Problem 63 - Calculus for Biology and Medicine

Question

(a) Show that the eigenvalues of the matrix \(A=\left[\begin{array}{ll}a & 0 \\\ 0 & c\end{array}\right]\) are \(\lambda_{1}=a\), and \(\lambda_{2}=c\). (b) Show that the corresponding eigenvectors are \(\mathbf{v}_{1}=\left[\begin{array}{l}1 \\ 0\end{array}\right]\) and \(\mathbf{v}_{2}=\left[\begin{array}{l}0 \\ 1\end{array}\right]\).

Step-by-Step Solution

Answer

The eigenvalues are \( \lambda_1 = a \) and \( \lambda_2 = c \). The eigenvectors are \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \).

Step 1: Understand the Eigenvalue Problem

To find the eigenvalues of the matrix \( A \), we need to solve the characteristic equation, which is given by \( \det(A - \lambda I) = 0 \). Here, \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues we are trying to find.

Step 2: Setup the Characteristic Equation

For the matrix \( A = \begin{bmatrix} a & 0 \ 0 & c \end{bmatrix} \), the characteristic polynomial is derived from \( A - \lambda I = \begin{bmatrix} a - \lambda & 0 \ 0 & c - \lambda \end{bmatrix} \). The determinant of this matrix is \((a-\lambda)(c-\lambda)\).

Step 3: Solve for Eigenvalues

Set the determinant from Step 2 equal to zero: \( (a-\lambda)(c-\lambda) = 0 \). Solving this equation gives two solutions: \( \lambda_1 = a \) and \( \lambda_2 = c \). These are the eigenvalues of the matrix \( A \).

Step 4: Find Eigenvectors for \( \lambda_1 = a \)

To find the eigenvector corresponding to \( \lambda_1 = a \), solve \((A - aI)\mathbf{v}_1 = 0\). Substituting \( \lambda_1 \) into \( A - \lambda I \), we get \( \begin{bmatrix} 0 & 0 \ 0 & c-a \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This simplifies to \( x_2 = 0 \), thus \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \).

Step 5: Find Eigenvectors for \( \lambda_2 = c \)

Similarly, to find the eigenvector corresponding to \( \lambda_2 = c \), solve \((A - cI)\mathbf{v}_2 = 0\). Substituting \( \lambda_2 \) into \( A - \lambda I \), we have \( \begin{bmatrix} a-c & 0 \ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This implies \( x_1 = 0 \), thus \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

Key Concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra and play a crucial role in understanding the behavior of matrices. Essentially, an eigenvalue is a scalar that tells you how much a matrix stretches or compresses vectors along a specific direction. In the context of the matrix \( A = \begin{bmatrix} a & 0 \ 0 & c \end{bmatrix} \), finding the eigenvalues means identifying the values of \( \lambda \) that satisfy the equation \( \det(A - \lambda I) = 0 \). This equation is known as the characteristic equation. For the given matrix, the eigenvalues can be directly computed to be \( \lambda_1 = a \) and \( \lambda_2 = c \). These values show that when vectors are transformed by matrix \( A \), they are scaled by factors of \( a \) and \( c \) along their respective axes.

Eigenvectors

An eigenvector is a vector that, when multiplied by a given matrix, results only in a scalar multiplication of the vector rather than a change in its direction. For the matrix \( A \), once we have the eigenvalues, the corresponding eigenvectors are found by solving the linear system \( (A - \lambda I)\mathbf{v} = \mathbf{0} \). For the eigenvalue \( \lambda_1 = a \), the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \) corresponds, and for \( \lambda_2 = c \), the eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \) corresponds. These vectors are already normalized to the standard basis vectors along the x and y axes, emphasizing that each direction is not altered but only scaled by the eigenvalue.

Characteristic Equation

The characteristic equation is essential to finding both the eigenvalues and eigenvectors of a matrix. It is formed by setting \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix and \( \lambda \) represents potential eigenvalues. For the matrix \( A = \begin{bmatrix} a & 0 \ 0 & c \end{bmatrix} \), this results in \( \det(A - \lambda I) = (a - \lambda)(c - \lambda) \). Solving \( (a - \lambda)(c - \lambda) = 0 \) yields the roots of the equation, which are the eigenvalues \( \lambda_1 = a \) and \( \lambda_2 = c \). The simplicity of the given matrix makes it straightforward to solve since it directly breaks into a product of linear terms.

Matrix Determinants

The determinant of a matrix provides important information regarding the matrix's properties, including whether the matrix is invertible and the volume scaling effect it has on space. In this exercise, the determinant also helps to derive the characteristic equation essential for finding eigenvalues. For a 2x2 matrix, the determinant is calculated as \( \det(A) = ad - bc \) for a matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \). However, in our case, since off-diagonal elements are zero, \( \det(A - \lambda I) \) simplifies to \( (a - \lambda)(c - \lambda) \). This formula reflects how the diagonal elements of the matrix influence the stretching along each principal direction, directly linking to the eigenvalues found earlier.