Problem 54 - 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 \(\mathrm{Av}_{2}\) $$A=\left[\begin{array}{rr}-1 & 0 \\ 4 & 3\end{array}\right]$$

Step-by-Step Solution

Answer

Eigenvalues are \( \lambda_1=3 \) and \( \lambda_2=-1 \). Eigenvectors are \( \mathbf{v}_1=\begin{bmatrix} 0 \\ 1 \end{bmatrix} \) and \( \mathbf{v}_2=\begin{bmatrix} 1 \\ 0 \end{bmatrix} \). Lines are y-axis and x-axis respectively.

Step 1: Find the Characteristic Polynomial

To find the eigenvalues, we first need to find the characteristic polynomial of the matrix \( A \). The characteristic polynomial \( p(\lambda) \) is given by \( \det(A - \lambda I) \), where \( I \) is the identity matrix. For \( A = \begin{bmatrix} -1 & 0 \ 4 & 3 \end{bmatrix} \), we form \( A - \lambda I = \begin{bmatrix} -1 - \lambda & 0 \ 4 & 3 - \lambda \end{bmatrix} \). The determinant is \(((-1 - \lambda)(3 - \lambda) - 0 \cdot 4) = \lambda^2 - 2\lambda - 3\).

Step 2: Solve the Characteristic Equation

Next, solve the characteristic equation \( \lambda^2 - 2\lambda - 3 = 0 \) for \( \lambda \). Factoring gives \((\lambda - 3)(\lambda + 1) = 0\). Therefore, the eigenvalues are \( \lambda_1 = 3 \) and \( \lambda_2 = -1 \).

Step 3: Find Eigenvector for \( \lambda_1 = 3 \)

Substitute \( \lambda = 3 \) back into the equation \( (A - \lambda I)\mathbf{v} = 0 \). This gives \( \begin{bmatrix} -1-3 & 0 \ 4 & 3-3 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \), simplifying to \( -4x = 0 \) and \( 4x = 0 \), which implies \( x = 0 \) and \( y \) is free. An eigenvector is \( \mathbf{v}_1 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

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

Now substitute \( \lambda = -1 \) into \( (A - \lambda I)\mathbf{v} = 0 \), resulting in \( \begin{bmatrix} 0 & 0 \ 4 & 4 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \), simplifying to \( 4y = 0 \), so \( y = 0 \) and \( x \) is free. An eigenvector is \( \mathbf{v}_2 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \).

Step 5: Determine Equations of Lines through the Origin

The equation of the line in the direction of \( \mathbf{v}_1 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \) is \( x = 0 \), which is the y-axis. For \( \mathbf{v}_2 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \), the line is \( y = 0 \), which is the x-axis.

Step 6: Graph Vectors and Lines

On a coordinate plane, plot the eigenvectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) along with lines \( x = 0 \) and \( y = 0 \). Compute \( A\mathbf{v}_1 = \begin{bmatrix} 0 \ 4 \end{bmatrix} \) and \( A\mathbf{v}_2 = \begin{bmatrix} -1 \ 4 \end{bmatrix} \) and plot these vectors, which illustrate the effect of matrix \( A \) on its eigenvectors.

Key Concepts

Matrix Algebra

Matrix algebra is a branch of mathematics that deals with matrices and their manipulation. It's like working with numbers, but instead, you're working with collections of numbers organized in rows and columns. In our problem, we're given a 2x2 matrix \( A = \begin{bmatrix} -1 & 0 \ 4 & 3 \end{bmatrix} \). To find eigenvalues and eigenvectors, we need to perform operations using this matrix.

In matrix algebra, certain operations are crucial, such as:

• Matrix addition and subtraction: Combine or compare the elements of matrices of the same dimension.
• Scalar multiplication: Multiply each element of a matrix by a scalar (a single number).
• Matrix multiplication: Multiply matrices to combine transformations they represent.
• Determinant calculation: Helps in solving systems of linear equations, finding eigenvalues, and checking matrix inversion.

For the given matrix, we use these operations to find its characteristic polynomial, which then helps in determining the eigenvalues.

Characteristic Polynomial

The characteristic polynomial is a key concept in finding eigenvalues of a matrix. It's the polynomial obtained from the determinant of the matrix \( A \) subtracted by \( \lambda \) times the identity matrix. This concept helps ascertain the eigenvalues by solving the polynomial equation derived from it.

For our matrix \( A = \begin{bmatrix} -1 & 0 \ 4 & 3 \end{bmatrix} \), the identity matrix is \( I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \). We subtract \( \lambda I \) from \( A \), giving:
\( A - \lambda I = \begin{bmatrix} -1-\lambda & 0 \ 4 & 3-\lambda \end{bmatrix} \).

The characteristic polynomial can be found by calculating the determinant:
\( \det(A - \lambda I) \).

Calculating this determinant gives:
\[ (-1 - \lambda)(3 - \lambda) - 0 \cdot 4 = \lambda^2 - 2\lambda - 3 \].

Solving the characteristic equation \( \lambda^2 - 2\lambda - 3 = 0 \) by factoring, we get the roots (eigenvalues): \( \lambda_1 = 3 \) and \( \lambda_2 = -1 \). This process reveals the eigenvalues and sets the stage for finding the eigenvectors.

Graphing Vectors

Graphing vectors is a visual representation that helps understand linear transformations and the action of matrices on vectors.

When dealing with eigenvectors and eigenvalues, plotting them helps observe how transformations either stretch or shrink vectors along certain lines called **eigenlines**. For the eigenvectors found, \( \mathbf{v}_1 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \):

• \( \mathbf{v}_1 \) lies along the y-axis, which corresponds to the line \( x = 0 \).
• \( \mathbf{v}_2 \) lies along the x-axis, which corresponds to the line \( y = 0 \).

Once we calculate \( A\mathbf{v}_1 = \begin{bmatrix} 0 \ 4 \end{bmatrix} \) and \( A\mathbf{v}_2 = \begin{bmatrix} -1 \ 4 \end{bmatrix} \), these vectors show the transformed positions after the matrix \( A \) acts on the eigenvectors. Plotting these vectors provides a dynamic view of the actions of matrix transformations in a two-dimensional plane. This visualization aids in comprehending the nature of eigenvectors as directions that stay invariant in span, only changing in magnitude by the corresponding eigenvalue.