How to find the eigenvalues of a block-diagonal matrix? The matrix $A$ below is a block diagonal matrix where each block $A_i$ is a  $4 \times 4$ matrix with known eigenvalues. 
$$A= \begin{pmatrix}A_1 & 0 & \cdots & 0 \\ 0 & A_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & A_n  \end{pmatrix}$$
How do I find the eigenvalues of the block diagonal matrix $A$? Does this mean that I will have $4 n$ eigenvalues? 
Am I correct in thinking that the eigenvalues of the block diagonal matrix $A$ above are just a list of the individual eigenvalues of each $A_i$ and not the product of everything? 

Since,
$\det(A-\lambda I)=\det(A_1-\lambda I)\det(A_2-\lambda I)...\det(A_n-\lambda I)$,
the eigenvalues of $A$ are just the list of eigenvalues of each $A_i$.
