The Eigenvalues of a block matrix It is known that all $B$, $C$ and $D$ are $3 \times 3$ matrices. And the eigenvalues of $B$ are $1, 2, 3$; $C$ are $4, 5, 6$; and $D$ are $7, 8, 9$. What are the eigenvalues of the $6 \times 6$ matrix 
$$\begin{pmatrix}
B &amp; C\\0 &amp; D
\end{pmatrix}$$
where $0$ is the $3 \times 3$ matrix whose entries are all $0$. 
From my intuition, I think the eigenvalues of the new $6 \times 6$ matrix are the eigenvalues of $B$ and $D$. But how can I show that?
By definition, an eigenvalue $\lambda$ of the block matrix $A$ satisfies

$$\det \begin{pmatrix} B-\lambda I &amp; C \\ 0 &amp; D-\lambda I \end{pmatrix} = 0.$$

Using a property of block matrix determinants, we have

$$\det \begin{pmatrix} B-\lambda I &amp; C \\ 0 &amp; D-\lambda I \end{pmatrix} = \det(B-\lambda I)\det(D-\lambda I) = 0$$

Thus the eigenvalues of $B,D$ are also the eigenvalues of $A$.
$BE=-CD^{-1}$
By your assumptions $B$ is invertible, hence we have:
$-B^{-1}CD^{-1}$