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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,
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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
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