Can I quickly determine the eigenvalues of this matrix? I am working on observability and detectability in controls and I ran across this example that I didn't understand. The author deliberately sought the form of this matrix, because of its "block-form" in order to quickly find the eigenvalues
\begin{bmatrix}
    l_{11}  & -1  & -1 & 0 & 0 \\
    0 & -1  & 0 & 0 & 0 \\
    0 & -1  & -1 & 0 & 0 \\
    0 & -.1 & -2 & l_{42} & -.1 \\
    0 &  1  &  2 & 0 & -.2
\end{bmatrix}
The author was then able to state the eigenvalues were $\{l_{11}, -1, l_{42}, -.2\}$ 
I was under the impression that I could only determine the eigenvalues via  a matrix diagonal if the matrix was upper/lower triangular? 
 A: This matrix is block lower triangular:
$$\left[\begin{array}{ccc|cc}
    l_{11}  & -1  & -1 & 0 & 0 \\
    0 & -1  & 0 & 0 & 0 \\
    0 & -1  & -1 & 0 & 0 \\
\hline
    0 & -.1 & -2 & l_{42} & -.1 \\
    0 &  1  &  2 & 0 & -.2
\end{array}\right]$$
From the bottom-right block we see, that eigenvalues are $l_{42}$ and $-0.2$. 
The upper-left block is again block upper triangular
$$\left[\begin{array}{c|cc}
    l_{11}  & -1  & -1\\
\hline
    0 & -1  & 0\\
    0 & -1  & -1 \\
\end{array}\right]$$
and has eigenvalue $l_{11}$ and double eigenvalue $-1$.
A: Note that the characteristic polynomial is easy to evaluate by using the Laplace expansion of a determinant (along the blue column or row):
\begin{align}&\det\begin{bmatrix}
    \color{Blue}{l_{11}-x}  & -1  & -1 & 0 & 0 \\
    \color{Blue}{0} & -1-x  & 0 & 0 & 0 \\
    \color{Blue}{0} & -1  & -1-x & 0 & 0 \\
    \color{Blue}{0} & -.1 & -2 & l_{42}-x & -.1 \\
    \color{Blue}{0} &  1  &  2 & 0 & -.2-x
\end{bmatrix}\\&=(l_{11}-x)\det\begin{bmatrix}
    \color{Blue}{-1-x}  & \color{Blue}{0} & \color{Blue}{0} & \color{Blue}{0} \\
    -1  & -1-x & 0 & 0 \\
    -.1 & -2 & l_{42}-x & -.1 \\
    1  &  2 & 0 & -.2-x
\end{bmatrix}
\\&=(l_{11}-x)(-1-x)\det\begin{bmatrix}
    \color{Blue}{-1-x}  & \color{Blue}{0} & \color{Blue}{0} \\
    -2 & l_{42}-x & -.1 \\
    2 & 0 & -.2-x
\end{bmatrix}
\\&=(l_{11}-x)(-1-x)^2\det\begin{bmatrix}
    \color{Blue}{l_{42}-x} & -.1 \\
   \color{Blue}{0} & -.2-x
\end{bmatrix}\\&=(l_{11}-x)(-1-x)^2(l_{42}-x)(-.2-x).
\end{align}
Therefore the eigenvalues of the given matrix, which are the roots of the characteristic polynomial, are $$\{l_{11}, -1, l_{42}, -.2\}.$$ 
A: Let $\sigma(A)$ denote the set of eigenvalues of a matrix $A$. We have
$$\sigma(A \oplus B)=\sigma(A) \cup \sigma(B) $$
where $\oplus$ is the matrix direct sum. This turns the original problem into two easier ones.
