Math 307

Spring, 2003Hentzel

Time: 1:10-2:00 MWFRoom: 1324 Howe Hall

Instructor: Irvin Roy HentzelOffice 432 Carver

Phone 515-294-8141E-mail: hentzel@iastate.edu

http://www.math.iastate.edu/hentzel/class.307.ICN

Text: Linear Algebra With Applications, Second Edition Otto Bretscher

• 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.

• 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.

• 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.

• 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.

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

• 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 |

• Check.

• | -2 b + 5 b || 1 | | 5|

• | -2 d +10 d || 2 | = |10 |

• 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.

• Find the Eigen values and vectors of

• | 2 -1 -1 |

• |-1 2 -1 |

• |-1 -1 2 |

• | 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 |

• | 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 |

• | 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 |

• 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)

• The eigen values are 3,3,0• • x=3 A-3I = | -1 -1 -1 |• | -1 -1 -1 |• | -1 -1 -1 |

• RCF(A-3I) = | 1 1 1 |• | 0 0 0 |• | 0 0 0 |

• [V1 V2 ] = | -1 -1 |• | 1 0 |• | 0 1 |

• 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 |

• 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 |

• Check:

• | 2 -1 -1 | | 1 | | 0 | | 1 |

• |-1 2 -1 | | 1 | = | 0 | = 0 | 1 |

• |-1 -1 2 | | 1 | | 0 | | 1 |

• Diagonalization:

• -1

• | -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 |

• | -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 |

• | -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 |

• Find a formula for the Fibonacci Numbers.

• fo = 1

• f1 = 1

• f2 = 2

• f3 = 3

• fn = fn-1+fn-2.

• | 0 1 | | fn | = | fn+1 | = | fn+1 |

• | 1 1 | | fn+1 | | fn+fn+1| | fn+2 |

•

• n

• F | 1 | = | fn |

• | 1 | | fn+1 |

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

• | 1 1-x |

• Let the polynomial factor into (x-a)(x-b) where

• 1+Sqrt[5]• a = -----------• 2

• 1-Sqrt[5]• b = ----------• 2

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 |

• | fn | = P | a^n 0 | P^(-1)

• | fn+1 | | 0 b^n |

• 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 |

• P = [V1 V2] = |-1+a -1+b | = |-b -a |

• | 1 1 | | 1 1 |

• P^(-1) = 1/(a-b) | 1 a |

• | -1 -b |

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

• | 1 1 | | 0 b^n | | -1 -b |

• | 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 |

• 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 |

•

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

• -----------------------------------------

• a-b