If the operator $(A+B)$ has an eigenvector $v$, is $v$ also the eigenvector of $A$ and $B$? I think the reverse is true (and I think the eigenvalue of $(A+B)$ corresponding to $v$ is also the sum of those of $A$ and $B$ corresponding to $v$. ) 
But I have trouble believing the statement in the question. Thank you.
 A: This is false. Consider
$$A = \begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix}$$
$$B = \begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}$$
Then
$$A + B = \begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix}$$
for which every nonzero vector is an eigenvector associated with eigenvalue $3$. But you can easily check that, for example, $v = \begin{pmatrix}1 \\ 1\end{pmatrix}$ is not an eigenvector of $A$ or $B$.
A: No, if you consider $A=\left[e_1,e_1\right]$ and $B= \left[0,e_2-e_1\right]$ then 
$A+B=I$ 
and, for example, the vector $v=(0,1)$ is an eigenvector for $A+B$, but 
$Av=(1,0)$ that is not linear depend with $v$
A: We should not expect this to be true: Given any collection $L$ of $n$ lines $\ell_1, \ldots, \ell_n$ in an $n$-dimensional real or complex vector space $\Bbb V$ in general position, we can find a linear transformation $T : \Bbb V \to \Bbb V$ whose eigenspaces are precisely those lines: If we pick a nonzero element $v_i$ respectively from each line $\ell_i$, then since the lines are in general position, $(v_i)$ is a basis of $\Bbb V$. Then, for any pairwise distinct scalars $\lambda_1, \ldots, \lambda_n$, the linear transformation characterized by $T(v_i) = \lambda_i v_i$ for all $i$ has the claimed property.
So, if $n \geq 2$, we can pick two such collections $L, L'$ such that $L \cap L' = \emptyset$, and respectively pick transformations $T, T'$ as above. Then, $T$ and $T'$ have no common eigenvectors, but all linear transformations have at least one eigenvector, and in particular $T + T'$ does.
For example, if we take $\Bbb V = \Bbb R^2$ and take $L$ to be the set containing the axes and $L'$ to be the set containing the two diagonal lines, we can take $T = \pmatrix{1&0\\0&0}$ and $T' = \pmatrix{1&1\\1&1}$.
