Question about matrices and eigenvectors For two $n\times n$ square matrices $A$ and $B$ satisfying $AB=A+B,\;$
suppose $A$ has an eigenvector $u.$
 How to show that $u$ is also and eigenvector of $B?$
 A: Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.
It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$
You should be able to continue from here since now $AB= BA$.
A: With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(\lambda,v)\;$ is an eigenpair of $A.$ 
Then $$Bv=(B-I)Av=(B-I)\lambda v=\lambda B v- \lambda v$$ hence $$(1-\lambda)Bv=-\lambda v.$$ This means that $(\frac{\lambda}{\lambda-1},v)$ is an eigenpair of $B.$
A: if two $n\times n$ matrix share a commutative association, then they would also share a common eigenvalues
$$A\cdot B = A+B$$
although the multiplication of matrix is not commutative but its addition is, so its easy to see that
$$ A + B = B+A = AB = BA$$
$$AB =BA$$
so your matrix can be an example of one with oriented elements, check here, then the eigenvalues of $A+B$ is the sum of the eigenvalues of $A$ and $B$
