Consider the following. 0 2 A = 3 -1 3 20 1. (a) Compute the characteristic polynomial of A. (b) Compute the eigenvalues and bases of the corresponding eigenspaces of A. (c) Compute the algebraic and geometric multiplicity of each eigenvalue.

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
icon
Related questions
Question
### Consider the following.

\[ A = 
\begin{bmatrix}
1 & 0 & 2 \\
3 & -1 & 3 \\
2 & 0 & 1
\end{bmatrix}
\]

(a) Compute the characteristic polynomial of \( A \).

(b) Compute the eigenvalues and bases of the corresponding eigenspaces of \( A \).

(c) Compute the algebraic and geometric multiplicity of each eigenvalue.
Transcribed Image Text:### Consider the following. \[ A = \begin{bmatrix} 1 & 0 & 2 \\ 3 & -1 & 3 \\ 2 & 0 & 1 \end{bmatrix} \] (a) Compute the characteristic polynomial of \( A \). (b) Compute the eigenvalues and bases of the corresponding eigenspaces of \( A \). (c) Compute the algebraic and geometric multiplicity of each eigenvalue.
Recall that the general method to find the eigenvalues of a matrix \( A \) is to find the solutions \( \lambda \) of the equation \( \text{det}(A - \lambda I) = 0 \). This is because \( \lambda \) is an eigenvalue of \( A \) if and only if \( A - \lambda I \) is noninvertible, which is the case exactly when \( \text{det}(A - \lambda I) = 0 \).

Notice that treating \( \lambda \) as an unknown variable, the determinant \( \text{det}(A - \lambda I) \) is an expression rather than a number and is called the characteristic polynomial.

First, set up the characteristic polynomial.

\[
\text{det}(A - \lambda I) = \text{det} \left( 
\begin{bmatrix}
1 & 0 & 2 \\
3 & -1 & 3 \\
2 & 0 & 1 
\end{bmatrix}
- 
\begin{bmatrix}
\lambda & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \lambda
\end{bmatrix}
\right) 
\]

\[
= 
\begin{vmatrix}
1 - \lambda & 0 & 2 \\
3 & -1 - \lambda & 3 \\
2 & 0 & 1 - \lambda
\end{vmatrix}
\]

To find the determinant, expand along the first row and substitute the appropriate expressions from the matrix.

\[
\text{det}(A - \lambda I) = a_{11} \text{det}(A_{11}) - a_{12} \text{det}(A_{12}) + a_{13} \text{det}(A_{13})
\]

\[
= (\underline{\hspace{0.5cm}}) \text{det}(A_{11}) - (0) \text{det}(A_{12}) + (\underline{\hspace{0.5cm}}) \text{det}(A_{13})
\]

Now, find the required submatrices that have nonzero coefficients in the above equation.

\[
\text{det}(A_{11}) = (-1 - \lambda) \left( \begin{vmatrix} \underline
Transcribed Image Text:Recall that the general method to find the eigenvalues of a matrix \( A \) is to find the solutions \( \lambda \) of the equation \( \text{det}(A - \lambda I) = 0 \). This is because \( \lambda \) is an eigenvalue of \( A \) if and only if \( A - \lambda I \) is noninvertible, which is the case exactly when \( \text{det}(A - \lambda I) = 0 \). Notice that treating \( \lambda \) as an unknown variable, the determinant \( \text{det}(A - \lambda I) \) is an expression rather than a number and is called the characteristic polynomial. First, set up the characteristic polynomial. \[ \text{det}(A - \lambda I) = \text{det} \left( \begin{bmatrix} 1 & 0 & 2 \\ 3 & -1 & 3 \\ 2 & 0 & 1 \end{bmatrix} - \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} \right) \] \[ = \begin{vmatrix} 1 - \lambda & 0 & 2 \\ 3 & -1 - \lambda & 3 \\ 2 & 0 & 1 - \lambda \end{vmatrix} \] To find the determinant, expand along the first row and substitute the appropriate expressions from the matrix. \[ \text{det}(A - \lambda I) = a_{11} \text{det}(A_{11}) - a_{12} \text{det}(A_{12}) + a_{13} \text{det}(A_{13}) \] \[ = (\underline{\hspace{0.5cm}}) \text{det}(A_{11}) - (0) \text{det}(A_{12}) + (\underline{\hspace{0.5cm}}) \text{det}(A_{13}) \] Now, find the required submatrices that have nonzero coefficients in the above equation. \[ \text{det}(A_{11}) = (-1 - \lambda) \left( \begin{vmatrix} \underline
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 2 images

Blurred answer
Recommended textbooks for you
Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:
9780134463216
Author:
Robert F. Blitzer
Publisher:
PEARSON
Contemporary Abstract Algebra
Contemporary Abstract Algebra
Algebra
ISBN:
9781305657960
Author:
Joseph Gallian
Publisher:
Cengage Learning
Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra And Trigonometry (11th Edition)
Algebra And Trigonometry (11th Edition)
Algebra
ISBN:
9780135163078
Author:
Michael Sullivan
Publisher:
PEARSON
Introduction to Linear Algebra, Fifth Edition
Introduction to Linear Algebra, Fifth Edition
Algebra
ISBN:
9780980232776
Author:
Gilbert Strang
Publisher:
Wellesley-Cambridge Press
College Algebra (Collegiate Math)
College Algebra (Collegiate Math)
Algebra
ISBN:
9780077836344
Author:
Julie Miller, Donna Gerken
Publisher:
McGraw-Hill Education