Section 3.7 The TraceDeterminant Plane
Objectives

To understand that the characteristic polynomial of a \(2 \times 2\) matrix can be written as
\begin{equation*} \lambda^2  T \lambda + D, \end{equation*}where \(T = \trace(A)\) and \(D = \det(A)\text{.}\) Furthermore, if a \(2 \times 2\) matrix \(A\) has eigenvalues \(\lambda_1\) and \(\lambda_2\text{,}\) then \(\trace(A)\) is \(\lambda_1 + \lambda_2\) and \(\det(A) = \lambda_1 \lambda_2\text{,}\) and the trace and determinant of a \(2 \times 2\) matrix are invariant under a change of coordinates.
To understand that the tracedeterminant plane is determined by the graph of the parabola \(D= T^2/4\) on the \(TD\)plane and that the tracedeterminant plane can be used to determine the phase portrait of a linear system.
To understand that the tracedeterminant plane is useful for studying bifurcations.
Suppose that we have two tanks, Tank \(A\) and Tank \(B\text{,}\) that both have a volume of \(V\) liters and are both filled with a brine solution. Suppose that pure water enters Tank \(A\) at a rate of \(r_{\text{in}}\) liters per minute, and a salt mixture enters Tank \(A\) from Tank \(B\) at a rate of \(r_B\) liters per minute. Brine also enters Tank \(B\) from Tank \(A\) at a rate of \(r_A\) liters per minute. Finally, brine is drained from Tank \(B\) at a rate of \(r_{\text{out}}\) so that the volume in each tank is constant (FigureĀ 3.7.1).
If \(x(t)\) and \(y(t)\) are the amounts of salt in Tank \(A\) and Tank \(B\text{,}\) respectively, then our problem can be modeled with a linear system of two equations,
Furthermore, \(r_A = r_B + r_{\text{out}}\text{,}\) since the volume in Tank \(B\) is constant. Consequently, our system now becomes
If we have initial conditions \(x(0) = x_0\) and \(y(0) = y_0\text{,}\) it is not too difficult to deduce that the amount of salt in each tank will approach zero as \(t \to \infty\text{,}\) and we will have a stable equilibrium solution at \((0, 0)\text{.}\) Determining the nature of the equilibrium solution is a more difficult question. For example, is it ever possible that the equilibrium solution is a spiral sink? One solution is provided by studying the tracedeterminant plane.
Subsection 3.7.1 The TraceDeterminant Plane
The key to solving the system
is determining the eigenvalues of \(A\text{.}\) To find these eigenvalues, we need to derive the characteristic polynomial of \(A\text{,}\)
Of course, \(D = \det(A) = ad bc\) is the determinant of \(A\text{.}\) The quantity \(T = a + d\) is the sum of the diagonal elements of the matrix \(A\text{.}\) We call this quantity the trace of \(A\) and write \(\trace(A)\text{.}\) Thus, we can rewrite the characteristic polynomial as
We can use the trace and determinant to establish the nature of a solution to a linear system.
Theorem 3.7.2.
If a \(2 \times 2\) matrix \(A\) has eigenvalues \(\lambda_1\) and \(\lambda_2\text{,}\) then the trace of \(A\) is \(\lambda_1 + \lambda_2\) and \(\det(A) = \lambda_1 \lambda_2\text{.}\)
Proof.
The proof follows from a direct computation. Indeed, we can rewrite the characteristic polynomial as
The eigenvalues of \(A\) are now given by
Hence, \(T = \lambda_1 + \lambda_2\) and \(D = \lambda_1 \lambda_2\text{.}\)
TheoremĀ 3.7.2 tells us that we can determine the determinant and trace of a \(2 \times 2\) matrix from its eigenvalues. Thus, we should be able to determine the phase portrait of a system \({\mathbf x}' = A {\mathbf x}\) by simply examining the trace and determinant of \(A\text{.}\) Since the eigenvalues of \(A\) are given by
we can immediately see that the expression \(T^2  4D\) determines the nature of the eigenvalues of \(A\text{.}\)
If \(T^2  4D > 0\text{,}\) we have two distinct real eigenvalues.
If \(T^2  4D \lt 0\text{,}\) we have two complex eigenvalues, and these eigenvalues are complex conjugates.
If \(T^2  4D = 0\text{,}\) we have repeated eigenvalues.
If \(T^2  4D = 0\) or equivalently if \(D = T^2/4\text{,}\) we have repeated eigenvalues. In fact, we can represent those systems with repeated eigenvalues by graphing the parabola \(D= T^2/4\) on the \(TD\)plane or tracedeterminant plane (FigureĀ 3.7.3). Therefore, points on the parabola correspond to systems with repeated eigenvalues, points above the parabola (\(D \gt T^2/4\) or equivalently \(T^2  4D \lt 0\)) correspond to systems with complex eigenvalues, and points below the parabola (\(D \lt T^2/4\) or equivalently \(T^2  4D \gt 0\)) correspond to systems with real eigenvalues.
