1. Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://www.math.iastate.edu/hentzel/class.307.ICN Text: Linear Algebra With Applications, Second Edition Otto Bretscher

2. Monday, Apr 28 Generalized Eigen Spaces To be assigned Main Idea: What happens when there are not enough Eigen Vectors. Key Words: Generalized Eigen Space, Jordan Canonical Form 0  V 1  V 2 ...  V i-2  V i-1  V i Goal: Learn how to handle the case when there are not enough Eigen Vectors.

3. Previous Assignment Friday, Apr 25 Chapter 8.2 Page 381 Problems 18,20,22

4. Page 381 Problem 18 9x 2 - 4 x y + 6 y 2 = 1 A = | 9 -2 | | -2 6 | Det[A-xI] = Det| 9-x -2 | | -2 6-x | = 54 -15 x + x 2 -4 = x 2 -15 x + 50 = (x-10)(x-5)

5. Eigen Values 5, 10 x=5 A-5 I = | 4 -2 | V 1 = | 1 | | -2 1 | | 2 | x=10 A-10 I = | -1 -2 | V 2 = | 2 | | -2 -4 | | -1 |

6. P = 1/Sqrt[5] | 1 2 | | 2 -1 | [u v] 1/5 |1 2 || 9 -2 || 1 2 | |u| = 1 |2 -1 ||-2 6 || 2 -1 | |v| [u v] 1/5 | 5 10 | | 1 2 ||u| = 1 [u v] 1/5 |25 0 | |u| = 1 |20 -10 | | 2 -1 ||v| | 0 50 | |v| [u v] | 5 0 | |u| = 1 | 0 10 | |v| 5u 2 + 10 v 2 = 1 u 2 v 2 -------- + -------- = 1 1/5 1/10

7. | (1,2) |. -----------------+----------------------------- |. (2,-1)

8. Page 381 Problem 20 -3 x 2 + 6 x y + 5 y 2 - 1 A = | -3 3 | | 3 5 | Det [A-x I ] = | -3-x 3 | = -15 -2x + x 2 -9 | 3 5-x | = x 2 -2x-24 = (x-6)(x+4) Eigen Values -4,6

9. x = -4 A+4I = | 1 3 | V 1 = | -3 | | 3 9 | | 1 | x = 6 A-6I = | -9 3 | V 2 = | 1 | | 3 -1 | | 3 |

10. P = 1/Sqrt[10] | -3 1 | | 1 3 | [u v] 1/10 | -3 1 | | -3 3 | | -3 1 | |u| = 1 | 1 3 | | 3 5 | | 1 3 | |v| [u v] 1/10 | 12 -4 | |-3 1 | |u| = 1 [u v] 1/10 | -40 0 | |u| = 1 | 6 18 | | 1 3 | |v| | 0 60 | |v| [u v] | -4 0 | |u| = 1 -4 u 2 + 6 v 2 = 1 | 0 6 | |v| u 2 v 2 - ------- + ------- = 1 1/4 1/6

11. |. v |. u. |. `. |. ---------`+-------- |

12. Page 381 Problem 22 -x 2 + y 2 -z 2 + 10 x z = 1 | -1 0 5 | A = | 0 1 0 | | 5 0 -1 | | -1-x 0 5 | Det[A-x I] = | 0 1-x 0 | | 5 0 -1-x | = (1-x) ( 1 + 2 x + x 2 - 25) = (1-x) ( x 2 +2 x -24) = (1-x)(x+6)(x-4) Eigen Values 1,4,-6

13. x = 1 | -2 0 5 | | 1 0 0 | | 0 | A-I = | 0 0 0 | ~ | 0 0 1 | V 1 = | 1 | | 5 0 -2 | | 0 0 0 | | 0 | x=4 | -5 0 5 | | 1 0 -1 | | 1 | A-4 I = | 0 -4 0 | ~ | 0 1 0 | V 2 = | 0 | | 5 0 -5 | | 0 0 0 | | 1 | x = -6 | 5 0 5 | | 1 0 1 | | -1 | A+6I = | 0 7 0 | ~ | 0 1 0 | V 3 = | 0 | | 5 0 5 | | 0 0 0 | | 1 |

