equality of $\dim V_{\lambda_i}$. Let a square matrix: $$Q = \left(\begin{array}{cccc} A&B\\0&C\end{array}\right)$$
$A,C$ are two square matrices such that $\lambda$ an eigenvalue of $A$ implies $\lambda$ isn't an eigenvalue of $C$ and vice-versa.
I already proved that the set of eigenvalues of $Q$ is a union of the sets of eigenvalues of $A$ and $C$.
Now I need to prove the following: Let $\lambda$, an eigenvalue of $Q$. Let's assume the eigenspace dimension of $\lambda$ is $n$. Then, the eigenspace dimension of the same $\lambda$ for $A$ or $C$ is also $n$.
A start:
Let (WLOG) $\lambda$, an eigenvalue of $A$ and let $v$ an eigenvector of $\lambda$. Then we have that $A-\lambda I(v) = 0$.
We look at:
 $$Q-\lambda I(v) = \left(\begin{array}{cccc} A-\lambda I&B\\0&C-\lambda I\end{array}\right)(v) = \left(\begin{array}{cccc} 0&B(v)\\0&C-\lambda I(v)\end{array}\right)$$
I am not so sure how to proceed from here. I think the fact that $\lambda$ isn't an eigenvalue of $C$ should be helpful in this point. 
 A: It is possible to show that $im$ $Q = L_1 + L_2$ and $Q|L_1 = A, Q|L_2 = C$ where $L_1$ and $L_2$ are invariant subspaces of $Q$. Since eigenspace of $\lambda$,  $W_\lambda \subseteq L_1$ $(W_\lambda\subseteq L_2)$ it is obvious that $n = dim$ $W_\lambda$ is eigenspace dimension of $\lambda$ of both $Q$ and $A(C)$.
A: *

*For clarity of concept, it is worth to note that, by writing $$Q = \left(\begin{array}{cccc} A&B\\0&C\end{array}\right)$$
$A,C$ are two square matrices, it implies that: 


1.a. $Q$ is a linear transformation from one vector space $V$ into itself. 
1.b. There $V_1$ and $V_2$ subspaces of $V$ such that $V=V_1\oplus V_2$. 
1.c. $A$ is a linear transformation from $V_1$ into $V_1$; $B$ is a linear transformation from $V_2$ into $V_1$ and $C$ is a linear transformation from $V_2$ into $V_2$. 


*I will show a solution using the approach you proposed. 


Let $\lambda$ be an eigenvalue of $Q$. 
Let (WLOG) suppose $\lambda$ is an eigenvalue of $A$. Let $E_\lambda$ be the eigenspace of A for $\lambda$. Then for every $v\in E_\lambda$, we have that $(A-\lambda I)v = 0$.
Note that $v\in V_1$, so $v$ itself can not be an eigenvector of $Q$. Let $i_1$ be the canonical inclusion of $V_1$ into $V$. Then we have $i_1(v)=(v,0)$.
So
 $$(Q-\lambda I)(v,0) = \left(\begin{array}{cc} A-\lambda I&B\\0&C-\lambda I\end{array}\right)\left(\begin{array}{c}v\\0\end{array}\right) = \left(\begin{array}{c} (A-\lambda I)v+(B)0\\0v+(C-\lambda I)0\end{array}\right)=\left(\begin{array}{c} 0\\0 \end{array}\right)$$
So $E_\lambda$, the $\lambda$-eingenspace of A, can be injectively included (by the canonical inclusion of $V_1$ into $V$) into the $\lambda$-eingenspace of $Q$. 
Now suppose that $Q$ has an eigenvector (v,w) for the eigenvalue $\lambda$. Then
$$(Q-\lambda I)(v,w) = \left(\begin{array}{cc} A-\lambda I&B\\0&C-\lambda I\end{array}\right)\left(\begin{array}{c}v\\w\end{array}\right) = \left(\begin{array}{c} (A-\lambda I)v+(B)w\\0v+(C-\lambda I)w\end{array}\right)=\left(\begin{array}{c} 0\\0 \end{array}\right)$$
Since we know that $\lambda$ is not an eigenvalue of $C$, we must have $w=0$. And so we have $(B)w=0$. And so $(A-\lambda I)v=0$, which means that $v\in E_\lambda$. 
So we have proved that  the $\lambda$-eingenspace of $Q$ is precisely $i_1(E_\lambda)$. That is the $\lambda$-eingenspace of $Q$ is precisely  the image, by the canonical inclusion, of the $\lambda$-eingenspace of $Q$. So they have the same dimension.  
