{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ch6 - CHAPTER 6 EIGENVALUES AND EIGENVECTORS SECTION 6.1...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Section 6.1 289 CHAPTER 6 EIGENVALUES AND EIGENVECTORS SECTION 6.1 INTRODUCTION TO EIGENVALUES In each of Problems 1–32 we first list the characteristic polynomial ( ) p λ λ = A I of the given matrix A , and then the roots of ( ) p λ — which are the eigenvalues of A . All of the eigenvalues that appear in Problems 1–26 are integers, so each characteristic polynomial factors readily. For each eigenvalue j λ of the matrix A , we determine the associated eigenvector(s) by finding a basis for the solution space of the linear system ( ) . j λ = A I v 0 We write this linear system in scalar form in terms of the components of [ ] . T a b = v In most cases an associated eigenvector is then apparent. If A is a 2 2 × matrix, for instance, then our two scalar equations will be multiples one of the other, so we can substitute a convenient numerical value for the first component a of v and then solve either equation for the second component b (or vice versa). 1. Characteristic polynomial: 2 ( ) 5 6 ( 2)( 3) p λ λ λ λ λ = + = Eigenvalues: 1 2 2, 3 λ λ = = With 1 2: λ = 2 2 0 0 a b a b = = 1 1 1   =     v With 2 3: λ = 2 0 2 0 a b a b = = 2 2 1   =     v 2. Characteristic polynomial: 2 ( ) 2 ( 1)( 2) p λ λ λ λ λ = = + Eigenvalues: 1 2 1, 2 λ λ = − = With 1 1: λ = − 6 6 0 3 3 0 a b a b = = 1 1 1   =     v With 2 2: λ = 3 6 0 3 6 0 a b a b = = 2 2 1   =     v
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
290 Chapter 6 3. Characteristic polynomial: 2 ( ) 7 10 ( 2)( 5) p λ λ λ λ λ = + = Eigenvalues: 1 2 2, 5 λ λ = = With 1 2: λ = 6 6 0 3 3 0 a b a b = = 1 1 1   =     v With 2 5: λ = 3 6 0 3 6 0 a b a b = = 2 2 1   =     v 4. Characteristic polynomial: 2 ( ) 3 2 ( 1)( 2) p λ λ λ λ λ = + = Eigenvalues: 1 2 2, 5 λ λ = = With 1 1: λ = 3 3 0 2 2 0 a b a b = = 1 1 1   =     v With 2 2: λ = 2 3 0 2 3 0 a b a b = = 2 3 2   =     v 5. Characteristic polynomial: 2 ( ) 5 4 ( 1)( 4) p λ λ λ λ λ = + = Eigenvalues: 1 2 1, 4 λ λ = = With 1 1: λ = 9 9 0 6 6 0 a b a b = = 1 1 1   =     v With 2 4: λ = 6 9 0 6 9 0 a b a b = = 2 3 2   =     v 6. Characteristic polynomial: 2 ( ) 5 6 ( 2)( 3) p λ λ λ λ λ = + = Eigenvalues: 1 2 1, 4 λ λ = = With 1 2: λ = 4 4 0 3 3 0 a b a b = = 1 1 1   =     v With 2 3: λ = 3 4 0 3 4 0 a b a b = = 2 4 3   =     v 7. Characteristic polynomial: 2 ( ) 6 8 ( 2)( 4) p λ λ λ λ λ = + = Eigenvalues: 1 2 2, 4 λ λ = =
Image of page 2
Section 6.1 291 With 1 2: λ = 8 8 0 6 6 0 a b a b = = 1 1 1   =     v With 2 4: λ = 6 8 0 6 8 0 a b a b = = 2 4 3   =     v 8. Characteristic polynomial: 2 ( ) 3 2 ( 2)( 1) p λ λ λ λ λ = + + = + + Eigenvalues: 1 2 2, 1 λ λ = − = − With 1 2: λ = − 9 6 0 12 8 0 a b a b = = 1 2 3   =     v With 2 1: λ = − 8 6 0 12 9 0 a b a b = = 2 3 4   =     v 9.
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern