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More Linear Transformations and Differential Equations

Session Material:

Lay: 5.4 and 5.7

Recap and Exercises

Session Notes

Session Material

Session Description

This session explores the deep connection between eigenvalues and the geometry of linear transformations, ultimately applying these tools to solve systems of Differential Equations. We begin by revisiting linear transformations through the lens of a basis. You will learn how to find the matrix of a transformation relative to a specific basis and how choosing the "right" basis—specifically one consisting of eigenvectors—can simplify a complex transformation into a simple diagonal matrix.

We then transition into the continuous world with Section 5.7, where we use these algebraic tools to solve systems of first-order linear differential equations (\(\mathbf{x}' = A\mathbf{x}\)). You will discover how the eigenvalues of matrix \(A\) determine the long-term behavior of a dynamic system. By "decoupling" the equations using eigenvectors, we can transform a tangled system of changing variables into a set of independent, solvable equations. Finally, we will visualize these solutions using phase portraits, learning to identify whether a system’s origin is an attractor, a repeller, or a saddle point.

Key Concepts

  • The Matrix of a Linear Transformation
  • Diagonal Matrix Representations (\([T]_{\mathcal{B}}\))
  • Similarity and Diagonalization
  • Systems of First-Order Differential Equations
  • Decoupling a Dynamical System
  • Fundamental Sets of Solutions
  • Phase Portraits (Attractors, Repellers, and Saddle Points)
  • Stability Analysis

Learning Objectives

  • Find the matrix of a linear transformation relative to a non-standard basis.
  • Understand how similarity transformations relate different matrix representations of the same mapping.
  • Solve systems of linear differential equations using eigenvalues and eigenvectors.
  • Decouple systems of equations to find general solutions for \(\mathbf{x}' = A\mathbf{x}\).
  • Construct and interpret phase portraits to describe the stability and trajectory of a system over time.
  • Relate the sign and magnitude of eigenvalues to the physical behavior of a dynamical system.

Exercises

Exercise 1 (5.4.1)

Let \(T: \mathbb{R}^2 \to \mathbb{R}^2\) be the linear transformation that rotates vectors counterclockwise by \(\pi / 4\) radians. Find the matrix of \(T\) relative to the standard basis \(\mathcal{E}_2\).

 

\(T=\left[\begin{array}{rr}\cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4}\end{array}\right]=\left[\begin{array}{rr}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{array}\right]\)

Exercise 2 (5.4.7)

Assume the mapping \(T: \mathbb{P}_2 \rightarrow \mathbb{P}_2\) defined by

\[ T\left(a_0+a_1 t+a_2 t^2\right)=3 a_0+\left(5 a_0-2 a_1\right) t+\left(4 a_1+a_2\right) t^2 \]

is linear. Find the matrix representation of \(T\) relative to the basis \(\mathcal{B}=\left\{1, t, t^2\right\}\).

 

\(\left[\begin{array}{rrr}3 & 0 & 0 \\ 5 & -2 & 0 \\ 0 & 4 & 1\end{array}\right]\)

Exercise 1 (5.7.1)

A particle moving in a planar force field has a position vector \(\mathbf{x}\) that satisfies \(\mathbf{x}^{\prime}=A \mathbf{x}\). The \(2 \times 2\) matrix \(A\) has eigenvalues 4 and 2 , with corresponding eigenvectors \(\mathbf{v}_1=\left[\begin{array}{r}-3 \\ 1\end{array}\right]\) and \(\mathbf{v}_2=\left[\begin{array}{r}-1 \\ 1\end{array}\right]\). Find the position of the particle at time \(t\), assuming that \(\mathbf{x}(0)=\left[\begin{array}{r}-6 \\ 1\end{array}\right]\).

