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

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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 λλ =− AI 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 of the matrix A , we determine the associated eigenvector(s) by finding a basis for the solution space of the linear system . j −= v 0 We write this linear system in scalar form in terms of the components of [ ] . T ab = v ± In most cases an associated eigenvector is then apparent. If A is a 22 × 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 λλλ + Eigenvalues: 12 2, 3 == With 1 2: = 0 0 1 1 1   =     v With 2 3: = 20 2 2 1 = v 2. Characteristic polynomial: 2 2 ( 1 ) ) p =+ Eigenvalues: 1, 2 = With 1 1: 66 0 33 0 1 1 1   =     v With 2 = 36 0 0 2 2 1 = v

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290 Chapter 6 3. Characteristic polynomial: 2 () 7 1 0 ( 2 ) ( 5 ) p λλλ λ =− +=− Eigenvalues: 12 2, 5 λλ == With 1 2: = 66 0 33 0 ab −= 1 1 1   =     v With 2 5: = 36 0 0 2 2 1 = v 4. Characteristic polynomial: 2 3 2 ( 1 ) ) p + Eigenvalues: 5 With 1 1: = 0 22 0 1 1 1   =     v With 2 = 23 0 0 2 3 2 = v 5. Characteristic polynomial: 2 5 4 ) ( 4 ) p + Eigenvalues: 1, 4 With 1 = 99 0 0 1 1 1   =     v With 2 4: = 69 0 0 2 3 2 = v 6. Characteristic polynomial: 2 5 6 ) ( 3 ) p + Eigenvalues: 4 With 1 = 44 0 0 1 1 1   =     v With 2 3: = 34 0 0 2 4 3 = v 7. Characteristic polynomial: 2 6 8 ( 2 ) ) p + Eigenvalues: 4
Section 6.1 291 With 1 2: λ = 88 0 66 0 ab −= 1 1 1   =     v With 2 4: = 68 0 0 2 4 3 = v 8. Characteristic polynomial: 2 () 3 2 ( 2 ) ( 1 ) p λλλ =+ + + Eigenvalues: 12 2, 1 λλ =− With 1 96 0 12 8 0 1 2 3 = v With 2 1: 86 0 12 9 0 2 3 4 = v 9. Characteristic polynomial: 2 7 1 2 ( 3 ) ( 4 ) p +=− Eigenvalues: 3, 4 == With 1 3: = 51 0 0 24 0 1 2 1 = v With 2 = 41 0 0 25 0 2 5 2 = v 10. Characteristic polynomial: 2 9 2 0 ) ( 5 ) p Eigenvalues: 4, 5 With 1 = 0 0 0 1 2 1 = v With 2 5: = 0 0 0 2 5 2 = v 11. Characteristic polynomial: 2 0 ) ) p Eigenvalues: 5 With 1 = 15 10 0 21 14 0 1 2 3 = v With 2 = 14 10 0 21 15 0 2 5 7 = v

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292 Chapter 6 12. Characteristic polynomial: 2 ( ) 7 212 ( 3)( 4) p λλλ λλ =− + Eigenvalues: 12 3, 4 == With 1 3: λ = 10 15 0 69 0 ab −= 1 3 2
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## This note was uploaded on 04/08/2008 for the course MATH 374 taught by Professor Zhu during the Spring '08 term at Western Michigan.

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

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