The stability of a system can be related to the eigenvalues of the system A matrix. These eigenvalues can be determined from the characteristic polynomial of the matrix. For a matrix A, its characteristic polynomial is
and the eigenvalues of the matrix A are the roots of the characteristic polynomial.
Example 1: We will calculate the eigenvalues of the following matrix
We first find the characteristic polynomial of A.
The roots of the characteristic polynomial are the values of s that satisfy the following relation
Therefore the roots of the characteristic polynomial are s = -1 and s = -2. Thus the eignevalues of A are -1 and -2. It is a fact that the eigenvalues of a diagonal matrix, such as the matrix A, are its diagonal elements. Notice that the A matrix of this example is the same as the A matrix of the unforced system studied in a previous example (see previous lecture). Notice also that the eigenvalues of A are the constants multiplying t in the exponentials of the unforced state solution of that example. Remember that
Because of this fact, it is easy to see that negative eigenvalues correspond to a stable system while non-negative eigenvalues will produce an unstable system.
Stability (asymptotic stability): A linear system of the form
is a stable system if all of the eigenvalues of matrix A have negative real part.
Example 2: We will analyze the stability of the following system
The eignenvalues of the A matrix are -1 and 2. Since one of the eigenvalues is non-negative, this system is unstable.
Exercise: Solve the differential equation of this example and note that the second state variable will blow up as t goes to infinity which produces an unstable system.
Example 3: Let us analyze the stability of a system whose A matrix is of size 3x3. Also, find the state transition matrix.
The characteristic polynomial is
The characteristic equation, 
has the roots s = -1, s = -2+4j and s = -2-4j . Since the real parts of all the roots are negative, we conclude that the system is stable.
