The Eigenvalues of a block matrix

By definition, an eigenvalue $\lambda$ of the block matrix $A$ satisfies

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

Using a property of block matrix determinants, we have

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

Thus the eigenvalues of $B,D$ are also the eigenvalues of $A$.

If $\lambda$ is an eigenvalue of $A$ with eigenvector $(x_1, x_2, x_3)^t$ then $(x_1,x_2,x_3,0,0,0)^t$ is an eigenvector of the block matrix. Similarly, for $D$ but you put three zeros at the beggining.

Hint: $$ \pmatrix{B & C\cr 0 & D\cr} \pmatrix{B^{-1} & E\cr 0 & D^{-1}\cr} = \pmatrix{I & BE + CD^{-1}\cr 0 & I\cr} $$ What $E$ will make $BE + CD^{-1} = 0$?