# Matrices commuting with the matrix of ones

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?

• 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. Mar 4 '19 at 13:01

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.

• If $V$ is spanned by doubly stochastic matrices, which are, in turn, spaned by permutation matrices, isn't $V$ also spaned by permutation matrices? Mar 4 '19 at 22:25
• @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. Mar 4 '19 at 22:51

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

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