If A is non-singular and B is nilpotent, with some additional properties, A-B is non-singular? Let $A$ and $B$ be two square matrices of size $n$ with the following properties:
$A$ is non-singular, $a_{ii}=1$ $\ \forall i$, if $a_{ij}\neq 0$ then $a_{ji}= 0$ $\ \forall i,j$ and $i\neq j$;
$B$ is singular, $b_{ii}=0$ $\ \forall i$, if $[b_{i,1},\dots,b_{i,n}]\neq [0,0,...,0]$ then $[b_{1,i},\dots,b_{n,i}]^T= [0,0,...,0]^T$ (i.e. for $i=1,...,n$, the $i$-th row or the $i$-th column or together is completely null).
Is it possible to prove that $A-B$ is non-singular?
What about the non-singularity of $A-B$ if $a_{ij}\geq 0$ and $b_{ij}\geq 0$ $\forall i,j$?
 A: What about $A$=$\left [\begin{array} {rr}
1  &  0\\
1  &  1\\
\end{array}\right]$
and $B$=$\left [\begin{array} {rr}
0  &  -1\\
0  &  0\\
\end{array}\right]$?
$A$ is non singular clearly nonsingular with $\det(A) = 1$, $B$ is nilpotent with $B^{2}=0$, but $A-B$=$\left [\begin{array} {rr}
1  &  1\\
1  &  1\\
\end{array}\right]$ is singular. I think these two matrices meet your criteria as well.
Edit: in response to your comment, I don't believe so. Take $A$ =$\left [\begin{array} {rr}
1  &  1 & 0 \\
0  &  1 & 1\\
0  &  0 & 1
\end{array}\right]$, and $B$=$\left [\begin{array} {rr}
0  &  0 & 0\\
0  &  0 & 0\\
1  &  0 & 0
\end{array}\right]$.
Then $\det(A)=1$, $B^{2} = 0$ and $A-B$=$\left [\begin{array} {rr}
1  &  1 & 0 \\
0  &  1 & 1\\
-1  &  0 & 1
\end{array}\right]$, which if I haven't miscomputed is indeed singular.
A: Consider $A$=$\left [\begin{array} {rr}
1  &  1\\
0  &  1\\
\end{array}\right]$
and $B$=$\left [\begin{array} {rr}
0  &  0\\
-1  &  0\\
\end{array}\right]$
$A$ is non singular $B$ is nilpotent but $A-B$ is singular
