Problem 52 - Calculus for Biology and Medicine

Question

Find the eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\) and corresponding eigenvectors \(\mathrm{v}_{1}\) and \(\mathrm{v}_{2}\) for each matrix A. Determine the equations of the lines through the origin in the direction of the eigenvectors \(\mathrm{v}_{1}\) and \(\mathrm{v}_{2}\), and graph the lines together with the eigenvectors \(\mathrm{v}_{1}\) and \(\mathrm{v}_{2}\) and the vectors \(\mathrm{Av}_{1}\) and \(\mathrm{Av}_{2}\) $$A=\left[\begin{array}{rr}-1 & 0 \\ 0 & 2\end{array}\right]$$

Step-by-Step Solution

Answer

Eigenvalues are \( \lambda_1 = -1 \), \( \lambda_2 = 2 \). Eigenvectors are \( \mathbf{v}_1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \), \( \mathbf{v}_2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \). Lines: \( x = 0 \), \( y = 0 \).

Step 1: Identify the Matrix

The given matrix is \( A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} \).

Step 2: Calculate the Characteristic Polynomial

The characteristic polynomial is derived from the determinant \( \det(A - \lambda I) \), where \( I \) is the identity matrix. This results in the following polynomial:\[\det \left( \begin{bmatrix} -1-\lambda & 0 \ 0 & 2-\lambda \end{bmatrix} \right) = (-1-\lambda)(2-\lambda)\]

Step 3: Solve the Characteristic Polynomial

By solving \( (-1-\lambda)(2-\lambda) = 0 \), we find the eigenvalues:1. For \( -1-\lambda = 0 \), solving gives \( \lambda_1 = -1 \).2. For \( 2-\lambda = 0 \), solving gives \( \lambda_2 = 2 \).

Step 4: Find Eigenvectors for \(\lambda_1 = -1\)

To find the eigenvector \( \mathbf{v}_1 \) for eigenvalue \( \lambda_1 = -1 \), solve:\[(A + I) \mathbf{v}_1 = \begin{bmatrix} 0 & 0 \ 0 & 3 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]This implies \( y \) can be any value, while \( x = 0 \). Choose \( y = 1 \), so \( \mathbf{v}_1 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

Step 5: Find Eigenvectors for \(\lambda_2 = 2\)

To find the eigenvector \( \mathbf{v}_2 \) for eigenvalue \( \lambda_2 = 2 \), solve:\[(A - 2I) \mathbf{v}_2 = \begin{bmatrix} -3 & 0 \ 0 & 0 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]This implies \( x = 0 \) while \( y \) can be any value. Choose \( y = 1 \), so \( \mathbf{v}_2 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \).

Step 6: Determine the Equations of Lines through Origin

The equations for the lines in the directions of the eigenvectors are:1. For \( \mathbf{v}_1 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \), the line is \( x = 0 \).2. For \( \mathbf{v}_2 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \), the line is \( y = 0 \).

Step 7: Graph the Lines and Vectors

Using a coordinate system, draw the lines \( x = 0 \) (vertical) and \( y = 0 \) (horizontal). Plot the eigenvectors \( \mathbf{v}_1 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \). Calculate the vectors \( A\mathbf{v}_1 = \begin{bmatrix} 0 \ 2 \end{bmatrix} \) and \( A\mathbf{v}_2 = \begin{bmatrix} -1 \ 0 \end{bmatrix} \), and plot them as well.

Key Concepts

Characteristic Polynomial

The characteristic polynomial is a fundamental concept in linear algebra, especially when working with eigenvalues and eigenvectors. It is crucial because it helps in finding the eigenvalues of a matrix. Given a square matrix \(A\), the characteristic polynomial is formed by taking the determinant of \(A - \lambda I\), where \(\lambda\) is a scalar, and \(I\) is the identity matrix. This gives us a polynomial equation in \(\lambda\), known as the characteristic polynomial.
To find the eigenvalues, you solve this polynomial equation for \(\lambda\). The roots of the characteristic polynomial correspond to the eigenvalues of the matrix \(A\). In our problem, solving the equation \((-1-\lambda)(2-\lambda) = 0\) provides the eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = 2\).
Understanding how to derive and solve the characteristic polynomial is essential because it directly leads to finding eigenvalues, which are key in various linear algebra applications.

Linear Algebra

Linear algebra is the branch of mathematics dealing with vectors, matrices, and linear transformations. It provides the foundation for analyzing and solving systems of linear equations. One of its primary applications is determining eigenvalues and eigenvectors, essential for simplifying matrix operations and understanding geometrical transformations.
In the context of our problem, linear algebra allows us to transform the matrix \(A\) into a form that highlights its eigenvalues and eigenvectors. By using methods such as finding the characteristic polynomial and solving the resulting equations, we can obtain the eigenvalues and corresponding eigenvectors.
The role of linear algebra extends beyond simple calculations; it includes visualizing vector spaces, understanding transformations, and solving real-world problems in fields such as engineering and computer science.

Matrix Transformation

Matrix transformations refer to the process of applying a matrix to alter or transform another vector or shape in a vector space. Essentially, matrices can be used to rotate, scale, skew, or translate objects, and understanding their properties is crucial in fields such as graphics and machine learning.
In our exercise, the matrix \(A\) transforms vectors according to the rules dictated by its eigenvectors and eigenvalues. With eigenvectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\), namely \(\begin{bmatrix} 0 \ 1 \end{bmatrix}\) and \(\begin{bmatrix} 1 \ 0 \end{bmatrix}\), the transformation can be visualized: the matrix scales vector \(\mathbf{v}_1\) by 2 and \(\mathbf{v}_2\) by -1 (these are the effects of the eigenvalues on these vectors).
Such matrix transformations illustrate how eigenvectors remain in their original span, only scaled by their corresponding eigenvalues. This property is particularly useful in understanding how linear transformations behave, as they often preserve directionality while only altering magnitude.