TITLE: If A has an eigenvector, a, and A = C+D then a is an eigenvector of C and D
QUESTION [0 upvotes]: Is the following true?
If  A has an eigenvector $\vec{a}$ and if A = C+D then $\vec{a}$ is an eigenvector of C and D. Furthermore if A$\vec{a}$ = $\lambda$$\vec{a}$ and C$\vec{a}$ = $\gamma$$\vec{a}$ and D$\vec{a}$ = $\beta$$\vec{a}$, then $\lambda$=$\gamma$+$\beta$.

REPLY [3 votes]: No, this is not true...
Let $A,C,D$: $\mathbb{R}^2 \to \mathbb{R}^2$ be the linear operators defined by 
$A(x,y)=(x,y)$ ,
$C(x,y)=(x,0)$ and
$D(x,y)=(0,y)$.. 
So $A=C+D$...
Then $(1,1)$ is an eigen vector of $A$ but $(1,1)$ is neither an eigen vector of $C$ nor an eigen vector of $D$
{"set_name": "stack_exchange", "score": 0, "question_id": 1463203}