Problem 28 - Calculus for Biology and Medicine

Question

We discuss the case of repeated eigenvalues. $$ \left[\begin{array}{l} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ (a) Show that $$ A=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right] $$ has the repeated eigenvalues \(\lambda_{1}=\lambda_{2}=1\). (b) Show that every eigenvector of \(A\) is of the form $$ c_{1}\left[\begin{array}{l} 1 \\ 0 \end{array}\right] $$ where \(c_{1}\) is a real number different from \(0 .\) (c) Show that $$ \mathbf{x}_{1}(t)=e^{t}\left[\begin{array}{l} 1 \\ 0 \end{array}\right] $$ is a solution of \((11.35)\).

Step-by-Step Solution

Answer

(a) Eigenvalues are \( \lambda_1 = \lambda_2 = 1 \). (b) Eigenvectors are \( c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} \). (c) \( \mathbf{x}_1(t) = e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) is a solution.

Step 1: Find the characteristic polynomial

To find the eigenvalues, we first find the characteristic polynomial of matrix \( A \). The characteristic polynomial is given by the determinant \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Write down the expression for \( A - \lambda I \): \[A - \lambda I = \begin{bmatrix} 1 - \lambda & 2 \ 0 & 1 - \lambda \end{bmatrix}\]Now, compute the determinant:\[\det(A - \lambda I) = (1 - \lambda)(1 - \lambda) - (0 \times 2) = (1 - \lambda)^2\]Set the determinant equal to zero and solve for \( \lambda \):\[(1 - \lambda)^2 = 0\]This yields the repeated eigenvalue \( \lambda_1 = \lambda_2 = 1 \).

Step 2: Find eigenvectors of A

Using \( \lambda = 1 \), solve \( (A - \lambda I)\mathbf{v} = \mathbf{0} \) to find the eigenvectors. Substitute \( \lambda = 1 \) into the equation:\[(A - I) = \begin{bmatrix} 0 & 2 \ 0 & 0 \end{bmatrix}\]We need to solve:\[\begin{bmatrix} 0 & 2 \ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]The first row gives \( 2x_2 = 0 \), implying \( x_2 = 0 \). The general solution for \( x_1 \) is any real number. Thus, the eigenvectors \( \mathbf{v} \) are of the form:\[\mathbf{v} = c_1 \begin{bmatrix} 1 \ 0 \end{bmatrix}\]where \( c_1 \) is a non-zero real number.

Step 3: Verify solution \(\mathbf{x}_1(t) = e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix}\)

We need to confirm that \( \mathbf{x}_1(t) = e^t \begin{bmatrix} 1 \ 0 \end{bmatrix} \) is a solution to the differential equation system. Compute the derivative:\[\frac{d}{dt}\mathbf{x}_1(t) = \frac{d}{dt} \left( e^t \begin{bmatrix} 1 \ 0 \end{bmatrix} \right) = e^t\begin{bmatrix} 1 \ 0 \end{bmatrix}\]Now, calculate \( A\mathbf{x}_1(t) \) using matrix multiplication:\[A \mathbf{x}_1(t) = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} e^t \ 0 \end{bmatrix} = \begin{bmatrix} e^t \ 0 \end{bmatrix}\]Since \( \frac{d}{dt}\mathbf{x}_1(t) = A\mathbf{x}_1(t) \), \( \mathbf{x}_1(t) = e^t \begin{bmatrix} 1 \ 0 \end{bmatrix} \) is indeed a solution.

Key Concepts

Characteristic Polynomial

In linear algebra, the characteristic polynomial is a crucial concept used to find the eigenvalues of a matrix. For a given square matrix \( A \), the characteristic polynomial is derived from the equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix of the same size as \( A \). Calculating the determinant of \( A - \lambda I \) allows us to find a polynomial in \( \lambda \). Solving this polynomial equation provides the eigenvalues of the matrix.

In the context of our original exercise, for matrix \( A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \), we set \( A - \lambda I \) as \( \begin{bmatrix} 1 - \lambda & 2 \ 0 & 1 - \lambda \end{bmatrix} \).
To compute the determinant, we perform the following operation:

• \[ \det(A - \lambda I) = (1 - \lambda)(1 - \lambda) - (0 \times 2) = (1 - \lambda)^2 \]

This determinant equation \( (1 - \lambda)^2 = 0 \) leads to the repeated eigenvalue \( \lambda_1 = \lambda_2 = 1 \). The characteristic polynomial \((1 - \lambda)^2\) indicates that there is only one distinct eigenvalue, suggesting a repeated root.

This polynomial is instrumental in understanding the behavior and properties of matrices in mathematics, specifically when analyzing stability and solutions to differential equations.

Eigenvectors

Eigenvectors are vectors that, when a linear transformation is applied, do not change direction. Instead, they are scaled by a corresponding eigenvalue. For a matrix \( A \) and its eigenvalue \( \lambda \), any non-zero vector \( \mathbf{v} \) that satisfies the equation \( A\mathbf{v} = \lambda\mathbf{v} \) is an eigenvector.

In our exercise, after establishing the eigenvalue \( \lambda = 1 \), we find the eigenvectors by solving \( (A - \lambda I)\mathbf{v} = \mathbf{0} \). With matrix \( A \), this problem boils down to:

• \[ (A - I) = \begin{bmatrix} 0 & 2 \ 0 & 0 \end{bmatrix} \]

The matrix simplification shows a purely upper triangular matrix, focusing the solution on:

• The equation \( 2x_2 = 0 \) from the first row, implying \( x_2 = 0 \).
• Consequently, \( x_1 \) can be any real number, indicating that the eigenvectors are of the form \( \mathbf{v} = c_1 \begin{bmatrix} 1 \ 0 \end{bmatrix} \) with \( c_1 eq 0 \).

This means the eigenvectors are aligned along the first axis in the 2D space, implying that any scalar multiple of \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \) is an eigenvector, provided the scalar is non-zero. Understanding eigenvectors is critical for exploring invariant subspaces and the structure of linear transformations.

Differential Equations

Differential equations involve unknown functions and their derivatives—the task is to find such a function given certain conditions. In the context of eigenvalues and eigenvectors, differential equations often come up when modeling dynamic systems.

The original differential equation in the problem is concerned with the matrix \( A \) acting on a vector function \( \mathbf{x}(t) \), where

• \( \frac{d}{dt}\mathbf{x}(t) = A\mathbf{x}(t) \)

This type of differential equation is an example of a linear system, where solutions involve exponential functions determined by the eigenvalues of \( A \).

To show that \( \mathbf{x}_1(t) = e^t \begin{bmatrix} 1 \ 0 \end{bmatrix} \) is indeed a solution, we differentiate with respect to \( t \):

• \( \frac{d}{dt}\mathbf{x}_1(t) = e^t\begin{bmatrix} 1 \ 0 \end{bmatrix} \)

Calculating \( A\mathbf{x}_1(t) \) confirms:

• \( A\mathbf{x}_1(t) = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} e^t \ 0 \end{bmatrix} = e^t\begin{bmatrix} 1 \ 0 \end{bmatrix} \)

This match ensures validity, meaning the proposed solution satisfies the differential equation. Such solutions are pivotal in understanding how systems evolve over time, providing insights into stability and behavior under various conditions.