Problem 58 $$ \begin{array}{l} \text { ... [FREE SOLUTION]

Chapter 9: Problem 58

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

Short Answer

Expert verified

The eigenvalues are $-7$ and $6$.

Step by step solution

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 $.

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} $.

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) $.

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 $6 - \lambda = 0$.

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 $.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

91影视

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

Short Answer

Step by step solution

Understanding Eigenvalues

Set Up the Determinant Equation

Calculate the Determinant

Solve the Characteristic Equation

Find the Eigenvalues

Key Concepts

Determinant

Characteristic Equation

Diagonal Matrix

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Biology Textbooks

Astrobiological Science

Ecology

Biology Experiments

Control of Gene Expression

Ecological Levels

Reproduction

Study anywhere. Anytime. Across all devices.