mn\neq0$ then $$\frac{1}{c-mn}(M-mJ_n),$$ is a matrix with nonnegative entries whose rows and columns sum to $1$, and hence a doubly stochastic matrix. This shows that $M$ is a linear combination of doubly stochastic matrices, because $$M=(c+mn)\left(\frac{1}{c-mn}(M-mJ_n)\right)+\frac{mn}{c-mn}\left(\frac{1}{n}J_n\right).$$

If $c-mn=0$ then $M-mJ_n=0$ and so $M=mJ_n=mn\left(\frac{1}{n}J_n\right)$, also a linear combination of doubly stochastic matrices.

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  • $\begingroup$ If $V$ is spanned by doubly stochastic matrices, which are, in turn, spaned by permutation matrices, isn't $V$ also spaned by permutation matrices? $\endgroup$
    – Fizikus
    Mar 4 '19 at 22:25
  • $\begingroup$ @Fizikus The set of doubly stochastic matrices isn't a vector space, so they aren't 'spanned' by any set. I don't know whether the space $V$ is spanned by the set of permutation matrices, possibly it is. $\endgroup$
    – Servaes
    Mar 4 '19 at 22:51
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We have that

$$(MJ)_{ij}=\sum_{k=1}^n M_{ik}J_{kj}=\sum_{k=1}^n M_{ik}$$ and

$$(JM)_{ij}=\sum_{k=1}^n J_{ik}M_{kj}=\sum_{k=1}^n M_{kj}.$$

So $$(MJ)_{ij}=(JM)_{ij} \iff \sum_{k=1}^n M_{ik}=\sum_{k=1}^n M_{kj}.$$ And

$$MJ=JM\iff \sum_{k=1}^n M_{ik}=\sum_{k=1}^n M_{kj}, \forall i,j\in\{1,\cdots,n\}.$$

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