Theorem 3.7.4.
The trace and determinant of a \(2 \times 2\) matrix are invariant under a change of coordinates. That is, \(\det(T^{1} A T) = \det(A)\) and \(\trace(T^{1} A T) = \trace(A)\) for any \(2 \times 2\) matrix \(A\) and any invertible \(2 \times 2\) matrix \(T\text{.}\)
Proof.
It is straightforward to verify that \(\det(AB) = \det(A) \det(B)\) and \(\det(T^{1}) = 1/\det(T)\) for \(2 \times 2\) matrices \(A\) and \(B\text{.}\) Therefore,
A direct computation shows that \(\trace(AB) = \trace(BA)\text{.}\) Thus,
Furthermore, the expression \(T^2  4D\) is not affected by a change of coordinates by TheoremĀ 3.7.4. That is, we only need to consider systems \({\mathbf x}' = A {\mathbf x}\text{,}\) where \(A\) is one of the following matrices:
The system
has eigenvalues \(\lambda = \alpha \pm i \beta\text{.}\) The general solution to this system is
The \(e^{\alpha t}\) factor tells us that the solutions either spiral into the origin if \(\alpha \lt 0\text{,}\) spiral out to infinity if \(\alpha \gt 0\text{,}\) or stay in a closed orbit if \(\alpha = 0\text{.}\) The equilibrium points are spiral sinks and spiral sources, or centers, respectively.
The eigenvalues of \(A\) are given by
If \(T^2  4D \lt 0\text{,}\) then we have a complex eigenvalues, and the type of equilibrium point depends on the real part of the eigenvalue. The sign of the real part is determined solely by \(T\text{.}\) If \(T \gt 0\) we have a source. If \(T \lt 0\text{,}\) we have a sink. If \(T = 0\text{,}\) we have a center. See FigureĀ 3.7.5.
The situation for distinct real eigenvalues is a bit more complicated. Suppose that we have a system
with distinct eigenvalues \(\lambda\) and \(\mu\text{.}\) We will have three cases to consider if none of our eigenvalues are zero:
Both eigenvalues are positive (source).
Both eigenvalues are negative (sink).
One eigenvalue is negative and the other is positive (saddle).
Our two eigenvalues are given by
If \(T \gt 0\text{,}\) then the eigenvalue
is positive and we need only determine the sign of the second eigenvalue
If \(D \lt 0\text{,}\) we have one positive and one zero eigenvalue. That is, we have a saddle if \(T \gt 0\) and \(D \lt 0\text{.}\)
If \(D \gt 0\text{,}\) then
Since we are considering the case \(T \gt 0\text{,}\) we have
and the value of the second eigenvalue \((T  \sqrt{T^2  4D}\,)/2\) is postive. Therefore, any point in the first quadrant below the parabola corresponds to a system with two positive eigenvalues and must correspond to a nodal source.
One the other hand, suppose that \(T \lt 0\text{.}\) Then the eigenvalue \((T  \sqrt{T^2  4D}\,)/2\) is always negative, and we need to determine if other eigenvalue is positive or negative. If \(D \lt 0\text{,}\) then \(T^2  4D \gt T^2\) and \(\sqrt{T^2  4D} \gt T\text{.}\) Therefore, the other eigenvalue \((T  \sqrt{T^2  4D}\,)/2\) is positive, telling us that any point in the fourth quadrant must correspond to a saddle. If \(D \gt 0\text{,}\) then \(\sqrt{T^2  4D} \lt T\) and the second eigenvalue is negative. In this case, we will have a nodal sink. We summarize our findings in FigureĀ 3.7.6.
For repeated eigenvalues, the analysis depends only on \(T\text{.}\) Since
the only eigenvalue is \(T/2\text{.}\) Thus, we have sources if \(T > 0\) and sinks if \(T \lt 0\) (FigureĀ 3.7.7).
Example 3.7.8.
Let us return to the mixing problem that we proposed at the beginning of this section. The problem could be modeled by the system of equations
The matrix corresponding to this system is
Computing the trace and determinant of the matrix yields \(T =  2 r_A/V\) and \(D = (r_A^2  r_A r_B)/V^2\text{,}\) where \(r_A\) and \(r_B\) are both positive. Certainly, \(T \lt 0\) and
Therefore, any solution must be stable. Finally, since
we are below the parabola in the tracedeterminant plane and know that our solution must be a nodal sink.
