It is a nice exercise to classify linear systems based on their position in
the trace-determinant plane. Consider the matrix
$$
A=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]
$$
a) Show that the characteristic polynomial of the matrix \(A\) is
\(p(\lambda)=\lambda^{2}-T \lambda+D\), where \(T=a+d\) is the trace of \(A\) and
\(D=\operatorname{det}(A)=a d-b c\) is the determinant of \(A\).
b) We know that the characteristic polynomial factors as
\(p(\lambda)=\left(\lambda-\lambda_{1}\right)\left(\lambda-\lambda_{2}\right)\),
where \(\lambda_{1}\) and \(\lambda_{2}\) are the eigenvalues. Use this and the
result of part (a) to show that the product of the eigenvalues is equal to the
determinant of matrix A. Note: This is a useful fact. For example, if the
determinant is negative, then you must have one positive and one negative
eigenvalue, indicating a saddle equilibrium point. Also, show that the sum of
the eigenvalues equals the trace of matrix \(A\).
c) Show that the eigenvalues of matrix \(A\) are given by the formula
$$
\lambda=\frac{T \pm \sqrt{T^{2}-4 D}}{2}
$$
Note that there are three possible scenarios. If \(T^{2}-4 D<0\), then there are
two complex eigenvalues. If \(T^{2}-4 D>0\), there are two real eigenvalues.
Finally, if \(T^{2}-4 D=0\), then there is one repeated eigenvalue of algebraic
multiplicity two.
d) Draw a pair of axes on a piece of poster board. Label the vertical axis \(D\)
and the horizontal axis \(T\). Sketch the graph of \(T^{2}-4 D=0\) on your poster
board. The axes and the parabola defined by \(T^{2}-4 D=0\) divide the trace-
determinant plane into six distinct regions, as shown in Figure 13.17.
e) You can classify any matrix \(A\) by its location in the trace-determinant
plane. For example, if
$$
A=\left[\begin{array}{rr}
1 & 2 \\
-3 & 2
\end{array}\right]
$$
then \(T=3\) and \(D=8\), so the point \((T, D)\) is located in the first quadrant.
Furthermore, \((3)^{2}-4(8)<\) 0 , placing the point \((3,8)\) above the parabola
\(T^{2}-4 D=0\). Finally, if you substitute \(T=3\) and \(D=8\) into the formula
\(\lambda=\left(T \pm \sqrt{T^{2}-4 D}\right) / 2\), then you get eigenvalues
that are complex with a positive real part, making the equilibrium point of
the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) a spiral source. Use pplane6 to
generate a phase portrait of this particular system and attach the printout to
the poster board at the point \((3,8)\).
f) Linear systems possess a small number of distinctive phase portraits. Each
of these is graphically different from the others, but each corresponds to the
pair of eigenvalues and their multiplicities. For each case, use pplane 6 to
construct a phase portrait, and attach a printout at its appropriate point
\((T, D)\) in your poster board trace-determinant plane. Hint: There are
degenerate cases on the axes and the parabola. For example, you can find
degenerate cases on the parabola in the first quadrant that separate nodal
sources from spiral sources. There are also a number of interesting degenerate
cases at the origin of the trace-determinant plane. One final note: We have
intentionally used the words "small number of distinctive cases"' so as to
spur argument amongst our readers when working on this activity. What do you
think is the correct number?
