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

Step-by-Step Solution

Answer

Eigenvalues are \( \lambda_1 = 0 \), \( \lambda_2 = -3 \). Eigenvectors: \( \mathbf{v}_1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix} \), \( \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \). Graph lines: \( y=\frac{1}{3}x \) and \( x=0 \).

Step 1: Calculate the Characteristic Polynomial

To find the eigenvalues, start by calculating the characteristic polynomial of matrix \( A \). The characteristic polynomial is given by \( \det(A - \lambda I) \), where \( I \) is the identity matrix. For matrix \( A = \begin{bmatrix} 0 & 0 \ 1 & -3 \end{bmatrix} \), we have:\[A - \lambda I = \begin{bmatrix} 0 & 0 \ 1 & -3 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix} = \begin{bmatrix} -\lambda & 0 \ 1 & -3 - \lambda \end{bmatrix}\]The determinant is calculated as:\[\det(A - \lambda I) = (-\lambda)(-3-\lambda) - (0)(1) = \lambda(3 + \lambda)\]So, the characteristic equation is \( \lambda^2 + 3\lambda = 0 \).

Step 2: Solve for Eigenvalues

Solve the characteristic equation \( \lambda^2 + 3\lambda = 0 \) to find the eigenvalues:\[\lambda(\lambda + 3) = 0\]This equation has solutions when \( \lambda = 0 \) or \( \lambda = -3 \). Thus, the eigenvalues are \( \lambda_1 = 0 \) and \( \lambda_2 = -3 \).

Step 3: Find Eigenvectors

For each eigenvalue, substitute back into \( (A - \lambda I) \mathbf{v} = 0 \) to find the corresponding eigenvector.**Eigenvalue \( \lambda_1 = 0 \):**\[(A - 0I) \mathbf{v} = \begin{bmatrix} 0 & 0 \ 1 & -3 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = 0\]Which gives the system:\[\begin{align*}0x + 0y &= 0 \1x - 3y &= 0\end{align*}\]From \( x - 3y = 0 \), choose \( y = t \), then \( x = 3t \). Thus, \( \mathbf{v}_1 = \begin{bmatrix} 3 \ 1 \end{bmatrix} \).**Eigenvalue \( \lambda_2 = -3 \):**\[(A + 3I) \mathbf{v} = \begin{bmatrix} 3 & 0 \ 1 & 0 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = 0\]Which gives the system:\[\begin{align*}3x &= 0 \1x + 0y &= 0\end{align*}\]This implies \( x = 0 \), choose \( y = t \). Thus, \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

Step 4: Determine Equations of Lines

The lines through the origin in the direction of the eigenvectors are determined by the parameterizations of those vectors.For \( \mathbf{v}_1 = \begin{bmatrix} 3 \ 1 \end{bmatrix} \), the line is \( \begin{bmatrix} x \ y \end{bmatrix} = t \begin{bmatrix} 3 \ 1 \end{bmatrix} \), which gives the line equation \( y = \frac{1}{3}x \).For \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \), the line is\( \begin{bmatrix} x \ y \end{bmatrix} = t \begin{bmatrix} 0 \ 1 \end{bmatrix} \), which gives the line \( x = 0 \) (the y-axis).

Step 5: Compute Av1 and Av2

Calculate \( A \mathbf{v}_1 \) and \( A \mathbf{v}_2 \) to graph these vectors as well.\( A \mathbf{v}_1 = A \begin{bmatrix} 3 \ 1 \end{bmatrix} = \begin{bmatrix} 0 \ 1 \end{bmatrix} \begin{bmatrix} 3 \ 1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \)\( A \mathbf{v}_2 = A \begin{bmatrix} 0 \ 1 \end{bmatrix} = \begin{bmatrix} 0 \ 1 \end{bmatrix} \begin{bmatrix} 0 \ 1 \end{bmatrix} = \begin{bmatrix} 0 \ -3 \end{bmatrix} \)

Step 6: Graph Results

Graph the lines \( y = \frac{1}{3}x \) and the y-axis \( x = 0 \). Plot the eigenvectors \( \mathbf{v}_1 = \begin{bmatrix} 3 \ 1 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \), as well as the vectors \( A \mathbf{v}_1 = \begin{bmatrix} 0 \ 0 \end{bmatrix} \) and \( A \mathbf{v}_2 = \begin{bmatrix} 0 \ -3 \end{bmatrix} \). The eigenvectors should be represented as directions, and \( A \mathbf{v}_2 \) indicates the vector transformation.

Key Concepts

Eigenvectors

When you think about eigenvectors, consider them as special directions in the space that remain unchanged (except for scaling) under the action of a matrix. In simpler terms, if you view a matrix as a machine doing some transformation on a vector, the eigenvectors are the directions that go straight through the machine without bending.

For the exercise provided, we have two directions that are of interest: one that aligns with the vector \(\mathbf{v}_1 = \begin{bmatrix} 3 \ 1 \end{bmatrix}\) and another with \(\mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix}\). Eigenvectors like these give us valuable insight into the nature of matrix transformations.

• They tell us which directions in space are inherently meaningful under the matrix operation.
• They reveal invariant lines under transformation.

These properties make understanding and finding eigenvectors crucial when analyzing matrices, such as those representing rotations or scaling in graphical applications.

Characteristic Polynomial

The characteristic polynomial is a key tool in linear algebra that helps us determine the eigenvalues of a matrix. Essentially, it expresses a matrix as an equation, from which eigenvalues are the solutions.

In the given problem, the characteristic polynomial is derived from the equation \(\det(A - \lambda I) = 0\). Calculating this for matrix \(A = \begin{bmatrix} 0 & 0 \ 1 & -3 \end{bmatrix}\), we obtained \(\lambda^2 + 3\lambda = 0\).

The steps generally involve:

• Subtracting \(\lambda\) times the identity matrix from the matrix \(A\).
• Finding the determinant of the resulting matrix.
• Setting up and solving the equation formed.

With this approach, we find the eigenvalues (or the specific values of \(\lambda\)), which signify how much the eigenvectors are scaled during the transformation.

The characteristic polynomial hence provides a bridge between the abstract matrix form and the concrete solutions we seek in problems involving transformations. It simplifies understanding of the behavior of the matrix system, especially for those tackling such topics for the first time.

Matrix Transformation

A matrix transformation is a function that takes a vector in one space and maps it to another vector, possibly in the same space, following specific rules encoded in the matrix itself. Think of it as the process where your matrix \(A\) acts on vectors to transform them into new directions or magnitudes.

Specifically in our example, transformation is evident when \(A\) acts on its eigenvectors. Here:
- Applying \(A\) to \(\mathbf{v}_1 = \begin{bmatrix} 3 \ 1 \end{bmatrix}\), surprisingly results in the zero vector, meaning \(\mathbf{v}_1\) becomes nullified through transformation.
- Conversely, \(A\) acts on \(\mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix}\) to yield \(\begin{bmatrix} 0 \ -3 \end{bmatrix}\), essentially scaling \(\mathbf{v}_2\) by \(-3\).

The eigenvectors and their corresponding eigenvalues efficiently showcase the structure and "behavior" of matrix transformations.

• This approach allows for easier visualization of complex operations.
• It is foundational for more applied scopes like computer graphics and system dynamics where understanding inputs and outputs under matrix operations is key.

Comprehending matrix transformations through eigenvectors and values helps in simplifying complex systems and predicting their behavior efficiently.