Subsection 3.7.2 Parameterized Families of Linear Systems
The tracedeterminant plane is an example of a parameter plane. We can adjust the entries of a matrix \(A\) and, thus, change the value of the trace and the determinant.
Recall that a harmonic oscillator can be modeled by the secondorder equation
where \(m > 0\) is the mass, \(b \geq 0\) is the damping coefficient, and \(k \gt 0\) is the spring constant. If we rewrite this equation as a firstorder system, we have
Thus, for the harmonic oscillator \(T = b/m\) and \(D= k/m\text{.}\) If we use the tracedeterminant plane to analyze the harmonic oscillator, we need only concern ourselves with the second quadrant (Figure FigureĀ 3.7.9).
If \((T, D) = (b/m, k/m)\) lies above the parabola, we have an underdamped oscillator. If \((T, D) = (b/m, k/m)\) lies below the parabola, we have an overdamped oscillator. If \((T, D) = (b/m, k/m)\) lies on the parabola, we have a critically damped oscillator. If \(b = 0\text{,}\) we have an undamped oscillator.
Example 3.7.10.
Now let us see what happens to our harmonic oscillator when we fix \(m = 1\) and \(k = 3\) and let the damping \(b\) vary between zero and infinity. We can rewrite our system as
Thus, \(T = b\) and \(D = 3\text{.}\) We can see how the phase portrait varies with the parameter \(b\) in Figure FigureĀ 3.7.11.
The line \(D = 3\) in the tracedeterminant plane crosses the repeated eigenvalue parabola, \(D = T^2/4\) if \(b^2 = 12\) or when \(b = 2 \sqrt{3}\text{.}\) If \(b = 0\text{,}\) we have purely imaginary eigenvalues. This is the undamped harmonic oscillator. If \(0 \lt b \lt 2 \sqrt{3}\text{,}\) the eigenvalues are complex with a nonzero real partāthe underdamped case. If \(b = 2 \sqrt{3}\text{,}\) the eigenvalues are negative and repeatedāthe critically damped case. Finally, if \(b \gt 2 \sqrt{3}\text{,}\) we have the overdamped case. In this case, the eigenvalues are real, distinct, and negative. A bifurcation occurs at \(b = 2 \sqrt{3}\text{.}\)
Activity 3.7.1. Harmonic Oscillator with a Varying Spring Constant.
Consider a harmonic oscillator modeled by the secondorder equation
where \(m = 2\) is the mass, \(b = 2\) is the damping coefficient, and \(k \gt 0\) is the spring constant.
(a)
Rewrite (3.7.1) as a system of firstorder differential equations, \(d\mathbf x/dt = A \mathbf x\text{.}\)
(b)
Calculate the trace and determinant of \(A\text{.}\)
(c)
Sketch a line in the tracedeterminant plane that parameterizes the family of equations \(d\mathbf x/dt = A \mathbf x\text{.}\)
(d)
For what values of \(k\) is the harmonic oscillator underdamped? Overdamped? For what value of \(k\) do we have a bifurcation?
Example 3.7.12.
Consider the system
The trace of \(A\) is always \(T = 2\text{,}\) but \(D = \det(A) = 2a\text{.}\) We are on the parabola if
Thus, a bifurcation occurs at \(a = 1/2\text{.}\) If \(a \gt 1/2\text{,}\) we have a spiral sink. If \(a \lt 1/2\text{,}\) we have a sink with real eigenvalues. Further more, if \(a \lt 0\text{,}\) our sink becomes a saddle (FigureĀ 3.7.13).
Activity 3.7.2. Parameterized Families of Linear Systems.
Consider the parameterized system of linear differential equations \(d\mathbf x/dt = A \mathbf x\text{,}\) where
(a)
Find the trace, \(T\text{,}\) and determinant, \(D\text{,}\) of \(A\text{.}\)
(b)
Calculate \(T^2 = 4D\text{.}\)
(c)
For what values of \(\alpha\) and \(\beta\) is the origin a spiral sink of \(d\mathbf x/dt = A \mathbf x\text{?}\) A spiral source? A center?
(d)
For what values of \(\alpha\) and \(\beta\) is the origin a nodal sink of \(d\mathbf x/dt = A \mathbf x\text{?}\) A nodal source? A saddle?
(e)
Identify all of the regions in the \(\alpha\beta\)plane where the system \(d\mathbf x/dt = A \mathbf x\) possesses a saddle, a sink, a spiral sink, and so on. Plot your results on the \(\alpha\beta\)plane.
Example 3.7.14.
