Problem 49 - Calculus for Biology and Medicine

Question

In Problems 49-56, 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 \(A \mathbf{v}_{2}\). $$ A=\left[\begin{array}{rr} 2 & 3 \\ 0 & -1 \end{array}\right] $$

Step-by-Step Solution

Answer

Eigenvalues: \( \lambda_1 = 2 \), \( \lambda_2 = -1 \); Eigenvectors: \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \), \( \mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \); Lines: \( y = 0 \), \( y = -x \).

Step 1: Find the Characteristic Polynomial

To find the eigenvalues, we first need the characteristic polynomial of matrix \( A \). The characteristic polynomial is given by \( \det(A - \lambda I) \), where \( I \) is the identity matrix of the same size as \( A \). For the matrix \( A = \begin{bmatrix} 2 & 3 \ 0 & -1 \end{bmatrix} \), the characteristic polynomial is formed as follows:\[\det\left(\begin{bmatrix} 2 & 3 \ 0 & -1 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix}\right) = \det\left(\begin{bmatrix} 2-\lambda & 3 \ 0 & -1-\lambda \end{bmatrix}\right)\]Calculating the determinant, we have:\[(2-\lambda)(-1-\lambda) - (3 \times 0) = \lambda^2 - \lambda - 2\]

Step 2: Solve the Characteristic Equation

Once we have the characteristic polynomial \( \lambda^2 - \lambda - 2 = 0 \), we solve it to find the eigenvalues. This can be factored as:\[(\lambda - 2)(\lambda + 1) = 0\]Thus, the eigenvalues are \( \lambda_1 = 2 \) and \( \lambda_2 = -1 \).

Step 3: Find Eigenvectors for \(\lambda_1 = 2\)

To find the eigenvector corresponding to \( \lambda_1 = 2 \), solve \((A - 2I)\mathbf{v}_1 = 0\):\[\begin{bmatrix} 0 & 3 \ 0 & -3 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]This simplifies to the condition: \(3y = 0\), thus \(y = 0\). Therefore the eigenvector is \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \).

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

To find the eigenvector corresponding to \( \lambda_2 = -1 \), solve \((A + I)\mathbf{v}_2 = 0\):\[\begin{bmatrix} 3 & 3 \ 0 & 0 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \]This simplifies to the condition: \(3x + 3y = 0\), or \(x = -y\). Therefore, the eigenvector can be expressed as \( \mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix} \).

Step 5: Equation of Lines through the Origin

The lines through the origin in the direction of eigenvectors correspond to their direction vectors. For \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \), the line is \( y = 0 \). For \( \mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix} \), the line is \( y = -x \).

Step 6: Graphing the Vectors and Lines

Draw axes and plot the eigenvectors: \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix} \). These vectors begin from the origin and point in their respective directions, \( x \)-axis and the line \( y = -x \). Then plot the lines through the origin along these directions: \( y = 0 \) and \( y = -x \). Apply the transformation \( A \) to the vectors to obtain \( A\mathbf{v}_1 = \begin{bmatrix} 2 \ 0 \end{bmatrix} \) and \( A\mathbf{v}_2 = \begin{bmatrix} -1 \ 0 \end{bmatrix} \). These transformed vectors are also plotted with their origin being the same as the original one at (0,0).

Step 7: Final Verification

Confirm that lines through eigenvectors and their image vectors \( A\mathbf{v}_1 \) and \( A\mathbf{v}_2 \) satisfy the transformation behavior of matrix \( A \). The vectors \( A\mathbf{v}_1 \) and \( A\mathbf{v}_2 \) should lie on the direction of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \), validating the correctness of the calculations.

Key Concepts

Characteristic Polynomial

To find eigenvalues of a matrix, we need to understand the concept of a characteristic polynomial. The characteristic polynomial of a matrix is a special polynomial which helps us identify its eigenvalues. Suppose we have a matrix \( A \). We form the characteristic polynomial by subtracting \( \lambda I \) from \( A \), where \( \lambda \) is a scalar and \( I \) is the identity matrix of the same dimensions as \( A \). To get the characteristic polynomial, we compute the determinant of this resulting matrix: \( \det(A - \lambda I) \). For our matrix \( A = \begin{bmatrix} 2 & 3 \ 0 & -1 \end{bmatrix} \), the polynomial becomes:

• Replace \( A \) with \( A - \lambda I \) leading to \( \begin{bmatrix} 2-\lambda & 3 \ 0 & -1-\lambda \end{bmatrix} \).
• Calculate the determinant: \((2-\lambda)(-1-\lambda) - (3 \times 0) = \lambda^2 - \lambda - 2 \).

Solving \( \lambda^2 - \lambda - 2 = 0 \) gives us the eigenvalues. These eigenvalues are critical because they tell us the scalar factors associated with the stretching or contracting happens when we transform vectors by \( A \).

Eigenvectors

Once we've determined the eigenvalues from the characteristic polynomial, the next step is to find their corresponding eigenvectors. An eigenvector is a non-zero vector that only changes by a scalar factor when a transformation is applied. To find these eigenvectors, we substitute each eigenvalue back into the expression \((A - \lambda I)\mathbf{v} = 0\) and solve for the vector \( \mathbf{v} \).

• For eigenvalue \( \lambda_1 = 2 \): \( (A - 2I)\mathbf{v}_1 = 0 \) simplifies to \( \begin{bmatrix} 0 & 3 \ 0 & -3 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \), resulting in the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \).
• For eigenvalue \( \lambda_2 = -1 \): \( (A + I)\mathbf{v}_2 = 0 \) leads to \( \begin{bmatrix} 3 & 3 \ 0 & 0 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \), giving the eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix} \).

Eigenvectors point in the specific directions that remain unchanged aside from scaling, thus enabling us to understand how the transformation reshaped the original plane.

Linear Transformation

Linear transformation refers to the operation that maps vectors from one space to another while preserving the operations of vector addition and scalar multiplication. In the context of matrices, each matrix can be seen as a linear transformation that reshapes spaces. Considering the matrix \( A \) given in the exercise, it transforms the eigenvectors and scales them by their respective eigenvalues. For instance, applying \( A \) to the eigenvectors demonstrates how they are stretched or shrunk along the lines they define:

• The transformation of \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \) by matrix \( A \) results in \( A\mathbf{v}_1 = \begin{bmatrix} 2 \ 0 \end{bmatrix} \), which is a stretching by factor 2. This shows that the transformation scales vectors on the x-axis.
• Similarly, the transformation of \( \mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix} \) results in \( A\mathbf{v}_2 = \begin{bmatrix} -1 \ 0 \end{bmatrix} \), indicating a direction change and magnitude scaling by -1 on the line \( y = -x \).

When we examine such transformations visually by plotting these vectors and their transformations, they offer an insightful view of how the whole input space is affected. This makes linear transformations a fundamental concept in fields like computer graphics and engineering.