Can I quickly determine the eigenvalues of this matrix?

Question

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?

Answer 1

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 $-1$.

Answer 2

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.

Answer 3

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 $$\{l_{11}, -1, l_{42}, -.2\}.$$