Problem 51 - 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 \(A \mathbf{v}_{2}\). $$ A=\left[\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right] $$

Step-by-Step Solution

Answer

Eigenvalues are 1 and -1. Lines are x-axis for \( \mathbf{v}_1 \) and y-axis for \( \mathbf{v}_2 \).

Step 1: Write the Characteristic Equation

Start by finding the characteristic equation of the matrix \( A \). The characteristic equation is given by \( ext{det}(A - \lambda I) = 0 \). For our matrix \( A = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \), this becomes:\[\text{det}\begin{pmatrix} 1-\lambda & 0 \ 0 & -1-\lambda \end{pmatrix} = 0.\]

Step 2: Simplify the Determinant

Calculate the determinant of the matrix \( A - \lambda I \):\[(1-\lambda)(-1-\lambda) = 0.\]

Step 3: Find the Eigenvalues

Solve the equation from Step 2 for \( \lambda \):\[(1 - \lambda)(-1 - \lambda) = 0\]This results in two solutions:\[\lambda_1 = 1, \, \lambda_2 = -1.\]Thus, the eigenvalues are \( \lambda_1 = 1 \) and \( \lambda_2 = -1 \).

Step 4: Find Eigenvectors

For each eigenvalue, solve \( (A - \lambda I)\mathbf{v} = 0 \) to find eigenvectors.- For \( \lambda_1 = 1 \):\[(A - I)\mathbf{v} = \begin{pmatrix} 0 & 0 \ 0 & -2 \end{pmatrix}\begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\]The solution is any vector of the form \( \mathbf{v}_1 = \begin{pmatrix} x \ 0 \end{pmatrix} \).- For \( \lambda_2 = -1 \):\[(A + I)\mathbf{v} = \begin{pmatrix} 2 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\]The solution is any vector of the form \( \mathbf{v}_2 = \begin{pmatrix} 0 \ y \end{pmatrix} \).

Step 5: Determine Equations of the Lines

Lines through the origin in the direction of the eigenvectors:- For \( \mathbf{v}_1 \): Line is along the x-axis, equation: \( y = 0 \).- For \( \mathbf{v}_2 \): Line is along the y-axis, equation: \( x = 0 \).

Step 6: Graphical Representation

Graphically, plot the lines, the vectors \( \mathbf{v}_1 \), \( \mathbf{v}_2 \), and the vectors resulting from the transformation \( A\mathbf{v}_1 \) and \( A\mathbf{v}_2 \).- The line for \( \mathbf{v}_1 \) is horizontal on the x-axis.- The line for \( \mathbf{v}_2 \) is vertical on the y-axis.- \( A\mathbf{v}_1 = \mathbf{v}_1 \) (since \( \lambda_1 = 1 \) does not change \( \mathbf{v}_1 \)).- \( A\mathbf{v}_2 = -\mathbf{v}_2 \) (since \( \lambda_2 = -1 \) reflects \( \mathbf{v}_2 \) across the origin).

Key Concepts

Characteristic Equation

The characteristic equation is a fundamental part of finding eigenvalues of a matrix. It acts as a bridge from the matrix to its eigenvalues. When you have a square matrix \( A \), you can determine its eigenvalues by solving an equation based on its determinant, known as the characteristic equation. This equation is expressed as \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix of the same size as \( A \), and \( \lambda \) represents the eigenvalues we seek. The formula suggests that we are looking for values of \( \lambda \) which, when subtracted from the diagonal entries of \( A \), make the matrix \( A - \lambda I \) singular (its determinant is zero).
This equation helps in solving for \( \lambda \) because a matrix is singular if and only if its determinant is zero. Each solution \( \lambda \) to this equation is an eigenvalue of the matrix \( A \). In simple terms, the characteristic equation allows us to find the special numbers (eigenvalues) that reveal important properties about the matrix, such as scaling effects and directions of stretching.

Determinant

The determinant of a matrix provides valuable insights into many mathematical properties of the matrix, including its invertibility and effect on space transformations. Put simply, it's a scalar value that tells us about the scaling factor applied by the matrix transformation. To find the determinant of a matrix \( A \), you calculate a special sum of products of the entries of the matrix. For a 2x2 matrix like \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as \( \text{det}(A) = ad - bc \).
In the context of eigenvalues, the determinant of \( (A - \lambda I) \) is especially crucial. When this determinant is zero, it means that \( A - \lambda I \) is non-invertible, making it a key part of solving the characteristic equation. The determinant highlights when a matrix transformation will collapse space to lower dimensions, as it becomes zero if any amount of scaling or collapsing happens. Understanding determinants is crucial when working with matrices, especially for transformations in linear algebra.

Matrix Transformation

Matrix transformations are powerful operations that allow us to move, stretch, rotate, or otherwise transform vectors in space. When you apply a matrix transformation \( A \) to a vector \( \mathbf{v} \), it results in a new vector \( A\mathbf{v} \). This transformation can depict multiple geometric alterations depending on the entries of \( A \).
Eigenvalues and eigenvectors play a significant role in understanding these transformations. An eigenvalue shows how a transformation affects the length of an eigenvector, while the direction of the eigenvector remains unchanged. For instance, with an eigenvalue of 1, the matrix doesn't alter the eigenvector's length or direction, while a negative eigenvalue flips the direction.
Analyzing how a matrix transforms its eigenvectors provides insight into the essence of the transformation. It helps visualize transformations like reflections, rotations, or scalings represented by the matrix. By understanding matrix transformations, you gain a better understanding of how vector spaces can be manipulated and how they evolve under certain operations.