I would like to know if there exists a matrix B to get $A+B$ full rank for all $A \in \mathbb{R}^{n \times m}$ (with $n \geq m$).
I tried a solution for $B \in \mathbb{R}^{n \times m}$ where $b_{i,i}=1$ for all $i \leq m$ and where the other elements are equal to $0$. For example, for $n=4$ and $m=3$, we have $B=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$. For $n=m$ it is obviously the case and when $n \neq m$ we have $B^TB=I_m$. Besides, I can not find a counterexample where $A+B$ is not full rank and I do not have either a proof to know if it is true.
EDIT: I think the problem is equivalent to finding a matrix $B \in \mathbb{R}^{n \times m}$ such that for all $A \in \mathbb{R}^{n \times m}$, $A^TA + B^TB + B^TA + (B^TA)^T \in \mathbb{R}^{m \times m}$ is full-rank. One way would be to show $B^TA + (B^TA)^T$ or $B^TA$ is positive-definite.