In the example you have given, I assume is the output.
For linear systems, exponentials (including complex exponentials) are the eigenfunctions. Therefore you should try a solution. Notice that , where exp(-zT) is a constant, and remember that . Then you can rewrite the original system equation as
which is satsfied at the points z which satisfy
(eigenvalue equation for this problem) The z value(s) which satisfy the equation above are the eigenvalues, because .
You may use Matlab's fsolve() to find the eigenvalues, which are the root(s) of the eigenvalue equation. You must give fsolve() an initial guess. There may be 0, 1, or 2 real solutions, for this problem, since a straight line can intersect the exponential curve at 0, 1, or 2 points.
T=1; A0=0; A1=1; zinit=0;
You can illustrate the eigenvalues by plotting the curve and the line . The z value(s) where the curve and line intersect are the solutions to the eigenvalue equation above. Here is an example of how to plot the curve and the line for three examples. The z-values where line crosses the curve are the eigenvalues.
A0=0; A1=1; y1=z/A1-A0/A1;
A0=1.5; A1=-1; y2=z/A1-A0/A1;
A0=0.8; A1=-1; y3=z/A1-A0/A1;
grid on; xlim([-1 3]); ylim([0 3]); xlabel('z');
legend('exp(-zT)','Ex.1 line','Ex.2 line','Ex.3 line','Ex.1 eigval','Ex.2 eigvals');
There will be complex solutions in Example 3. This means the eigenfunctions, exp(zt), will be damped or expanding sinusoids. Try function NEWTZERO from the Matlab file exchange to find the complex roots.