Let $A$ be a zero diagonal $n\times n$ bistochastic matrix. Let $B$ be a sub-matrix of A by deleting row and column $j$. we want to prove \begin{align*} (I - B)^{-1} 1_{n-1} < n \cdot 1_{n-1} \end{align*} (i.e. the max row sum of $(I - B)^{-1}$ is smaller than $n$)
UPDATE 1:
I was thinking about doing to following: Define norm $N(X)$ as
\begin{align*} N(X) = \max_{i} \sum_j X_{ij} \end{align*}
Then I want to prove \begin{align*} N((I - B)^{-1}) = N (I + B + B^2 + B^3 + ...)< n \end{align*}
and we know $N (I + B + B^2 + B^3 + ...) \leq N(I) + N(B) + N(B)^2 + ... <= \frac{1}{1 - N(B)}$. However, this does not complete the proof because $N(B)$ can be 1.
UPDATE 2:
I will also be glad someone can point out if the statement is not correct. From my simulation, it appears that I can confirm the above statement
N = 6;
mi = 01;
A=rand(N).*not(eye(N));
B=bistoch(A,1E-5)
inv(eye(N - mi) - B(1:N- mi,1:N- mi)) * ones(N- mi,1)
function a=bistoch(a,tol)
% return bistochastic matrix from input nonnegative matrix
if nargin<2, tol=0.001; end
s1=sum(a,2);
s2=sum(a);
while ((abs(max(s1)-1))>tol) | ((abs(max(s2)-1))>tol)
a=a./s1;
s1=sum(a,2);
s2=sum(a);
a=a./s2;
end
end
This is not true. E.g. when $0<t<1$ and $$ A=\left[\begin{array}{c|ccc} 0&t&1-t&0\\ \hline t&0&0&1-t\\ 1-t&0&0&t\\ 0&1-t&t&0 \end{array}\right], $$ we have $$ \begin{bmatrix}y_1\\ y_2\\ y_3\end{bmatrix} :=(I-B)^{-1}\mathbf1 =\begin{bmatrix}1&0&t-1\\ 0&1&-t\\ t-1&-t&1\end{bmatrix}^{-1} \begin{bmatrix}1\\ 1\\ 1\end{bmatrix} =\begin{bmatrix}\frac{t+1}{t}\\ \frac{2-t}{1-t}\\ \frac{1}{t(1-t)}\end{bmatrix}. $$ Therefore $(I-B)^{-1}\mathbf1$ is unbounded. In fact, none of its entries is bounded above, because $\lim_{t\to0}y_1=\infty$, $\lim_{t\to1}y_2=\infty$ and $\lim_{t\to0}y_3=\lim_{t\to1}y_3=\infty$.