Let $\mspace{10mu}A,B\in M^{4\times 4}(\mathbb{R})\mspace{10mu}, $
$A=\begin{pmatrix} 1 &0 &2 &-1 \\ 0 &1 &-1 &1 \\ 1 &1 &1 &0 \\ 1 &-1 &3 &-2 \end{pmatrix} \mspace{10mu} \mathrm{and} \mspace{10mu} \mathrm{rk}(B)=1$
What values may have $\mspace{10mu}\mathrm{rk}(A+B)\mspace{10mu}?$
I only figured out that $\mathrm{rk}(A)=2$.
By SVD, we see that \begin{align} A = U\Sigma V^T = \sigma_1 u_1v_1^T +\sigma_2 u_2v_2^T. \end{align} since $\text{rank}(A) = 2$. Since $B$ is rank 1, then we see that $B = u v^T$ for some vectors $u, v$. Hence \begin{align} A+B = \sigma_1 u_1v_1^T +\sigma_2 u_2v_2^T+uv^T. \end{align} If $B = \lambda u_1v_1^T$ then we see that $A+B = (\lambda+\sigma_1)u_1v_1^T+\sigma_2u_2v_2^T$ which means $\text{rank}(A+B) = 2$ if $\lambda \neq -\sigma_1$. Otherwise, rank equals 1.
In the case $uv^T$ is neither $u_1v_1^T$ or $u_2v_2^T$ then $A+B$ is rank 3.
In general it holds that $$ \operatorname{rank}(A+B) \leq \operatorname{rank}(A) + \operatorname{rank}(B) $$ and similarly, $$ \operatorname{rank}(-B+(A+B)) \leq \operatorname{rank}(B) + \operatorname{rank}(A+B) \implies \\ \operatorname{rank}(A) \leq \operatorname{rank}(B) + \operatorname{rank}(A+B). $$ It follows that the rank of $A+B$ must be 1, 2, or 3.
It cannot be $A = -B$ since $\operatorname{rank}(A) = 2$ and $\operatorname{rank}(B) = 1$ so $A+B \ne 0$. Therefore by the subadditivity of rank we have $$1 \le \operatorname{rank}(A+B) \le \operatorname{rank}(A) + \operatorname{rank}(B) = 3$$ so $\operatorname{rank}(A+B) \in \{1,2,3\}$.
All possibilities are indeed attainable, consider $B$ equal to $$\begin{pmatrix} 0 &0 &0 &0 \\ 0 &0 &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &0 \end{pmatrix}, \quad\begin{pmatrix} 0 &0 &0 &0 \\ 1 &1 &1 &1 \\ 1 &1 &1 &1 \\ -1 &-1 &-1 &-1 \end{pmatrix}, \quad\begin{pmatrix} 0 &0 &0 &0 \\ 0 &-1 &1 &-1 \\ 0 &-1 &1 &-1 \\ 0 &1 &-1 &1 \end{pmatrix}$$ respectively.
