Question

Find the eigenvalues of the given matrix. For each eigenvalue, give an eigenvector. $$\left[\begin{array}{ccc}-1 & 18 & 0 \\ 1 & 2 & 0 \\ 5 & -3 & -1\end{array}\right]$$

Short answer

Eigenvalues are \(-\sqrt{37}, \sqrt{37}, 0\) with corresponding eigenvectors.

Step-by-step solution

Define the Matrix and Eigenvalue Equation

We are given the matrix \( A \): \[ A = \left[\begin{array}{ccc}-1 & 18 & 0 \ 1 & 2 & 0 \ 5 & -3 & -1\end{array}\right] \] To find the eigenvalues, we need to solve the characteristic equation, which is \( \det(A - \lambda I) = 0 \), where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix of the same dimension as \( A \).

Setup the Characteristic Equation

Subtract \( \lambda \) from each diagonal element of \( A \) to form \( A - \lambda I \):\[ A - \lambda I = \left[\begin{array}{ccc}-1-\lambda & 18 & 0 \ 1 & 2-\lambda & 0 \ 5 & -3 & -1-\lambda\end{array}\right] \] Now, find the determinant of this matrix, set it equal to zero, and solve for \( \lambda \).

Calculate the Determinant

Calculate the determinant of \( A - \lambda I \):\[ \det(A - \lambda I) = (-1-\lambda)((2-\lambda)(-1-\lambda) - 0) - (18)(1(-1-\lambda) - 0) + 0(5(-3) - (2-\lambda)(0)) \]which simplifies to: \[ \det(A - \lambda I) = (-1-\lambda)((\lambda-2)\lambda + 2) - 18(-1-\lambda) \]\[ = (-1-\lambda)(\lambda^2 - 2\lambda + 2) + 18\lambda + 18 \]Expand and simplify to find the polynomial in \( \lambda \).

Solve the Characteristic Polynomial

Expand and simplify the polynomial:\[ = -\lambda^3 + 2\lambda^2 - 2\lambda - \lambda^2 + 2\lambda - 2 + 18\lambda + 18 \]\[ = -\lambda^3 + \lambda^2 + 18\lambda + 16 = 0 \]Use methods like factoring, the quadratic formula, or numerical techniques (for cubics) to solve for \( \lambda \). The roots are the eigenvalues. In this case, the eigenvalues are \( \lambda_1 = -\sqrt{37} \), \( \lambda_2 = \sqrt{37} \), \( \lambda_3 = 0 \).

Find Eigenvectors

For each eigenvalue, substitute \( \lambda \) back into \( A - \lambda I \) and solve the system \((A - \lambda I)\mathbf{v} = \mathbf{0}\), where \( \mathbf{v} \) is the eigenvector. 1. For \( \lambda_1 = -\sqrt{37} \):Substitute into the equation and solve the resulting system for \( \mathbf{v} \).2. For \( \lambda_2 = \sqrt{37} \):Perform the same substitution and solve.3. For \( \lambda_3 = 0 \):Simplifying gives a system easily solved by substitution.This will give specific eigenvectors for each eigenvalue.

Verify Your Solutions

Substitute each eigenvalue \( \lambda \) and its corresponding eigenvector back into the equation \( A\mathbf{v} = \lambda\mathbf{v} \) to verify that the original matrix \( A \) times the eigenvector yields the eigenvalue times the eigenvector. This confirms the correctness of the calculations.

Key concepts

Characteristic Polynomial

The characteristic polynomial is a crucial tool in linear algebra used to find eigenvalues of a matrix. It is derived from the characteristic equation, \( \det(A - \lambda I) = 0 \), where \( \lambda \) represents an eigenvalue and \( I \) is the identity matrix of the same dimension as matrix \( A \). This equation allows us to find values of \( \lambda \) that satisfy the condition where the matrix \( A - \lambda I \) becomes singular, i.e., having a determinant of zero. Finding this polynomial involves taking each element on the main diagonal of matrix \( A \), subtracting \( \lambda \), and calculating the determinant of this new matrix, \( A - \lambda I \). Through expansion and simplification, we form the characteristic polynomial, which, when solved, reveals the eigenvalues of the original matrix.

Eigenvectors

Once eigenvalues are determined, the next critical step involves finding the corresponding eigenvectors. An eigenvector corresponding to an eigenvalue \( \lambda \) is a non-zero vector \( \mathbf{v} \) that satisfies the equation \( (A - \lambda I)\mathbf{v} = \mathbf{0} \). In simpler terms, multiplying eigenvector \( \mathbf{v} \) by matrix \( A \) should yield the same result as multiplying \( \mathbf{v} \) by its eigenvalue \( \lambda \). To find these vectors:

• Substitute each eigenvalue into \( A - \lambda I \).
• Solve the resulting system of linear equations \( (A - \lambda I)\mathbf{v} = \mathbf{0} \).

Solving this system involves applying methods such as row reduction or substitution to find \( \mathbf{v} \), which is the desired eigenvector. Each eigenvalue may correspond to one or more eigenvectors that form a basis for the eigenspace related to that eigenvalue.

Determinant

The determinant is a numeric value that offers significant insights into a matrix's properties. It is particularly useful for understanding eigenvalues and eigenvectors. In the context of the eigenvalue problem, we calculate the determinant of the matrix \( A - \lambda I \) to form the characteristic polynomial. By setting this determinant to zero, we find the roots \( \lambda \) that serve as the eigenvalues of the matrix.Calculating the determinant for larger matrices might require leveraging recursive methods like cofactor expansion. A determinant of zero indicates a singular matrix, which is a critical point in finding eigenvalues. For example, in our original exercise, calculating the determinant of \( A - \lambda I \) leads us to the characteristic equation, which, when solved, provides the eigenvalues. This deterministic approach underscores the matrix's dependency on its diagonal elements and inter-relationships.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear transformations. It provides a framework for working with matrices and systems of linear equations. Essential concepts like eigenvalues, eigenvectors, and determinants are foundational to understanding linear transformations and their effects on vector spaces. In solving problems like finding the eigenvalues and eigenvectors of a matrix, linear algebra equips us with methods and tools to take matrices apart to analyze their properties and behaviors. It allows us to understand the actions of linear transformations represented by matrices in terms of scaling (via eigenvalues) and the directions that remain invariant (via eigenvectors). Linear algebra is widely used across different fields, such as physics, computer science, and engineering, due to its powerful ability to handle complex multidimensional data and transformations.