/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 In Problems, determine whether t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 14

In Problems, determine whether the given matrix \(\mathbf{A}\) is diagonalizable. If so, find the matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix \(\mathbf{D}\) such that \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rrr} 0 & -9 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right) $$

Short Answer

Expert verified
\(\mathbf{A}\) is diagonalizable with \(\mathbf{P} = \begin{pmatrix} 1 & 0 & 9 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\).

Step by step solution

01

Identify the Matrix

The matrix given is \( \mathbf{A} = \begin{pmatrix} 0 & -9 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{pmatrix} \). We need to determine if \( \mathbf{A} \) is diagonalizable.
02

Find Characteristic Polynomial

Calculate the characteristic polynomial by solving \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This gives: \[ \mathbf{A} - \lambda \mathbf{I} = \begin{pmatrix} -\lambda & -9 & 0 \ 1 & -\lambda & 0 \ 0 & 0 & 1-\lambda \end{pmatrix} \]. The determinant is: \(-\lambda((1-\lambda)(-\lambda) - (-9 \cdot 0)) = 0\), and simplifying we get \(\lambda^3 - \lambda^2 = 0\).
03

Solve Characteristic Equation

Factor the characteristic polynomial \( \lambda^2(\lambda - 1) = 0 \). The roots are \( \lambda_1 = 0 \) with a multiplicity of 2, and \( \lambda_2 = 1 \).
04

Determine Eigenvectors for \(\lambda = 0\)

For \( \lambda = 0 \), solve \((\mathbf{A} - 0 \cdot \mathbf{I})\mathbf{x} = 0\). This gives the system: \( -9y = 0 \), \( x = 0 \). The eigenvector is \( \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} \). However, since the algebraic and geometric multiplicity need to match for diagonalizability, check for another vector, finding another eigenvector \( \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \).
05

Determine Eigenvectors for \(\lambda = 1\)

For \( \lambda = 1 \), solve \((\mathbf{A} - 1 \mathbf{I})\mathbf{x} = 0\). The equation becomes \( \begin{pmatrix} -1 & -9 & 0 \ 1 & -1 & 0 \ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \). This determines the eigenvector \( \begin{pmatrix} 9y \ 9 \ z \end{pmatrix} \), or simplified \( \begin{pmatrix} 9 \ 1 \ 0 \end{pmatrix} \).
06

Verify Linearly Independent Eigenvectors

The eigenvectors are \( \mathbf{v_1} = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \), \( \mathbf{v_2} = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} \), and \( \mathbf{v_3} = \begin{pmatrix} 9 \ 1 \ 0 \end{pmatrix} \). Check that they are linearly independent.
07

Construct \(\mathbf{P}\) and \(\mathbf{D}\)

With linearly independent eigenvectors, form \( \mathbf{P} \) using these vectors as columns: \( \mathbf{P} = \begin{pmatrix} 1 & 0 & 9 \ 0 & 1 & 1 \ 0 & 0 & 0 \end{pmatrix} \). Place eigenvalues on the diagonal of \( \mathbf{D} \): \( \mathbf{D} = \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{pmatrix} \).
08

Conclusion

\( \mathbf{A} \) is diagonalizable since there are enough linearly independent eigenvectors (3). The matrix \( \mathbf{P} \) is as found and \( \mathbf{D} \) has the eigenvalues.

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.

