Sum of matrices value 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 $\mathrm{rk}(A)=2$.
 A: 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. 
A: 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.
A: 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.
