Problem 58 - Calculus for Biology and Medicine

Question

$$ \begin{array}{l} \text { In Problems , find the eigenvalues } \lambda_{1} \text { and } \lambda_{2} \text { for each matrix }\\\ A \end{array} $$ $$ A=\left[\begin{array}{rr} -7 & 0 \\ 0 & 6 \end{array}\right] $$

Step-by-Step Solution

Answer

The eigenvalues are \(-7\) and \(6\).

Step 1: Understanding Eigenvalues

Eigenvalues are the scalars \( \lambda \) such that for a square matrix \( A \), there exists a nonzero vector \( \mathbf{v} \) where \( A\mathbf{v} = \lambda \mathbf{v} \). To find these eigenvalues, we solve \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix of the same size as \( A \).

Step 2: Set Up the Determinant Equation

For the given matrix \( A = \begin{bmatrix} -7 & 0 \ 0 & 6 \end{bmatrix} \), we subtract \( \lambda I \), where \( I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \). This gives us the matrix \( A - \lambda I = \begin{bmatrix} -7 - \lambda & 0 \ 0 & 6 - \lambda \end{bmatrix} \).

Step 3: Calculate the Determinant

The determinant of a diagonal matrix \( \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix} \) is simply \( ab \). Here, we have \( \text{det}(A - \lambda I) = (-7 - \lambda)(6 - \lambda) \).

Step 4: Solve the Characteristic Equation

We need the determinant to be zero: \((-7 - \lambda )(6 - \lambda) = 0\). Solving each factor, we find \(-7 - \lambda = 0\) and \(6 - \lambda = 0\).

Step 5: Find the Eigenvalues

From \(-7 - \lambda = 0\), we have \( \lambda = -7 \). From \( 6 - \lambda = 0 \), we have \( \lambda = 6 \). Therefore, the eigenvalues are \( \lambda_1 = -7 \) and \( \lambda_2 = 6 \).

Key Concepts

Determinant

The determinant is a special number that can be calculated from a square matrix. In the context of eigenvalues, the determinant plays a crucial role in the characteristic equation. For a matrix \( A \), the determinant of \( A - \lambda I \) must be zero to find the eigenvalues. This is always true when dealing with an eigenvalue equation.

• Determinant provides important properties about a matrix, such as whether the matrix is invertible. If the determinant is zero, the matrix does not have an inverse.
• For a \( 2 \times 2 \) matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as \( ad - bc \).
• For a diagonal matrix, calculating the determinant becomes much simpler. It is simply the product of the diagonal entries.

For the matrix given in the exercise, \( A - \lambda I \) is transformed to a diagonal form, which makes the calculation of the determinant straightforward: \((-7 - \lambda)(6 - \lambda)\). Understanding how the determinant is used will help in solving for eigenvalues efficiently.

Characteristic Equation

The characteristic equation is derived from the determinant equation of a matrix minus \( \lambda I \). It is this equation that provides us the eigenvalues of the matrix. In mathematical terms, this is expressed as \( \text{det}(A - \lambda I) = 0 \).

• The characteristic equation is polynomial, where the roots of the polynomial are the eigenvalues of the matrix.
• For the given matrix, simplifying \( \text{det}(A - \lambda I) = (-7 - \lambda)(6 - \lambda) = 0 \), reveals that the equation consists of two factors.
• Each factor \((-7 - \lambda)\) and \((6 - \lambda)\) when solved for zero gives the eigenvalues \( \lambda = -7 \) and \( \lambda = 6 \) respectively.

The beauty of the characteristic equation lies in its simplicity, especially when dealing with diagonal matrices, making the finding of eigenvalues as simple as solving basic algebraic equations.

Diagonal Matrix

A diagonal matrix is a matrix in which the entries outside the main diagonal are all zero. This form is significant because it greatly simplifies matrix operations, especially computation of determinants and eigenvalues.

• Due to its structure, the determinant of a diagonal matrix is simply the product of its diagonal elements.
• In the context of eigenvalues, if a matrix \( A \) is diagonal, the diagonal elements themselves are the eigenvalues of the matrix with corresponding eigenvectors as standard basis vectors.
• In this exercise, \( A = \begin{bmatrix} -7 & 0 \ 0 & 6 \end{bmatrix} \) is already a diagonal matrix, which implies the eigenvalues are readily apparent from the matrix as \( -7 \) and \( 6 \).

Recognizing a diagonal matrix and understanding its properties can make matrix problems much easier to tackle, saving time and reducing potential calculation errors.