Problem 53 - 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 \(\mathrm{Av}_{2}\) $$A=\left[\begin{array}{rr}-1 & 2 \\ 4 & 1\end{array}\right]$$

Step-by-Step Solution

Answer

Eigenvalues: \( \lambda_1 = \frac{-1 + \sqrt{41}}{2} \) and \( \lambda_2 = \frac{-1 - \sqrt{41}}{2} \); Eigenvectors: \( \mathrm{v}_1 = \begin{bmatrix} 2 \\ 1 + \lambda_1 \end{bmatrix} \) and \( \mathrm{v}_2 = \begin{bmatrix} 2 \\ 1 + \lambda_2 \end{bmatrix} \).

Step 1: Finding the Characteristic Polynomial

To find the eigenvalues \( \lambda \), we need to compute the characteristic polynomial \( \text{det}(A - \lambda I) \). The matrix \( A \) is \[ \begin{bmatrix} -1 & 2 \ 4 & 1 \end{bmatrix} \]. The identity matrix \( I \) is \[ \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]. Thus, \( A - \lambda I \) becomes \[ \begin{bmatrix} -1 - \lambda & 2 \ 4 & 1 - \lambda \end{bmatrix} \]. The determinant is \((-1 - \lambda)(1 - \lambda) - 8 \).

Step 2: Solving the Characteristic Equation

Expand the determinant: \((-1 - \lambda)(1 - \lambda) - 8 = \lambda^2 + \lambda - 2 - 8 = \lambda^2 + \lambda - 10 \). Set the characteristic polynomial equal to zero: \( \lambda^2 + \lambda - 10 = 0 \). Solve this quadratic equation using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=1, b=1, \) and \( c=-10 \).

Step 3: Calculating Eigenvalues

Plugging into the quadratic formula, we have: \( \lambda = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 40}}{2} = \frac{-1 \pm \sqrt{41}}{2} \). The eigenvalues are \( \lambda_1 = \frac{-1 + \sqrt{41}}{2} \) and \( \lambda_2 = \frac{-1 - \sqrt{41}}{2} \).

Step 4: Finding Eigenvectors for \(\lambda_1\)

To find the eigenvector corresponding to \( \lambda_1 = \frac{-1 + \sqrt{41}}{2} \), solve \( (A - \lambda_1 I)x = 0 \): \[ \begin{bmatrix} -1 - \lambda_1 & 2 \ 4 & 1 - \lambda_1 \end{bmatrix} \cdot \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \]. Solving the system provides us with the eigenvector \( \mathrm{v}_1 = \begin{bmatrix} 2 \ 1 + \lambda_1 \end{bmatrix} \).

Step 5: Finding Eigenvectors for \(\lambda_2\)

Similarly, for \( \lambda_2 = \frac{-1 - \sqrt{41}}{2} \), solve \( (A - \lambda_2 I)x = 0 \): \[ \begin{bmatrix} -1 - \lambda_2 & 2 \ 4 & 1 - \lambda_2 \end{bmatrix} \cdot \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \]. Solving this system provides us with the eigenvector \( \mathrm{v}_2 = \begin{bmatrix} 2 \ 1 + \lambda_2 \end{bmatrix} \).

Step 6: Equations of Lines through the Origin

The equation of a line through the origin in the direction of an eigenvector \( \mathrm{v}_1 = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \) is given by the parametric form \( \begin{bmatrix} x \ y \end{bmatrix} = t \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \). For \( \mathrm{v}_1 = \begin{bmatrix} 2 \ 1 + \lambda_1 \end{bmatrix} \), the line is \( y = \frac{1 + \lambda_1}{2}x \). Similarly, for \( \mathrm{v}_2 = \begin{bmatrix} 2 \ 1 + \lambda_2 \end{bmatrix} \), the line is \( y = \frac{1 + \lambda_2}{2}x \).

Step 7: Graphing Eigenvectors and Resultant Vectors

To graph, plot the eigenvectors \( \mathrm{v}_1 = \begin{bmatrix} 2 \ 1 + \lambda_1 \end{bmatrix} \) and \( \mathrm{v}_2 = \begin{bmatrix} 2 \ 1 + \lambda_2 \end{bmatrix} \) from the origin. Then, compute \( A\mathrm{v}_1 \) and \( A\mathrm{v}_2 \) by multiplying \( A \) with each eigenvector. Plot \( A\mathrm{v}_1 \) and \( A\mathrm{v}_2 \) on the same graph showing their directions along the respective lines discovered in the previous step.

