Convergence of Positive definite matrix I sign up this website for this question. Suppose I have a vector 1 (all elements are 1) with $11^T-B$ positive definite, where $B$ is symmetric non-negative matrix (all elements are greater or equal to zero, but may not have inverse). Let $B\rightarrow 0$, can we say something about $1^T(11^T-B)^{-1}1$? I am supposing it converges to 1, but have no ideas. Any ideas would be appreciate.
 A: This is false. Consider the case of $2\times 2$ matrices. If we write
$$
A=11^T-B = \begin{pmatrix} 1-a & 1-b \\ 1-b & 1-c \end{pmatrix} ,
$$
then a straightforward calculation shows that the quantity in question equals
$$
\frac{2b-a-c}{\det A} = \frac{2b-a-c}{2b-a-c+ac-b^2} ,
$$
and this will go to $1$ if and only if
$$
\frac{ac-b^2}{2b-a-c}\to 0 . \quad\quad\quad\quad (1)
$$
Now it's easy to build a counterexample. We can take $a=t^{10}$, $c=2t$, $b=t+t^2$, and send $t\to 0+$. Notice that $\det A= t^2+O(t^3)>0$, so this satisfies the assumptions. However, (1) fails: the limit equals $-1/2$.
A: I believe the answer is yes, this converges to 1. It's a little subtle because if $B$ is small the eigenvalues of ${\bf 11}^T - B$ will be close to the eigenvalues of ${\bf 11}^T$, namely $1, 0, \ldots, 0$, and the eigenvector for the eigenvalue close to 1 will be close to the vector ${\bf 1}$, but after inverting $A = {\bf 11}^T - B$ its other eigenvalues will be large and it's not obvious how that affects ${\bf 1}^TA^{-1}{\bf 1}$.
Since this geometric picture didn't answer the question for me, I decided to look at a straight computation. The matrix entries of $A$ are $1 - b_{ij}$ and $A^{-1}$ has entries $\frac{1}{{\rm det}(A)}A_{ij}$ where $A_{ij}$ is the $i,j$-cofactor of $A$. This is relevant because ${\bf 1}^T A^{-1}{\bf 1}$ is just the sum of the entries of $A^{-1}$. So the question is what is $\frac{1}{{\rm det}(A)} \sum_{i,j} A_{ij}$? It seems to me that if the entries of $A$ are close to $1$ then both ${\rm det}(A)$ and $\sum_{i,j} A_{ij}$ vanish to order $n-2$ in the $b_{ij}$ where $n$ is the dimension and they are equal at order $n-1$. So as $B$ goes to zero the value of ${\bf 1}^T A^{-1}{\bf 1}$ will go to 1. But I won't try to write out the verification, which seems tedious but not difficult.
