Problem 51 - 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 \(\mathrm{Av}_{2}\) $$A=\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]$$

Step-by-Step Solution

Answer

Eigenvalues are 1 and -1; eigenvectors are \( \mathrm{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) and \( \mathrm{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \). Lines: \( y = 0 \) and \( x = 0 \).

Step 1: Find the Characteristic Equation

To find the eigenvalues of matrix \( A \), we need to solve the characteristic equation \( \det(A - \lambda I) = 0 \). For \( A = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \) and \( I \) the identity matrix, \( A - \lambda I = \begin{bmatrix} 1-\lambda & 0 \ 0 & -1-\lambda \end{bmatrix} \), so \( \det(A - \lambda I) = (1-\lambda)(-1-\lambda) \).

Step 2: Solve for Eigenvalues

Setting the determinant found in Step 1 equal to zero yields: \((1-\lambda)(-1-\lambda) = 0\). This gives us the eigenvalues: \( \lambda_1 = 1 \) and \( \lambda_2 = -1 \).

Step 3: Find Eigenvectors for \( \lambda_1 = 1 \)

Substitute \( \lambda_1 = 1 \) back into \( A - \lambda I = \begin{bmatrix} 0 & 0 \ 0 & -2 \end{bmatrix} \). Solve \( (A - 1I)x = 0 \). This results in the equation \( 0 \cdot x_1 + 0 \cdot x_2 = 0 \), which is satisfied by any vector of the form \( \begin{bmatrix} x \ 0 \end{bmatrix} \). Thus, the eigenvector is \( \mathrm{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \).

Step 4: Find Eigenvectors for \( \lambda_2 = -1 \)

Substitute \( \lambda_2 = -1 \) back into \( A - \lambda I = \begin{bmatrix} 2 & 0 \ 0 & 0 \end{bmatrix} \). Solve \( (A + I)x = 0 \). This results in the equation \( 2x_1 = 0 \), which is satisfied when \( x_1 = 0 \), and \( x_2 \) is free. The eigenvector is \( \mathrm{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

Step 5: Determine the Equations of the Lines

The lines through the origin in the direction of the eigenvectors \( \mathrm{v}_1 \) and \( \mathrm{v}_2 \) can be expressed as \( y = 0 \) and \( x = 0 \) respectively, because \( \mathrm{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \) and \( \mathrm{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

Step 6: Graph the Lines and Vectors

On a 2D coordinate system, draw the line \( y = 0 \) (horizontal line) and line \( x = 0 \) (vertical line). Plot the eigenvectors \( \mathrm{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \) and \( \mathrm{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \) starting from the origin. Plot \( A\mathrm{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \) and \( A\mathrm{v}_2 = \begin{bmatrix} 0 \ -1 \end{bmatrix} \), showing the direction and scaling effect of \( A \) on these vectors.

Key Concepts

Characteristic Equation

The characteristic equation is a crucial element in finding the eigenvalues of a matrix. It emerges from an important concept in linear algebra related to how a matrix interacts with its eigenvectors. An eigenvalue \( \lambda \) informs us how much a specific vector (the eigenvector) is "stretched" or "squeezed" when the matrix acts on it. To compute these values, we need to solve the characteristic equation: \( \text{det}(A - \lambda I) = 0 \).

In this equation:

• \( A \) stands for the matrix in question.
• \( \lambda \) is the eigenvalue we are trying to find.
• \( I \) is the identity matrix of the same size as \( A \).

The process involves subtracting \( \lambda I \) from \( A \), finding the determinant of the resulting matrix, and setting it to zero. By doing this, we derive a polynomial where the solutions are the eigenvalues.

While it may sound overwhelming, think of the characteristic equation as a formula that reveals how the matrix alters vectors. This alteration is quantified by the eigenvalues.

Matrix Algebra

Matrix algebra is not just about adding or multiplying matrices but also understanding their properties and effects on space. Matrices are fundamental components in linear transformations, and they help us express complex relationships simply and elegantly.

When we explore eigenvalues and eigenvectors, we dive deep into a fundamental aspect of matrix algebra. In this context, matrices represent linear transformations that scale vectors.

• An eigenvector of a matrix is a non-zero vector that changes only in scale when that matrix is applied to it.
• Its corresponding eigenvalue tells us the factor of this scaling.

Given matrix \( A = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \), this matrix performs a scaling effect—specifically stretching and compressing along particular directions denoted by the eigenvectors. For instance, an eigenvalue of 1 means that the vector keeps its original length, while an eigenvalue of -1 signifies a directional flip or reflection.

By mastering matrix algebra, we can predict how transformations impact points in-space, enhancing our ability to solve practical problems and contribute to fields like computer graphics, data analysis, and more.

Linear Transformations

Linear transformations are the key actions that matrices represent. They map vectors from one place to another while preserving vector addition and scalar multiplication. In simpler terms, these transformations shift and scale vectors in a predictable, uniform way.

The core idea is that a matrix can be seen as an operation that transforms every vector in its domain through a systematic rule. Understanding the underlying mechanics involves analyzing eigenvectors and eigenvalues because:

• Eigenvectors define the directions that remain constant through this transformation, although they may be stretched or compressed.
• Eigenvalues tell us how much these vectors are scaled, giving insight into the transformation's nature.

Consider our example with matrix \( A = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \). This matrix performs a linear transformation by reflecting vectors across the x-axis. It keeps vectors lying on the x-axis unchanged and flips those on the y-axis. This understanding helps illustrate the symmetrical action of matrix transformations.

Linear transformations offer profound insights into geometry and can be applied in various computational fields, allowing for transformations of graphics, data manipulation, and beyond. By mastering these concepts, we gain powerful tools to interpret and shape the digital world.