 

\(c_1=5 / 2, c_2=-3 / 2\), and \(\mathbf{x}(t)=\frac{5}{2}\left[\begin{array}{r}-3 \\ 1\end{array}\right] e^{4 t}-\frac{3}{2}\left[\begin{array}{r}-1 \\ 1\end{array}\right] e^{2 t}\)

Exercise 2 (5.7.2)

Let \(A\) be a \(2 \times 2\) matrix with eigenvalues -3 and -1 and corresponding eigenvectors \(\mathbf{v}_1=\left[\begin{array}{r}-1 \\ 1\end{array}\right]\) and \(\mathbf{v}_2=\left[\begin{array}{l}1 \\ 1\end{array}\right]\). Let \(\mathbf{x}(t)\) be the position of a particle at time \(t\). Solve the initial value problem \(\mathbf{x}^{\prime}=A \mathbf{x}, \mathbf{x}(0)=\left[\begin{array}{l}2 \\ 3\end{array}\right]\).

 

\(c_1=1 / 2, c_2=5 / 2\), and \(\mathbf{x}(t)=\frac{1}{2}\left[\begin{array}{r}-1 \\ 1\end{array}\right] e^{-3 t}+\frac{5}{2}\left[\begin{array}{l}1 \\ 1\end{array}\right] e^{-t}\)

Exercise 3 (5.7.3-5.7.6)

Solve the initial value problem \(\mathbf{x}^{\prime}(t)=A \mathbf{x}(t)\) for \(t \geq 0\), with \(\mathbf{x}(0)=(3,2)\). Classify the nature of the origin as an attractor, repeller, or saddle point of the dynamical system described by \(\mathbf{x}^{\prime}=A \mathbf{x}\). Find the directions of greatest attraction and/or repulsion. When the origin is a saddle point, sketch typical trajectories.

  1. \(A=\left[\begin{array}{rr}2 & 3 \\ -1 & -2\end{array}\right]\)
  2. \(A=\left[\begin{array}{rr}-2 & -5 \\ 1 & 4\end{array}\right]\)
  3. \(A=\left[\begin{array}{rr}7 & -1 \\ 3 & 3\end{array}\right]\)
  4. \(A=\left[\begin{array}{ll}1 & -2 \\ 3 & -4\end{array}\right]\)
 
  1. \(c_1=-5 / 2, c_2=9 / 2\), and \(\mathbf{x}(t)=-\frac{5}{2}\left[\begin{array}{r}-3 \\ 1\end{array}\right] e^t+\frac{9}{2}\left[\begin{array}{r}-1 \\ 1\end{array}\right] e^{-t}\). Since one eigenvalue is positive and the other is negative, the origin is a saddle point of the dynamical system described by \(\mathbf{x}^{\prime}=A \mathbf{x}\). The direction of greatest attraction is the line through \(\mathbf{v}_2\) and the origin. The direction of greatest repulsion is the line through \(\mathbf{v}_1\) and the origin.
  2. \(c_1=13 / 4, c_2=-5 / 4\), and \(\mathbf{x}(t)=\frac{13}{4}\left[\begin{array}{r}-1 \\ 1\end{array}\right] e^{3 t}-\frac{5}{4}\left[\begin{array}{r}-5 \\ 1\end{array}\right] e^{-t}\). Since one eigenvalue is positive and the other is negative, the origin is a saddle point of the dynamical system described by \(\mathbf{x}^{\prime}=A \mathbf{x}\). The direction of greatest attraction is the line through \(\mathbf{v}_2\) and the origin. The direction of greatest repulsion is the line through \(\mathbf{v}_1\) and the origin.
  3. \(c_1=-1 / 2, c_2=7 / 2\), and \(\mathbf{x}(t)=-\frac{1}{2}\left[\begin{array}{l}1 \\ 3\end{array}\right] e^{4 t}+\frac{7}{2}\left[\begin{array}{l}1 \\ 1\end{array}\right] e^{6 t}\). Since both eigenvalues are positive, the origin is a repellor of the dynamical system described by \(\mathbf{x}^{\prime}=A \mathbf{x}\). The direction of greatest repulsion is the line through \(\mathbf{v}_2\) and the origin.
  4. \(c_1=-1, c_2=5\), and \(\mathbf{x}(t)=-\left[\begin{array}{l}2 \\ 3\end{array}\right] e^{-2 t}+5\left[\begin{array}{l}1 \\ 1\end{array}\right] e^{-t}\). Since both eigenvalues are negative, the origin is an attractor of the dynamical system described by \(\mathbf{x}^{\prime}=A \mathbf{x}\). The direction of greatest attraction is the line through \(\mathbf{v}_1\) and the origin.

