Problem 61 - Calculus for Biology and Medicine

Question

Find the eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\) for $$ A=\left[\begin{array}{ll} a & 0 \\ c & b \end{array}\right] $$

Step-by-Step Solution

Answer

The eigenvalues are \(\lambda_1 = a\) and \(\lambda_2 = b\).

Step 1: Understand the Eigenvalue Equation

The eigenvalue equation for a matrix is given by \(Av = \lambda v\), where \(A\) is the matrix, \(v\) is the eigenvector, and \(\lambda\) is the eigenvalue. To find the eigenvalues, we need to solve \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix of the same size as \(A\).

Step 2: Write the Matrix \(A - \lambda I\)

The identity matrix \(I\) for a 2x2 matrix is \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\). Calculate \(A - \lambda I\):\[A - \lambda I = \begin{bmatrix} a & 0 \ c & b \end{bmatrix} - \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix} = \begin{bmatrix} a - \lambda & 0 \ c & b - \lambda \end{bmatrix}\]

Step 3: Calculate the Determinant

The determinant of the matrix \(\begin{bmatrix} a - \lambda & 0 \ c & b - \lambda \end{bmatrix}\) is:\[\det(A - \lambda I) = (a - \lambda)(b - \lambda) - c \cdot 0 = (a - \lambda)(b - \lambda)\]

Step 4: Solve the Characteristic Equation

Now, solve the characteristic equation \(\det(A - \lambda I) = 0\):\[(a - \lambda)(b - \lambda) = 0\]This gives us the solutions \(\lambda_1 = a\) and \(\lambda_2 = b\).

Step 5: Summary of the Solution

The eigenvalues of the matrix \(A\) are \(\lambda_1 = a\) and \(\lambda_2 = b\). This result shows that the structure of the matrix yields its diagonal elements as the eigenvalues.

Key Concepts

Matrix Determinant

The determinant of a matrix is a special number that can be calculated from a square matrix. It helps in identifying whether a matrix is invertible or singular (not invertible). The determinant is essential when computing eigenvalues because it is involved in the process of solving the equation \( \det(A - \lambda I) = 0 \). Here, \( A \) is the matrix, \( \lambda \) is a scalar value (eigenvalue), and \( I \) is the identity matrix.Calculating the determinant for a 2x2 matrix like \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) involves a simple formula:

• \( ad - bc \)

The determinant basically helps in assessing the matrix's properties, such as its ability to transform vectors.

Characteristic Equation

The characteristic equation is derived from the determinant of \( A - \lambda I \) and is a key step in finding the eigenvalues of a matrix. The equation is called 'characteristic' because it uses properties intrinsic to the matrix. In the exercise, it is given by

• \( (a - \lambda)(b - \lambda) = 0 \)

This equation is quadratic in nature for a 2x2 matrix, and solving it provides the eigenvalues.The solutions to this equation give us the roots, or the scalar \( \lambda \), that maintain the relationship \( Av = \lambda v \), fulfilling the condition needed for an eigenvalue.

Eigenvectors

Eigenvectors are the vectors that, when multiplied by the matrix \( A \), only get scaled by the eigenvalue without changing direction. Once the eigenvalues are found, finding the corresponding eigenvectors involves substituting each eigenvalue into the equation \( (A - \lambda I)v = 0 \) and solving for \( v \), which is the eigenvector.For each eigenvalue, there is potentially a corresponding eigenvector or set of eigenvectors. It's important to note that these vectors are not unique and can be scaled by any non-zero scalar to obtain other valid eigenvectors.