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Let $A$ be a nonnegative square matrix, and let $U$, respectively $V$, be obtained from $A$ by normalizing the rows, respectively columns of $A$ so that they sum to one. That is, $U = (A_{ij} / \sum_{r} A_{ir})$ and $V = (A_{ij} / \sum_{k} A_{kj})$ assuming the sums are nonzero.

Let us further assume that $1-U_{kk} \le \alpha$ and $1 - V_{kk} \le \alpha$ for all $k$. Can we obtain a bound on the deviations of the column sums of $U$ from one: say $$\sum_{i} U_{ij} \le 1 + c \,\alpha, \forall j$$ for some constant $c > 0$ (independent of the dimension of the matrices)? We can assume $\alpha$ is sufficiently small if need be.

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Consider the $n\times n$ matrix $$A=\left[\begin{array}{ccccc}1-\alpha&0&0&\cdots&0\\ \alpha/(n-1)&\epsilon&0&\cdots&0\\\alpha/(n-1)&0&\epsilon&\ddots&0\\\vdots&\vdots&\ddots&\ddots&\vdots\\\alpha/(n-1)&0&0&\cdots&\epsilon\end{array}\right],$$ and let $\alpha'=\alpha/(n-1).$ Then $V$ satisfies $1-V_{k,k}\leq\alpha$ for all $k,$ since $V_{1,1}=1-\alpha,$ and $V_{k,k}=1$ for all $k\geq 2.$ We have $$U=\left[\begin{array}{ccccc}1&0&0&\cdots&0\\\alpha'/(\alpha'+\epsilon)&\epsilon/(\alpha'+\epsilon)&0&\cdots&0\\\alpha'/(\alpha'+\epsilon)&0&\epsilon/(\alpha'+\epsilon)&\ddots&0\\\vdots&\vdots&\ddots&\ddots&\vdots\\\alpha'/(\alpha'+\epsilon)&0&0&\cdots&\epsilon/(\alpha'+\epsilon)\end{array}\right].$$ Then if we choose $\epsilon=(1-\alpha)/(n-1),$ we have that $U_{k,k}=1-\alpha$ for all $k\geq2,$ so $1-U_{k,k}\leq\alpha$ for all $k.$ Note that $U_{k,1}=\alpha'/(\alpha'+\epsilon)=\alpha$ for all $k\geq2,$ so we have that $\sum_{i=1}^{n}U_{i,1}=1+(n-1)\alpha.$ This shows that it is impossible for there to exist such a $c$ which is independent of $n.$

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    Thanks for the nice example. The question now is whether this is the worst case? Can we show that the bound holds with $c = n-1$ for example? – 2017-02-21
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    Suppose $U$ is row-normalized and subject to $1-U_{k,k}\leq \alpha$ for all $k.$ WLOG, take the first column to be the one with the max column sum. Then $U_{1,1}\leq 1,$ and $U_{k,1}\leq\alpha$ for each $2\leq k\leq n,$ so $\sum_{i}U_{i,1}\leq 1+(n-1)\alpha,$ so this is the worst case. – 2017-02-21