14. P = | 0 w -w | | 1 0 0 | w = 1/Sqrt[2] | 0 w w | | 0 1 0 | |-1 0 5 | | 0 w -w | | x | [x y z ] | w 0 w | | 0 1 0 | | 1 0 0 | | y | = 1 | -w 0 w | | 5 0 -1 | | 0 w w | | z | | 0 1 0 | | 0 w -w | | x | [x y z ] |4w 0 4w | | 1 0 0 | | y | = 1 |6w 0 -6w | | 0 w w | | z | | 1 0 0 | | x | [x y z ] | 0 4 0 | | y | = 1 | 0 0 -6 | | z |

15. x 2 + 4 y 2 -6z 2 = 1 x 2 y 2 z 2 --------- + ------- - -------- = 1 1 1/4 1/6

16. New Material: We now show a way to create "extra eigen vectors“ for those matrices who do not have an eigen basis. Certainly we cannot really get more true eigen vectors, but we can get the next best thing.

17. | -4 -6 -10 | A = | 4 5 7 | | -1 -1 -1 | Det[A-x I] = -x 3 The eigen values are 0,0,0 x = 0 | 1 1 1 | | 1 0 -2 | | 2 | A-0 I ~ | 0 -2 -6 | ~ | 0 1 3 | V 1 = |-3 | | 0 1 3 | | 0 0 0 | | 1 | One vector is not enough for a basis. We obtain another vector by Solving AX = V 1

18. | -4 -6 -10 | 2 | [A|V 1 ] = | 4 5 7 | -3 | | -1 -1 -1 | 1 | | 1 1 1 -1 | ~ | 0 -2 -6 -2 | | 0 1 3 1 | | 1 0 -2 -2 | | -2 | | 2 | ~ | 0 1 3 1 | V 2 = | 1 | + a | -3 | | 0 0 0 0 | | 0 | | 1 |

19. Now do it again: | -4 -6 -10 | -2 | [A | V 2 ] = | 4 5 7 | 1 | | -1 -1 -1 | 0 | | 1 1 1 0 | ~ | 0 -2 -6 -2 | | 0 1 3 1 | | 1 0 -2 -1 | |-1 | | 2 | ~ | 0 1 3 1 | V 3 = | 1 | + a | -3 | | 0 0 0 0 | | 0 | | 1 |

20. We have enough vectors. If we try it again, we will not have any solution. | -4 -6 -10 | -1 | [ A | V 3 ] = | 4 5 7 | 1 | | -1 -1 -1 | 0 | | 1 1 1 0 | ~ | 0 -2 -6 -1 | | 0 1 3 1 | | 1 0 -2 -1 | ~ | 0 1 3 1 | | 0 0 0 1 | Since there is a stair step one in the last column, there is no solution.

21. We have a chain of vectors 0  V 1  V 2  V 3. If we use these vectors for our matrix P V 3 V 2 V 1 | 2 -2 -1 | P = |-3 1 1 | | 1 0 0 |

22. Then P -1 A P is the best we can do. | 0 1 0 | P -1 A P = | 0 0 1 |. | 0 0 0 |

23. The principle is this. (1) Find the eigen values. (2) For each Eigen Value x. Find as many generalized eigen vectors as the multiplicity of x in the characteristic polynomial. You do this by solving [A-x I ] V 1 = 0 and then continuing with [A-x I ] V n+1 = V n again and again. (e) In total one will get n generalized eigen vectors. Collect the generalized eigen vectors together in a matrix P = [ V 1 V 2... V n-2 V n-1 V n ] and then (f) P -1 A P is the next best thing to diagonalization.

24. Hand-In-Assignment: 1.Find the Generalized Eigen Vectors for | 1 4 -1 | A = | 0 -2 1 | | -1 -4 1 | Compute P -1 A P where P is the matrix of Generalized Eigen Vectors.

25. 2. Find the Generalized Eigen Vectors for | 0 -4 2 | A = | 1 4 -1 | | 0 4 -2 | Compute P -1 A P where P is the matrix of Generalized Eigen Vectors.

26. 3. Find the Generalized Eigen Vectors for | -2 -1 -1 | A = | 2 1 1 | | 2 1 1 | Compute P -1 A P where P is the matrix of Generalized Eigen Vectors.