Although the tracedeterminant plane gives us a great deal of information about our system, we can not determine everything from this parameter plane. For example, the matrices
both have the same trace and determinant, but the solutions to \({\mathbf x}' = A {\mathbf x}\) wind around the origin in a clockwise direction while those of \({\mathbf x}' = B{\mathbf x}\) wind around in a counterclockwise direction.
Subsection 3.7.3 Important Lessons

The characteristic polynomial of a \(2 \times 2\) matrix can be written as
\begin{equation*} \lambda^2  T \lambda + D, \end{equation*}where \(T = \trace(A)\) and \(D = \det(A)\text{.}\)
If a \(2 \times 2\) matrix \(A\) has eigenvalues \(\lambda_1\) and \(\lambda_2\text{,}\) then \(\trace(A)\) is \(\lambda_1 + \lambda_2\) and \(\det(A) = \lambda_1 \lambda_2\text{.}\)
The trace and determinant of a \(2 \times 2\) matrix are invariant under a change of coordinates.
The tracedeterminant plane is determined by the graph of the parabola \(D= T^2/4\) on the \(TD\)plane. Points on the tracedeterminant plane correspond to the trace and determinant of a linear system \({\mathbf x}' = A {\mathbf x}\text{.}\) Since the trace and the determinant of a matrix determine the eigenvalues of \(A\text{,}\) we can use the tracedeterminant plane to parameterize the phase portraits of linear systems.
The tracedeterminant plane is useful for studying bifurcations.
Reading Questions 3.7.4 Reading Questions
1.
What is the trace of a matrix?
2.
Explain what information the tracedeterminant plane provides about a \(2 \times 2\) linear system.
Exercises 3.7.5 Exercises
Classifiying Equilibrium Points.
Classify the equilibrium points of the system \(\mathbf x' = A \mathbf x\) based on the position of \((T, D)\) in the tracedeterminant plane in Exercise GroupĀ 3.7.5.1ā8. Sketch the phase portrait by hand and then use Sage to verify your result.
1.
\(A = \begin{pmatrix} 1 \amp 2 \\ 3 \amp 4 \end{pmatrix}\)
2.
\(A = \begin{pmatrix} 4 \amp 2 \\ 3 \amp 2 \end{pmatrix}\)
3.
\(A = \begin{pmatrix} 3 \amp 8 \\ 4 \amp 6 \end{pmatrix}\)
4.
\(A = \begin{pmatrix} 4 \amp 5 \\ 3 \amp 2 \end{pmatrix}\)
5.
\(A = \begin{pmatrix} 11 \amp 10 \\ 4 \amp 5 \end{pmatrix}\)
6.
\(A = \begin{pmatrix} 5 \amp 3 \\ 8 \amp 6 \end{pmatrix}\)
7.
\(A = \begin{pmatrix} 4 \amp 15 \\ 3 \amp 8 \end{pmatrix}\)
8.
\(A = \begin{pmatrix} 4 \amp 11 \\ 8 \amp 3 \end{pmatrix}\)
OneParameter Families and Bifurcations.
Each of the following matrices in Exercise GroupĀ 3.7.5.9ā14 describes a family of differential equations \(\mathbf x' = A \mathbf x\) that depends on the parameter \(\alpha\text{.}\) For each oneparameter family sketch the curve in the tracedeterminant plane determined by \(\alpha\text{.}\) Identify any values of \(\alpha\) where the type of system changes. These values are bifurcation values of \(\alpha\text{.}\)
9.
\(A = \begin{pmatrix} \alpha \amp 3 \\ 1 \amp 0 \end{pmatrix}\)
10.
\(A = \begin{pmatrix} \alpha \amp 3 \\ \alpha \amp 0 \end{pmatrix}\)
11.
\(A = \begin{pmatrix} \alpha \amp 2 \\ \alpha \amp \alpha \end{pmatrix}\)
12.
\(A = \begin{pmatrix} 1 \amp 2 \\ \alpha \amp 0 \end{pmatrix}\)
13.
\(A = \begin{pmatrix} \alpha \amp 1 \\ 1 \amp \alpha  1 \end{pmatrix}\)
14.
\(A = \begin{pmatrix} 0 \amp 1 \\ \alpha \amp \sqrt{1  \alpha^2} \end{pmatrix}\)
15.
Consider the twoparameter family of linear systems
Identify all of the regions in the \(\alpha\beta\)plane where this system possesses a saddle, a sink, a spiral sink, and so on.
16.
Consider the twoparameter family of linear systems
Identify all of the regions in the \(\alpha\beta\)plane where this system possesses a saddle, a sink, a spiral sink, and so on.
17.
Consider the twoparameter family of linear systems
Identify all of the regions in the \(\alpha\beta\)plane where this system possesses a saddle, a sink, a spiral sink, and so on.