How to solve the matrix inequality $(A+B)^{-1}v\ge0$ for a singular matrix $B$? I have to calculate $B$ from a matrix inequality such that
$$(A+B)^{-1}v\ge0$$
where $A$ is invertible, $B$ is non-invertible, $v$ is a vector and "$\ge$" means an entrywise comparison.
I have seen the identity for matrix inversion given by H. V. Henderson: 
$$(A+B)^{-1}=A^{-1}-A^{-1}B(I+A^{-1}B)^{-1}A^{-1}.$$
But unable to move forward for calculating the $B$.
 A: If $v=0$, simply pick $B=0$.
If $v\ne0$, let $Q$ be a real orthogonal matrix whose first column is $\frac1{\|v\|_2}v$ (e.g. you may construct $Q$ using Householder reflection or Gram-Schmidt orthogonalisation). Now, for any real square matrix $M$, $Mv\ge0$ if and only if every row of $M$ is in the form of $pv+v^\perp$ for some nonnegative scalar $p$ and some vector $v^\perp$ orthogonal to $v$. It follows that $M=[\mathbf p|X]\,Q^T$, where $\mathbf p$ is a nonnegative vector.
Therefore, if $M=(A+B)^{-1}$, the problem boils down to finding an invertible augmented matrix $[\mathbf p|X]$ such that $\mathbf p\ge0$ and
$$
B = Q\,[\mathbf p|X]^{-1} - A\tag{1}
$$
is singular, meaning that $1$ is an eigenvalue of
$$
Q^TA\,[\mathbf p|X]).
$$
There may be multiple ways to solve this problem. For example, let the first row of $Q^TA$ be $(a,w^T)$ where $a$ is the first entry and $w^T$ consists of the other entries. If $a\ne0$, then
$$
\underbrace{\pmatrix{a&w^T\\ \ast&\ast}}_{Q^TA}
\pmatrix{1&-\frac1aw^T\\ 0&I}=\pmatrix{a&0\\ \ast&\ast}
$$
is block lower triangular. If $a=0$, then $w$ cannot be zero (because $Q^TA$ is invertible) and hence there exists some vector $q\ge0$ such that $w^Tq\ne0$. Hence
\begin{align}
&\underbrace{\pmatrix{0&w^T\\ \ast&\ast}}_{Q^TA}
\pmatrix{1&0\\ q&I}
\pmatrix{1&-\frac1{w^Tq}w^T\\ 0&I}\\
=&\underbrace{\pmatrix{0&w^T\\ \ast&\ast}}_{Q^TA}
\pmatrix{1&-\frac1{w^Tq}w^T\\ q&I-\frac1{w^Tq}qw^T}
=\pmatrix{w^Tq&0\\ \ast&\ast}
\end{align}
is also block lower triangular. In short, there always exists an invertible matrix $[\mathbf p|Y]$ such that $\mathbf p\ge0$ and $Q^TA\,[\mathbf p|Y])$ is block lower triangular.
Now, let $Q^TA\,[\mathbf p|Y])=\pmatrix{\ast&0\\ \ast&Z}$. Then
$$
Q^TA\,[\mathbf p|YZ^{-1}])=\pmatrix{\ast&0\\ \ast&I}
$$
has $1$ as an eigenvalue. Substitute $[\mathbf p|X]=[\mathbf p|YZ^{-1}]$. into $(1)$, we get a solution $B$.
