Eigenvalues of $A+B$ in this special case Let $A$ and $B$ are real, square matrices with the same dimension. We know that $\text{rank } A = 1$ and we know the eigenvalues of $A$. Furthermore, we know that $B$ has only zeros in the diagonal, but we don't know anything else about $B$.
Can we say something about the eigenvalues of $A+B$ ?
Or in general, can we say anything about the eigenvalues of sum of two arbitrary non-symmetric  matrices?
 A: First let us note that since $\text{rank}(A)=1$, $0$ is an eigenvalue of $A$ of multiplicity $n-1$. Then the eigenvalues of $A$ must be $0,0,0,..,0, \text{tr}(A)$.
Therefore, since $\text{rank}(A)=1$, saying that we know the eigenvalues of $A$ is equivalent to saying that we know $\text{tr}(A)$. 
Moreover, as $\text{tr}(A+B)=\text{tr}(A)+\text{tr}(B)=\text{tr}(A)$, the sum of eigenvalues of $A+B$ must be $\text{tr}(A)$.
Let $a_1,\cdots,a_n\in\mathbb{R}:a_1+\cdots+a_n =\text{tr}(A)$. Then, $a_1,\cdots,a_n$ are the eigenvalues of $A+B$, where 
$$A= \begin{bmatrix} a_1 & a_1 & \cdots & a_1  \\
a_2 & a_2 & \cdots & a_2 \\
\vdots & \vdots & \ddots & \vdots \\
a_n & a_n & \cdots & a_n \end{bmatrix}  \\
B= \begin{bmatrix} 0 & -a_1 & \cdots & -a_1  \\
-a_2 &0 & \cdots & -a_2 \\
\vdots & \vdots & \ddots & \vdots \\
-a_n & -a_n & \cdots &0\end{bmatrix} $$
Moreover, because of the above comments, $A$ has the given eigenvalues.
We finish by covering the case  $a_1=\cdots=a_n=0$. They are eigenvalues of $A+B$, where
$$A= \begin{bmatrix} 0 & 0 & \cdots & 1  \\
0 & 0 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 0 \end{bmatrix}  \\
B= 0 $$
A: We note that generally we have that:
$$\operatorname{Tr}(\mathbf{M})=\sum_{i=1}^{n}\lambda_{i},$$
Where $\lambda_{i}$ is the $i$th eigenvalue of $\mathbf{M}$ (a proof can be found here). Therefore in your case we have:
$$\operatorname{Tr}(\mathbf{A}+\mathbf{B})=\sum_{i=1}^{n}\lambda_{i}$$
Where $\lambda_{i}$ is the $i$th eigenvalue of $\mathbf{A}$. This is the case because the sum of eigenvalues of $\mathbf{B}$ must be $0$ as $\operatorname{Tr}(\mathbf{B})=0$ as we are told.
