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. Friday, April 18 Chapter 7.2 Page 310 Problems 6,8,10,20 Main Idea: How do you tell what a matrix is going to do? Key Words: Eigen Value, Eigen Vector, Characteristic Polynomial Goal: Introduction to eigenvalues and eigen vectors.

3. Previous Assignment. Page 300 Problem 2 Let A be an invertible nxn matrix and V an eigenvector of A with associated eigen value c If V is an eigenvector of A^(-1) ? If So, what is its eigenvalue. If A stretches V by a factor of c, then A^(-1) must shrink V by a factor of 1/c.

4. Page 300 Problem 4 Let A be an invertible nxn matrix and V an eigenvector of A with associated eigen value c Is V an eigen vector of 7A? IF so, what is the eigenvalue? If A streches V by a factor of c, then 7 A stretches V by a factor of 7 c.

5. Page 300 Problem 6 If a vector V is an eigenvector of both A and B, is V necessarily an eigen vector of AB? Let A V = a V and B V = b V. A B V = A b V = b A V = ba V V is an eigen vector of AB and the eigenvalue is ab.

6. Page 300 Problem 10 Find all 2x2 matrices for which | 1 | | 2 | is an eigen vector for eigen value 5 | a b | | 1 | = | 5 | | c d | | 2 | |10| a+2b = 5 c+2d = 10

7. a b c d 1 2 0 0 5 0 0 1 2 10 | a | | -2| | 0 | | 5 | | b | = b | 1| + d | 0 | + | 0 | | c | | 0| |-2 | | 10 | | d | | 0| | 1 | | 0 | | -2 b + 5 b | | -2 d +10 d |

8. Check. | -2 b + 5 b || 1 | | 5| | -2 d +10 d || 2 | = |10 |

9. Page 300 Problem 40 Suppose that V is an eigenvector of the nxn matrix A, with eigen value 4. Explain why V is an eigenvector of A^2 + 2A + 3 In. What is its associated eigenvalue. (A^2 + 2 A + 3I)V = A(AV) + 2 AV + 3 V = (16+8+3)V = 27 V.

10. Find the Eigen values and vectors of | 2 -1 -1 | |-1 2 -1 | |-1 -1 2 |

11. | 2-x -1 -1 | Det[A-xI = | -1 2-x -1 | | -1 -1 2-x | | 2-x -1 -1 | Det[A-xI = |-3+x 3-x 0 | | -1 -1 2-x |

12. | 2-x -1 -1 | Det[A-xI = | -1 2-x -1 | | -1 -1 2-x | | 2-x -1 -1 | Det[A-xI = |-3+x 3-x 0 | | -1 -1 2-x |

13. | 2-x -1 -1 | Det[A-xI =(x-3) | 1 -1 0 | | -1 -1 2-x | | 2-x -1 -1 | Det[A-xI =(x-3) |-1+x 0 1 | |-3+x 0 3-x |

14. Det[A-xI =(x-3) |-1+x 1 | |-3+x 3-x | Det[A-xI =(x-3)^2 |-1+x 1 | | 1 -1 | Det[A-xI =(x-3)^2 (-x)

15. The eigen values are 3,3,0 x=3 A-3I = | -1 -1 -1 | | -1 -1 -1 | RCF(A-3I) = | 1 1 1 | | 0 0 0 | [V1 V2 ] = | -1 -1 | | 1 0 | | 0 1 |

16. Check: | 2 -1 -1 | | -1 -1 | | -3 -3 | | -1 -1 | |-1 2 -1 | | 1 0 | = | 3 0 | = 3| 1 0 | |-1 -1 2 | | 0 1 | | 0 3 | | 0 1 |

17. x = 0 | 2 -1 -1 | | 1 1 -2 | | 1 0 -1 | |-1 2 -1 | ~ | 0 -3 3 | ~ | 0 1 -1 | |-1 -1 2 | | 0 3 -3 | | 0 0 0 | | 1 | V3 = | 1 | | 1 |

18. Check: | 2 -1 -1 | | 1 | | 0 | | 1 | |-1 2 -1 | | 1 | = | 0 | = 0 | 1 | |-1 -1 2 | | 1 | | 0 | | 1 |

19. Diagonalization: | -1 -1 1 | | 2 -1 -1 | | -1 -1 1 | | 1 0 1 | | -1 2 -1 | | 1 0 1 | | 0 1 1 | | -1 -1 2 | | 0 1 1 |

20. | -1 2 -1 | | 2 -1 -1 | | -1 -1 1 | 1/3 | -1 -1 2 | | -1 2 -1 | | 1 0 1 | | 1 1 1 | | -1 -1 2 | | 0 1 1 |

21. | -1 2 -1 | | -1 -1 1 | | -1 -1 2 | | 1 0 1 | | 0 0 0 | | 0 1 1 | | 3 0 0 | | 0 3 0 | | 0 0 3 |

22. Find a formula for the Fibonacci Numbers. fo = 1 f1 = 1 f2 = 2 f3 = 3 fn = fn-1+fn-2.

23. | 0 1 | | fn | = | fn+1 | = | fn+1 | | 1 1 | | fn+1 | | fn+fn+1| | fn+2 | n F | 1 | = | fn | | 1 | | fn+1 |

24. Det[ F-xI ] = | -x 1 | = x^2 - x - 1 | 1 1-x |

25. Let the polynomial factor into (x-a)(x-b) where 1+Sqrt[5] a = ----------- 2 1-Sqrt[5] b = ---------- 2

26. There exist matrices P such that P^(-1) F P = | a 0 | | 0 b | F = P | a 0 | P^(-1) | 0 b | F^n = P | a^n 0 | P^(-1) | 0 b^n |

27. | fn | = P | a^n 0 | P^(-1) | fn+1 | | 0 b^n |

28. So we have to compute P. | -a 1 | ~ | 1 1-a | | 1 1-a | | 0 0 | | -b 1 | ~ | 1 1-b | | 1 1-b | | 0 0 |

29. P = [V1 V2] = |-1+a -1+b | = |-b -a | | 1 1 | | 1 1 | P^(-1) = 1/(a-b) | 1 a | | -1 -b |

30. F^n = 1/(a-b) | -b -a | | a^n 0 | | 1 a | | 1 1 | | 0 b^n | | -1 -b |

31. | n n n n | | -(a b) + a b a b (a - b ) | | -------------- -(-------------) | | a - b a - b | | n n 1 + n 1 + n | | a - b a - b | | ------- --------------- | | a - b a - b |

32. F^n | 1 | = | fn | | 1 | | fn+1 | | n 1 + n n | | -(a b) - a b + a b (1 + b) | | --------------------------------- | | a - b | | | | n 1 + n n | | a + a - b (1 + b) | | ------------------------ | | a - b |

33. So fn = -a^n b - a^(1+n) b + a b^n (1+b) ----------------------------------------- a-b