Key Concepts

Characteristic Polynomial

In linear algebra, finding the eigenvalues of a matrix begins with the characteristic polynomial. This polynomial is derived from the matrix minus a lambda-scaled identity matrix. Essentially, it helps us find values (eigenvalues) by which the matrix can be multiplied to retain its linear direction.

For a matrix \( A \), the characteristic polynomial is found by calculating \( \text{det}(A - \lambda I) \), where \( \lambda \) is a scalar and \( I \) is the identity matrix of the same order as \( A \). The determinant equation equates to a polynomial, which, when set to zero, provides the eigenvalues of the matrix. Each eigenvalue represents a specific scaling factor of the matrix.

In our example, the matrix \( A \) provided is \[ \begin{bmatrix} -1 & 2 \ 4 & 1 \end{bmatrix} \]. To compute \( A - \lambda I \), we subtract \( \lambda \) times the identity matrix, \[ I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \] from \( A \), giving us \[ \begin{bmatrix} -1-\lambda & 2 \ 4 & 1-\lambda \end{bmatrix} \]. The determinant \( \,|A - \lambda I| = (-1-\lambda)(1-\lambda) - 8 \) simplifies to form the quadratic polynomial \( \lambda^2 + \lambda - 10 \).

Quadratic Formula

Once you have the characteristic polynomial, the next step is solving it to find the eigenvalues. Our characteristic polynomial \( \lambda^2 + \lambda - 10 = 0 \) is a quadratic equation. The quadratic formula is a handy tool for this.

The formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) allows us to find the roots of quadratic equations. Here, \( a = 1 \), \( b = 1 \), and \( c = -10 \). Substituting these into the formula yields \[ \lambda = \frac{-1 \pm \sqrt{1 + 40}}{2} = \frac{-1 \pm \sqrt{41}}{2} \].

The expression inside the square root, \( b^2 - 4ac \), is called the discriminant. For positive discriminants, the formula gives two real distinct solutions, marking the two distinct eigenvalues in this case, \( \lambda_1 = \frac{-1 + \sqrt{41}}{2} \) and \( \lambda_2 = \frac{-1 - \sqrt{41}}{2} \). Each eigenvalue results in unique eigenvectors.

Parametric Equation of a Line

To visualize vectors, we use the parametric equation of a line. This is crucial in linear algebra to understand how a vector scales and moves in space.

The parametric equation allows us to express a line in terms of a direction vector and a parameter, often \( t \), like \( \begin{bmatrix} x \ y \end{bmatrix} = t \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \). This setup describes every point on the line through the origin and simplifies many linear algebra calculations.

For eigenvectors \( \mathrm{v}_1 \) and \( \mathrm{v}_2 \), these represent lines in the direction they point. For example, if \( \mathrm{v}_1 = \begin{bmatrix} 2 \ 1 + \lambda_1 \end{bmatrix} \), then the line equation becomes \( y = \frac{1 + \lambda_1}{2}x \). Similarly, for \( \mathrm{v}_2 = \begin{bmatrix} 2 \ 1 + \lambda_2 \end{bmatrix} \), the corresponding line is \( y = \frac{1 + \lambda_2}{2}x \). These form the basis for exploring eigenvectors and their graphic representation.

Linear Algebra

Linear algebra is a fundamental branch of mathematics that deals with vectors, matrices, and linear transformations. One of its powerful tools is eigenvectors and eigenvalues, which reveal much about the underlying geometry of systems.

This area focuses on vector spaces and linear mappings between those spaces. The study involves various operations, such as matrix addition, scalar multiplication, and particularly, matrix multiplication, which is crucial for finding eigenvectors and eigenvalues.

Solutions in linear algebra often emerge from equations like \( A\mathbf{v} = \lambda\mathbf{v} \), where \( A \) is a matrix, \( \mathbf{v} \) is an eigenvector, and \( \lambda \) is the eigenvalue. Understanding these can help solve systems of linear equations, analyze dynamic systems, and even in simplifying complex vector spaces.

Matrix Multiplication

Matrix multiplication is at the heart of many calculations in linear algebra. It's how we apply transformations and find results like eigenvectors in systems represented by matrices.

Each element of the product matrix is found by taking the dot product of rows and columns, helping us transition from one linear system to another. This operation respects the linearity of transformations, preserving the vector space's structure.

For example, multiplying the matrix \( A \) by an eigenvector \( \mathrm{v} \) yields another vector that is a scaled version of \( \mathrm{v} \), or \( \lambda \mathrm{v} \). This property is essential when checking eigenvectors or forming resultant vectors like \( A\mathrm{v}_1 \) and \( A\mathrm{v}_2 \). These transformations can be visualized graphically, showing how matrices rotate, scale, or shear spatial vectors.