Let $B$ be a real invertible $N\times N$ matrix, and $O$ the $N\times N$ matrix with all the entries equal to 1.
Define the $\epsilon$ dependent matrix $A(\epsilon)$ as
$$
A(\epsilon) = O + \epsilon B,\quad \epsilon\in\Bbb R.
$$
$A(\epsilon)$ is invertible for each $\epsilon\neq0$, so we can let $C(\epsilon)$ be the inverse of $A(\epsilon)$ for all $\epsilon\neq0$.
As $\epsilon\to0$ the matrix $A(\epsilon)$ converges to $O$, so clearly $C(\epsilon)$ diverges as $\epsilon\to0$. Anyways, it happens that both the sum of the rows and the sum of the columns of $C(\epsilon)$ converge as $\epsilon\to0$. I have clear numerical evidence of this fact, but I don't know how to formally prove it. Can you help me?
Best Answer
Notice that $O = ee^T$, where $e$ is the all ones vector. As a consequence, if you suppose everything is invertible, you can use the Sherman-Morrison formula and get $$ (O + \epsilon B)^{-1}e = (\epsilon B)^{-1}e - \frac{(\epsilon B)^{-1}ee^T(\epsilon B)^{-1}}{1+e^T(\epsilon B)^{-1}e}e \\= \frac 1{1+e^T(\epsilon B)^{-1}e} (\epsilon B)^{-1}e \\ =\frac 1{\epsilon+e^TB^{-1}e} B^{-1}e $$ This quantity explodes only if $e^TB^{-1}e = 0$, so to get a counterexample, take for example $$ B = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} $$ and you will get sum of row/columns of the order $O(1/\epsilon)$.