Question

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?

Short answer

The characteristic polynomial is \( \lambda^2 - T\lambda + D \). Eigenvalues sum to trace and product to determinant. Graph \( T^2 - 4D = 0 \). Classify based on \( T^2 - 4D \).

Step-by-step solution

Find the Characteristic Polynomial

The characteristic polynomial of a matrix \( A \) is given by \( p(\lambda) = \det(A - \lambda I) \). For the matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the equation is \( A - \lambda I = \begin{bmatrix} a - \lambda & b \ c & d - \lambda \end{bmatrix} \). Thus, \( p(\lambda) = (a - \lambda)(d - \lambda) - bc = \lambda^2 - (a+d)\lambda + (ad-bc) \). So, \( T = a + d \) (trace) and \( D = ad - bc \) (determinant). This proves \( p(\lambda) = \lambda^2 - T\lambda + D \).

Relate Eigenvalues to Trace and Determinant

Let \( \lambda_1 \) and \( \lambda_2 \) be the eigenvalues such that \( p(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2) \). Expanding this, \( p(\lambda) = \lambda^2 - (\lambda_1 + \lambda_2)\lambda + \lambda_1\lambda_2 \). Comparing coefficients with the polynomial from Step 1, we find \( \lambda_1 + \lambda_2 = T \) and \( \lambda_1\lambda_2 = D \). Thus, the sum of the eigenvalues is the trace and the product of the eigenvalues is the determinant.

Find the Formula for Eigenvalues

The roots of the characteristic polynomial \( \lambda^2 - T\lambda + D = 0 \) are solutions to \( \lambda = \frac{T \pm \sqrt{T^2 - 4D}}{2} \). This follows directly from the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -T \), and \( c = D \).

Analyze Different Cases of the Discriminant

Based on \( T^2 - 4D \):- If \( T^2 - 4D < 0 \), the eigenvalues are complex conjugates, which implies a spiral.- If \( T^2 - 4D > 0 \), the eigenvalues are real and distinct.- If \( T^2 - 4D = 0 \), there is a real repeated eigenvalue, indicating a degenerate or nodal case.

Graph Trace-Determinant Plane

Draw axes with \( T \) as the horizontal and \( D \) as the vertical. The curve \( T^2 - 4D = 0 \) is a parabola dividing the plane into regions representing different eigenvalue types.

Classify Matrix \( A \) as per Example

For \( A = \begin{bmatrix} 1 & 2 \ -3 & 2 \end{bmatrix} \), calculate \( T = 3 \), \( D = 8 \). Since \( 3^2 - 4 \times 8 < 0 \), \( (3, 8) \) is above the parabola, indicating complex eigenvalues with positive real parts, thus a spiral source.

Key concepts

Linear Systems

Linear systems are a fundamental concept in mathematics and engineering, characterized by equations or functions where each term is either a constant or the product of a constant and a single variable. In the context of matrices, linear systems refer to a set of equations expressed in matrix form, which can be solved to determine various system properties. These systems can often be visualized using the trace-determinant plane, which offers insights into the system behavior.

When studying linear systems, one of the key components is the matrix that represents the system. For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the system's behavior is heavily influenced by the trace \( T = a + d \) and the determinant \( D = ad - bc \). These two values help us understand how the system behaves over time, predict system stability, and identify equilibrium points.

Characteristic Polynomial

The characteristic polynomial is a crucial part of understanding a matrix in a linear system. It is derived from the determinant of the matrix minus an identity matrix multiplied by a variable \( \lambda \). For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the characteristic polynomial is:

• \( p(\lambda) = \det(A - \lambda I) \)
• \( = (a - \lambda)(d - \lambda) - bc \)
• \( = \lambda^2 - (a + d)\lambda + (ad - bc) \)

Simplifying results in \( p(\lambda) = \lambda^2 - T\lambda + D \), where \( T \) is the trace and \( D \) is the determinant of \( A \).

The characteristic polynomial provides the roots which are the eigenvalues of the matrix. These roots reveal a lot about the system's dynamics and indicate whether solutions (eigenvectors) will diverge, converge, or oscillate around an equilibrium point.

Eigenvalues

Eigenvalues are pivotal in determining a system's behavior. They are the roots of the characteristic polynomial of a matrix. For the matrix \( A \), the eigenvalues \( \lambda_1 \) and \( \lambda_2 \) are solutions to the equation:

• \( \lambda = \frac{T \pm \sqrt{T^2 - 4D}}{2} \)

Here, \( T \) is the trace, and \( D \) is the determinant of the matrix \( A \).

The value \( T^2 - 4D \) is called the discriminant and dictates the nature of the eigenvalues:

• If \( T^2 - 4D < 0 \), the eigenvalues are complex and conjugate pairs, often resulting in spiral motions in the system.
• If \( T^2 - 4D = 0 \), the eigenvalues are real and repeated, indicating a node or degenerate case.
• If \( T^2 - 4D > 0 \), the eigenvalues are real and distinct, leading to saddle points or nodal behaviors.

These characteristics help classify the system's stability and predict its long-term behavior.

Phase Portrait

The phase portrait is a graphical representation of the trajectories that solutions follow over time in a dynamical system. It visualizes how eigenvalues influence the behavior of a system of linear equations.

In the trace-determinant plane, different regions signal distinct phase portraits based on the eigenvalues:

• For complex eigenvalues, the system may exhibit spiral patterns, where trajectories circle around a point.
• For real and distinct eigenvalues, trajectories either converge to or diverge from an equilibrium point, forming nodes or saddle points.
• If the eigenvalues are real and equal, the phase portrait shows a degenerate, sometimes forming nodal patterns.

Each type of eigenvalue configuration results in a unique phase portrait, which can tell us much about the system's stability and how initial conditions evolve.

Designing these portraits is integral to understanding the qualitative behavior of the system and allows one to connect mathematical analysis with intuitive graphical interpretations.