Problem 54 - Calculus for Biology and Medicine

Question

In Problems , find the eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\) and corresponding eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) for each matrix A. Determine the equations of the lines through the origin in the direction of the eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\), and graph the lines together with the eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) and the vectors \(A \mathbf{v}_{1}\) and \(A \mathbf{v}_{2}\). $$ A=\left[\begin{array}{rr} 3 & 6 \\ -1 & -4 \end{array}\right] $$

Step-by-Step Solution

Answer

The eigenvalues are \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \). The eigenvectors are \( \mathbf{v}_1 = \begin{bmatrix} -6 \\ 1 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix} \).

Step 1: Write the Eigenvalue Equation

The eigenvalues \( \lambda \) of a matrix \( A \) are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \). For the matrix given: \[ A = \begin{bmatrix} 3 & 6 \ -1 & -4 \end{bmatrix} \] Substitute \( A \) and \( I \) (identity matrix) into the equation:\[ \det\left( \begin{bmatrix} 3-\lambda & 6 \ -1 & -4-\lambda \end{bmatrix} \right) = 0 \]

Step 2: Calculate the Determinant

Calculate the determinant of the matrix to find the eigenvalues:\[ (3-\lambda)(-4-\lambda) - (6)(-1) = 0 \] Solve for \( \lambda \):\[ (3-\lambda)(-4-\lambda) + 6 = 0 \] \[ \lambda^2 + \lambda - 6 = 0 \]

Step 3: Solve the Quadratic Equation

Factor the quadratic equation, if possible, or use the quadratic formula to find the roots:\[ \lambda^2 + \lambda - 6 = (\lambda - 2)(\lambda + 3) = 0 \] Thus, \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \).

Step 4: Find Corresponding Eigenvectors

For each eigenvalue, substitute back into the equation \( (A - \lambda I) \mathbf{v} = \mathbf{0} \) to find the associated eigenvector:**For \( \lambda_1 = 2 \):**\[ \begin{bmatrix} 1 & 6 \ -1 & -6 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \] This reduces to the system \( x + 6y = 0 \), so one eigenvector is \( \mathbf{v}_1 = \begin{bmatrix} -6 \ 1 \end{bmatrix} \).**For \( \lambda_2 = -3 \):**\[ \begin{bmatrix} 6 & 6 \ -1 & -1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \] This results in the system \( 6x + 6y = 0 \), providing the eigenvector \( \mathbf{v}_2 = \begin{bmatrix} -1 \ 1 \end{bmatrix} \).

Step 5: Determine Equations of the Lines

The equations of the lines through the origin aligned with \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are given by the direction vectors of \( \mathbf{v}_1 = \begin{bmatrix} -6 \ 1 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} -1 \ 1 \end{bmatrix} \). Hence, the equations are:- \( y = \frac{1}{6}x \) for \( \mathbf{v}_1 \)- \( y = x \) for \( \mathbf{v}_2 \).

Step 6: Graph the Lines and Vectors

To visualize, plot the lines \( y = \frac{1}{6}x \) and \( y = x \) through the origin. Plot the eigenvectors \( \mathbf{v}_1 = \begin{bmatrix} -6 \ 1 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} -1 \ 1 \end{bmatrix} \). Also compute \( A\mathbf{v}_1 \) and \( A\mathbf{v}_2 \) to show transformation: For \( A\mathbf{v}_1 \):\[ A\mathbf{v}_1 = \begin{bmatrix} 3 & 6 \ -1 & -4 \end{bmatrix} \begin{bmatrix} -6 \ 1 \end{bmatrix} = \begin{bmatrix} -18 \ 3 \end{bmatrix} \] For \( A\mathbf{v}_2 \):\[ A\mathbf{v}_2 = \begin{bmatrix} 3 & 6 \ -1 & -4 \end{bmatrix} \begin{bmatrix} -1 \ 1 \end{bmatrix} = \begin{bmatrix} 3 \ -3 \end{bmatrix} \].Graph these resulting vectors as well.

Key Concepts

Matrices

Matrices are like grids of numbers formed in rows and columns. In linear algebra, they are fundamental tools used to represent and solve systems of linear equations. Think of them as a compact way to display a collection of numbers. Each entry in the matrix is called an element. For example, in the matrix given in the exercise, \[A = \begin{bmatrix} 3 & 6 \ -1 & -4 \end{bmatrix}\]A has two rows and two columns, making it a 2 x 2 matrix. This specific size is common when working with simple transformations and eigenvectors, where you can visualize changes in two-dimensional space.

Characteristic Equation

To find eigenvalues, you must solve the characteristic equation of a matrix. This equation arises from the expression \( \det(A - \lambda I) = 0 \), where \( A \) is your matrix, \( \lambda \) represents eigenvalues, and \( I \) is the identity matrix. When computing this equation, you replace \( A \) with the matrix from your problem and \( I \) with the identity matrix of matching size:\[A = \begin{bmatrix} 3 & 6 \ -1 & -4 \end{bmatrix}, \quad I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]Then, substitute \( \lambda \) into the diagonal of \( A \), leading to:\[\det \begin{bmatrix} 3-\lambda & 6 \ -1 & -4-\lambda \end{bmatrix} = 0\]This determinant is expanded to form a quadratic equation. Solving the quadratic gives you the eigenvalues \( \lambda_1 \) and \( \lambda_2 \). These values are crucial in determining how the matrix behaves in transformations.

Linear Algebra

Linear algebra is a branch of mathematics centered around vectors and matrices and their various transformations. It holds a critical role in fields like computer science, physics, and engineering. In the problem, linear algebra helps in finding eigenvalues and eigenvectors, which are key in understanding transformations.- **Eigenvalues** tell you the magnitude of transformation for each direction given by the eigenvectors.- **Eigenvectors** represent directions that are dialed in by a matrix \( A \), only scaled by the eigenvalues after transformation.The equations related to these allow solving not just mathematical problems but optimizing real-world systems like networks and structural engineering frameworks.

Transformation and Geometry

Transformations involve changing the position, size, and orientation of vectors in space. By observing eigenvectors and eigenvalues, transformations can be visualized geometrically. In the provided example with matrix \( A \):- **Eigenvectors** show the directions in which stretching or compression occurs without rotation. For the given matrix, they were \( \begin{bmatrix} -6 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} -1 \ 1 \end{bmatrix} \).- **Eigenvalues** describe how much the eigenvectors stretch or compress. In this instance, the eigenvalues were 2 and -3.When you multiply \( A \) by each eigenvector, the product lies along the original direction scaled by the respective eigenvalue. Graphically, this visualization helps in understanding how matrices influence shapes and patterns in geometry, providing insights into both mathematical and real-world systems.