Exercise 4 (5.7.7-5.7.8)

Make a change of variable that decouples the equation \(\mathbf{x}^{\prime}=A \mathbf{x}\). Write the equation \(\mathbf{x}(t)=P \mathbf{y}(t)\) and show the calculation that leads to the uncoupled system \(\mathbf{y}^{\prime}=D \mathbf{y}\), specifying \(P\) and \(D\).

  1. \(A=\left[\begin{array}{rr}7 & -1 \\ 3 & 3\end{array}\right]\)
  2. \(A=\left[\begin{array}{ll}1 & -2 \\ 3 & -4\end{array}\right]\)
 

  1. \(\left[\begin{array}{l}y_1^{\prime}(t) \\ y_2^{\prime}(t)\end{array}\right]=\left[\begin{array}{ll}4 & 0 \\ 0 & 6\end{array}\right]\left[\begin{array}{l}y_1(t) \\ y_2(t)\end{array}\right]\)

  2. \(\left[\begin{array}{l}y_1^{\prime}(t) \\ y_2^{\prime}(t)\end{array}\right]=\left[\begin{array}{rr}-2 & 0 \\ 0 & -1\end{array}\right]\left[\begin{array}{l}y_1(t) \\ y_2(t)\end{array}\right]\)

Exercise 6 (5.7.19)

[M] Find formulas for the voltages \(v_1\) and \(v_2\) (as functions of time \(t\) ) for the circuit in Example 1, assuming that \(R_1=1 / 5\) ohm, \(R_2=1 / 3\) ohm, \(C_1=4\) farads, \(C_2=3\) farads, and the initial charge on each capacitor is 4 volts.

 

The general solution is \(\mathbf{x}(t)=c_1\left[\begin{array}{l}1 \\ 2\end{array}\right] e^{-.5 t}+c_2\left[\begin{array}{r}-3 \\ 2\end{array}\right] e^{-2.5 t}\).

\(\left[\begin{array}{l}v_1(t) \\ v_2(t)\end{array}\right]=\mathbf{x}(t)=\frac{5}{2}\left[\begin{array}{l}1 \\ 2\end{array}\right] e^{-.5 t}-\frac{1}{2}\left[\begin{array}{r}-3 \\ 2\end{array}\right] e^{-2.5 t}\)

Exercise 7 (5.7.20)

[M] Find formulas for the voltages \(v_1\) and \(v_2\) for the circuit in Example 1, assuming that \(R_1=1 / 15 \mathrm{ohm}, R_2=1 / 3 \mathrm{ohm}\), \(C_1=9\) farads, \(C_2=2\) farads, and the initial charge on each capacitor is 3 volts.

 

The general solution is thus \(\mathbf{x}(t)=c_1\left[\begin{array}{l}1 \\ 3\end{array}\right] e^{-t}+c_2\left[\begin{array}{r}-2 \\ 3\end{array}\right] e^{-2.5 t}\).

\(\left[\begin{array}{l}v_1(t) \\ v_2(t)\end{array}\right]=\mathbf{x}(t)=\frac{5}{3}\left[\begin{array}{l}1 \\ 3\end{array}\right] e^{-t}-\frac{2}{3}\left[\begin{array}{r}-2 \\ 3\end{array}\right] e^{-2.5 t}\)