Eigenvalues
Eigenvalues are special numbers associated with a matrix that provide insight into its properties, such as whether it can be diagonalized. To discover them, we solve the characteristic equation. This involves finding the determinant of the matrix \( \mathbf{A} - \lambda \mathbf{I} \), where \( \lambda \) are the eigenvalues we need.
  • In our exercise, the characteristic polynomial resulted in \( \lambda^3 - \lambda^2 = 0 \).
  • This polynomial can be factored to identify the eigenvalues as \( \lambda_1 = 0 \) (with multiplicity of 2) and \( \lambda_2 = 1 \).
Finding eigenvalues helps us determine the structure of the matrix, and is a crucial step towards diagonalization.
Eigenvectors
Eigenvectors are vectors that indicate directions in which a linear transformation acts by stretching or shivining, without rotating. They are associated with specific eigenvalues. To find them, solve the equation:\( (\mathbf{A} - \lambda \mathbf{I})\mathbf{x} = \mathbf{0} \).
  • For \( \lambda = 0 \), the eigenvectors are \( \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \) and \( \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} \).
  • For \( \lambda = 1 \), the eigenvector is \( \begin{pmatrix} 9 \ 1 \ 0 \end{pmatrix} \).
These vectors form a basis for the subspace associated with their eigenvalue, allowing for the construction of the essential matrices for diagonalization.
Characteristic Polynomial
The characteristic polynomial is a crucial tool in linear algebra, representing how matrix transformation behaves. It is obtained when we compute the determinant of the matrix \( \mathbf{A} - \lambda \mathbf{I} \). For our exercise:
  • The characteristic polynomial derived was \( \lambda^3 - \lambda^2 = 0 \).
  • Factoring gives \( \lambda^2(\lambda - 1) = 0 \), providing our eigenvalues.
This polynomial encapsulates information about the eigenvalues, and each term's degree indicates the potential multiplicity of corresponding eigenvalues. Understanding it is fundamental for steps like determining matrix diagonalizability.
Algebraic Multiplicity
Algebraic multiplicity refers to how many times an eigenvalue appears as a root in the characteristic polynomial. It tells us how many times a specific eigenvalue contributes to the count of the matrix's characteristic equation.
  • For eigenvalue \( \lambda_1 = 0 \), the algebraic multiplicity is 2.
  • For eigenvalue \( \lambda_2 = 1 \), the multiplicity is 1.
This concept is important because it impacts diagonalizability. A matrix is diagonalizable if the sum of eigenvectors for each distinct eigenvalue equals the dimension of the matrix 鈥� a condition enriched by the algebraic multiplicity.
Geometric Multiplicity
Geometric multiplicity measures the number of linearly independent eigenvectors associated with an eigenvalue. It tells us about the size of the eigenvector space for each eigenvalue.
  • For \( \lambda_1 = 0 \), the geometric multiplicity is 2, matching its algebraic multiplicity.
  • For \( \lambda_2 = 1 \), it is 1.
A key requirement for a matrix to be diagonalizable is that for each eigenvalue, algebraic multiplicity must not exceed geometric multiplicity. This ensures a sufficient number of eigenvectors are available, confirming the "freeness" from dependency needed for matrix diagonalization.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In an experiment the following correspondence was found between temperature \(T\) (in \({ }^{\circ} \mathrm{C}\) ) and electrical resistance \(R\) (in M\Omega): $$ \begin{array}{|lrrrrrr|} \hline \boldsymbol{T} & 400 & 450 & 500 & 550 & 600 & 650 \\ \hline R & 0.47 & 0.90 & 2.0 & 3.7 & 7.5 & 15 \\ \hline \end{array} $$ Find the least squares line \(R=a T+b\). Use this line toestimate the resistance at \(T=700\).

Write the system in the form \(\mathbf{A} \mathbf{X}=\mathbf{B}\). Use \(\mathbf{X}=\mathbf{A}^{-1} \mathbf{B}\) to solve the system for each matrix \(\mathbf{B}\). $$ \begin{aligned} x_{1}+2 x_{2}+5 x_{3} &=b_{1} \\ 2 x_{1}+3 x_{2}+8 x_{3} &=b_{2} \\ -x_{1}+x_{2}+2 x_{3} &=b_{3} \\ \mathbf{B}=\left(\begin{array}{r} -1 \\ 4 \\ 6 \end{array}\right), \quad \mathbf{B} &=\left(\begin{array}{l} 3 \\ 3 \\ 3 \end{array}\right), & \mathbf{B}=\left(\begin{array}{r} 0 \\ -5 \\ 4 \end{array}\right) \end{aligned} $$

Determine whether the given message is a code word in the Hamming \((7,4)\) code. If it is, decode it. If it is not, correct the single error and decode the corrected message. $$ \left.\begin{array}{llllllll} (1 & 1 & 0 & 0 & 0 & 0 & 0 \end{array}\right) $$

Determine whether the given message is a code word in the Hamming \((7,4)\) code. If it is, decode it. If it is not, correct the single error and decode the corrected message. $$ \left.\begin{array}{lllllll} (1 & 1 & 1 & 1 & 1 & 1 & 1 \end{array}\right) $$

In Problems, determine whether the given matrix \(\mathbf{A}\) is diagonalizable. If so, find the matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix \(\mathbf{D}\) such that \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rrr} 1 & 3 & -1 \\ 0 & 2 & 4 \\ 0 & 0 & 1 \end{array}\right) $$
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Particle Model of Matter

Read Explanation

Radiation

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Electromagnetism

Read Explanation

Circular Motion and Gravitation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.