Problem 50 - 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 \(\mathrm{Av}_{2}\) $$A=\left[\begin{array}{rr}0 & 0 \\ 1 & -3\end{array}\right]$$

Step-by-Step Solution

Answer

Eigenvalues: \( \lambda_1 = 0 \), \( \lambda_2 = -3 \). Eigenvectors: \( \mathbf{v}_1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix} \), \( \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \).

Step 1: Set up the Characteristic Equation

The eigenvalues of a matrix \( A \) are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. For matrix \( A = \begin{bmatrix} 0 & 0 \ 1 & -3 \end{bmatrix} \), we set up the equation \( \begin{vmatrix} 0 - \lambda & 0 \ 1 & -3 - \lambda \end{vmatrix} = 0 \).

Step 2: Calculate the Determinant

Calculate the determinant of the matrix \( A - \lambda I \) which is \( (0 - \lambda)(-3 - \lambda) - (0 \cdot 1) \). Simplifying, this gives \( \lambda(\lambda + 3) = 0 \).

Step 3: Solve the Characteristic Equation

Solve \( \lambda(\lambda + 3) = 0 \) for \( \lambda \). The solutions are \( \lambda = 0 \) and \( \lambda = -3 \). Thus, the eigenvalues are \( \lambda_1 = 0 \) and \( \lambda_2 = -3 \).

Step 4: Find Eigenvectors for \( \lambda_1 = 0 \)

Substitute \( \lambda = 0 \) back into \( A - \lambda I \) to find the eigenvector. This gives the system \( 0\cdot x + 0 \cdot y = 0 \) and \( 1 \cdot x - 3 \cdot y = 0 \). Solving \( x = 3y \), we choose \( y = 1 \), hence \( \mathbf{v}_1 = \begin{bmatrix} 3 \ 1 \end{bmatrix} \).

Step 5: Find Eigenvectors for \( \lambda_2 = -3 \)

Substitute \( \lambda = -3 \) back into \( A - \lambda I \), leading to \( 0\cdot x + 0 \cdot y = 0 \) and \( 1\cdot x + 0 \cdot y = 0 \). This simplifies to \( x = 0 \), giving \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

Step 6: Equation of Lines Through the Origin

The lines through the origin in the direction of the eigenvectors are given by \( y = \frac{1}{3}x \) for \( \mathbf{v}_1 \) and \( x = 0 \) for \( \mathbf{v}_2 \).

Step 7: Graphical Representation

Graph the lines \( y = \frac{1}{3}x \) and \( x = 0 \). Include the eigenvectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) by plotting \( (3, 1) \) and \( (0, 1) \). Calculate \( A\mathbf{v}_1 = \begin{bmatrix} 0 \cdot 3 + 0 \cdot 1 \ 1 \cdot 3 - 3 \cdot 1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \) and \( A\mathbf{v}_2 = \begin{bmatrix} 0 \ -3 \end{bmatrix} \), plotting these points in the graph.

Key Concepts

Characteristic Equation

When you want to find the eigenvalues of a matrix, the key starting point is the characteristic equation. This equation is about solving for lambda (\( \lambda \)), which represents the eigenvalues of the matrix. Mathematically, we set up the equation \( \det(A - \lambda I) = 0 \). Here, \( A \) is your matrix, \( \lambda \) is our unknown eigenvalue, and \( I \) is the identity matrix of the same size as \( A \). This equation originates from the fact that eigenvectors \( \mathbf{v} \) satisfy \( A\mathbf{v} = \lambda\mathbf{v} \), showing that matrix \( A \) transforms vector \( \mathbf{v} \) by merely scaling it by \( \lambda \).
In our example with matrix \( A = \begin{bmatrix} 0 & 0 \ 1 & -3 \end{bmatrix} \), we subtract \( \lambda \) from the diagonal elements of \( A \) and find the determinant, resulting in the equation \( \lambda(\lambda + 3) = 0 \). Solving this characteristic equation gives the eigenvalues \( \lambda_1 = 0 \) and \( \lambda_2 = -3 \). These values tell us how vectors along certain directions are stretched or squished when multiplied by the matrix \( A \).

Matrix Algebra

Matrix algebra is a fundamental tool in linear algebra, which involves doing calculations with matrices, like additions, multiplications, and finding determinants. To work with matrices, it's essential to understand how matrix operations correlate with solving systems of equations and transformations.
For eigenvalues and eigenvectors, the process involves matrix operations to find \( (A - \lambda I) \), where we subtract a scalar \( \lambda \) from the diagonal elements of matrix \( A \). Once we've set up this new matrix, we calculate its determinant. The determinant is a special number that can tell us information about the matrix, such as whether it's invertible or determining its eigenvalues.
In the provided problem, matrix calculations allowed us to derive the system of equations to solve for eigenvectors. When \( \lambda = 0 \), the matrix became \( \begin{bmatrix} 0 & 0 \ 1 & -3 \end{bmatrix} \), leading to the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 3 \ 1 \end{bmatrix} \). For \( \lambda = -3 \), the matrix was \( \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix} \), giving the eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \). The ability to carry out these operations confidently is what allows us to solve similar problems efficiently.

Graphical Representation of Vectors

A graphical representation of vectors helps in easily visualizing the behavior of eigenvectors and eigenvalues. This representation is crucial in understanding concepts like direction and scaling. Vectors are typically represented as arrows pointing in a particular direction, starting from the origin. The length of the arrow shows the magnitude (or length) of the vector.
For eigenvectors \( \mathbf{v}_1 = \begin{bmatrix} 3 \ 1 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \), we can graphically draw them as arrows. \( \mathbf{v}_1 \) points along the line \( y = \frac{1}{3}x \), and \( \mathbf{v}_2 \) directly upwards along the y-axis. These vectors not only indicate directions but their scaling by \( A \) reflects how much they are altered by the transformation.
Graphing also involves plotting the lines corresponding to the eigenvectors and their transformations by matrix \( A \). In this case, lines are represented by \( y = \frac{1}{3}x \) for \( \mathbf{v}_1 \) and \( x = 0 \) for \( \mathbf{v}_2 \). The transformation of vectors is visualized with points like \( A\mathbf{v}_1 = \begin{bmatrix} 0 \ 0 \end{bmatrix} \) and \( A\mathbf{v}_2 = \begin{bmatrix} 0 \ -3 \end{bmatrix} \), which express how the original vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) react to the transformation described by matrix \( A \). This visualization strengthens the understanding of the algebraic solutions.