Problem 60 - Calculus for Biology and Medicine

Question

Find the eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\) for each matrix \(A\). $$A=\left[\begin{array}{rr}-1 & 4 \\ 0 & -2\end{array}\right]$$

Step-by-Step Solution

Answer

The eigenvalues are \(\lambda_1 = -1\) and \(\lambda_2 = -2\).

Step 1: Identify the Eigenvalue Equation

To find the eigenvalues of a matrix \(A\), we start with the eigenvalue equation \(A\mathbf{v} = \lambda \mathbf{v}\), where \(\mathbf{v}\) is an eigenvector and \(\lambda\) is an eigenvalue. This can be rewritten as \((A - \lambda I)\mathbf{v} = \mathbf{0}\), where \(I\) is the identity matrix of the same size as \(A\).

Step 2: Construct the Characteristic Equation

The eigenvalues \(\lambda\) are the solutions to the characteristic equation, given by \(\det(A - \lambda I) = 0\). For the provided matrix \(A = \left[\begin{array}{rr}-1 & 4 \ 0 & -2\end{array}\right]\), and identity matrix \(I = \left[\begin{array}{rr}1 & 0 \ 0 & 1\end{array}\right]\), we have:\[(A - \lambda I) = \left[\begin{array}{rr}-1 - \lambda & 4 \ 0 & -2 - \lambda\end{array}\right]\].

Step 3: Compute the Determinant

Calculate the determinant of \((A - \lambda I)\):\[\det\left(\begin{array}{cc}-1 - \lambda & 4 \0 & -2 - \lambda\end{array}\right) = (-1 - \lambda)(-2 - \lambda) - (4 \cdot 0) = (-1 - \lambda)(-2 - \lambda)\]Which simplifies to:\[\det(A - \lambda I) = (\lambda + 1)(\lambda + 2)\]

Step 4: Solve the Characteristic Equation

Set the determinant equal to zero to find the eigenvalues:\[(\lambda + 1)(\lambda + 2) = 0\]The solutions to this equation are \(\lambda + 1 = 0\) and \(\lambda + 2 = 0\).

Step 5: Solve for the Eigenvalues

Solving the equations \(\lambda + 1 = 0\) and \(\lambda + 2 = 0\) gives the eigenvalues:\[\lambda_1 = -1, \quad \lambda_2 = -2\]

Key Concepts

Characteristic Equation

When dealing with matrices, a key step in identifying eigenvalues is formulating the characteristic equation. This equation provides solutions that are the eigenvalues of a matrix. It is derived from rearranging the eigenvalue equation. Given a matrix \( A \) and its identity matrix \( I \), you would express it as \( \det(A - \lambda I) = 0 \). Here, \( \lambda \) denotes the eigenvalues you are trying to find.This equation essentially functions as a polynomial whose roots are the eigenvalues. Solving this polynomial is essential in determining the nature and number of solutions. For instance, a quadratic polynomial like \( (\lambda + 1)(\lambda + 2) = 0 \) provides two real solutions, equating to potential eigenvalues of the matrix. This might seem complex initially, but understanding this step-by-step helps break down the larger picture of matrix analysis. Using the characteristic equation, solutions to the matrix question are unveiled, leading to understanding the behavior of linear transformations.
Remember:

• The characteristic equation provides a bridge between matrix algebra and polynomial roots.
• Solving \( \det(A - \lambda I) = 0 \) yields the eigenvalues.

Determinant

The determinant is a central component in solving for eigenvalues. It reflects specific properties of the matrix, helping to analyze solutions and transformations. When we have the equation \( \det(A - \lambda I) = 0 \), we're using the determinant to help solve for \( \lambda \), the eigenvalues.Now, what exactly is a determinant? It’s a scalar value that can be computed from the elements of a square matrix, encapsulating essential information about the matrix. You can think of it as a tool that tests whether a matrix is invertible or if linear transformations apply to it effectively. In 2x2 matrices, it's calculated with this basic formula: \( ad - bc \), where \( a, b, c, \) and \( d \) are the elements of the matrix.For our exercise, the determinant of \((A - \lambda I)\) simplifies into a product of terms like \( (\lambda + 1) \) and \( (\lambda + 2) \). This polynomial of \( \lambda \) sets up the characteristic equation. The process reveals how apparent traits in the matrix, like its rank and transformations, behave when subjected to linear operations. It's fascinating how determinants unfold the entire potential of a matrix, aiding in decoding eigenvalues.

Eigenvector

Eigenvectors offer vital insight into the influence a matrix has in vector space. An eigenvector of a matrix \( A \) coincides with an eigenvalue and carries substantial geometric meaning. When you have \( A\mathbf{v} = \lambda \mathbf{v} \), \( \mathbf{v} \) is the eigenvector corresponding to the eigenvalue \( \lambda \). In essence, eigenvectors are special vectors whose direction remains unchanged when subjected to the transformation represented by \( A \). This makes them crucial in understanding how matrices scale and stretch within vector space. Essentially, determining whether a certain linear transformation has fixed-dimension scaling factors directly relates to its eigenvectors.To find an eigenvector, after deriving the eigenvalues, use them to solve \((A - \lambda I)\mathbf{v} = \mathbf{0}\). Here, you're seeking nonzero solutions \( \mathbf{v} \) that satisfy this equation. These solutions give you the eigenvectors corresponding to each eigenvalue previously determined. Understanding eigenvectors is fundamental:

• They determine axes along which linear transformations act without changing direction.
• They help decompose complex transformations into understandable components.

Grasping the concept of eigenvectors empowers you to analyze systems of equations and matrix behavior in detailed and meaningful ways.