I need to study the Lyapunov stability of these matrices.
Is there a criterion that does not need to calculate the eigenvalues?
For example if the trace of a matrix is > 0 then it is unstable.
$$
\begin{bmatrix}
-1 & 2 & 0 & -5 & 3 \\
0 & -1 & 1 & 6 & -4 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 0 & 2 \\
0 & 0 & 0 & -1 & -2 \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
1 & 2 & 0 & 0 & 0 \\
-5 & -8 & 0 & 0 & 0 \\
-1 & 3 & -12 & 1 & 1 \\
1 & -3 & 0 & -3 & 0 \\
-11 & 12 & 0 & 3 & -2 \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
-1 & 0 & 0 & 0 & 0 \\
-5 & -1 & 0 & 0 & 0 \\
-1 & 3 & -0.8 & 0 & 0 \\
1 & -3 & 0 & 3 & -2.1 \\
-11 & 12 & 0 & 3 & -2 \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
-1 & 0 & 0 & 0 & 0 \\
-5 & -2 & 0 & 0 & 0 \\
-1 & 3 & -0.8 & 0 & 0 \\
1 & -3 & 0 & 3 & -2.1 \\
-11 & 12 & 0 & 3 & -2 \\
\end{bmatrix}
$$
Best Answer
Note that these matrices are block triangular. Thus, each matrix is stable iff its blocks on the diagonal are each (separately) stable.
For example, the second matrix can be partitioned as $$ A = \left[\begin{array}{cc|ccc} 1 & 2 & 0 & 0 & 0 \\ -5 & -8 & 0 & 0 & 0 \\ \hline -1 & 3 & -12 & 1 & 1 \\ 1 & -3 & 0 & -3 & 0 \\ -11 & 12 & 0 & 3 & -2 \\ \end{array} \right] $$ Thus, $A$ is stable iff each of the matrices $$ A_1 = \pmatrix{1&2\\-5&-8}, \quad A_2 = \pmatrix{-12&1&1\\0&-3&0\\0&3&-2} $$ is stable. From there, we can see that $A_2$ is itself block triangular, since it can be partitioned as $$ A_2 = \left[\begin{array}{c|cc}-12&1&1\\ \hline 0&-3&0\\0&3&-2 \end{array}\right]. $$ Thus, $A_2$ is stable iff the $1\times1$ matrix $-12$ and the $2\times2$ matrix $\pmatrix{-3&0\\3&-2}$ are stable, and it is easy to see that they are (the $2 \times 2$ matrix is upper triangular, so its diagonal entries are its eigenvalues). All that remains is to decide whether $A_1$ is stable. We can confirm that $A_1$ is stable either by computing its eigenvalues or by finding its characteristic polynomial and noting that all coefficients are positive (this trick is necessary and sufficient for $2 \times 2$ matrices; larger matrices would require the Routh-Hurwitz criterion).
Thus, the matrix $A$ is stable.