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|
|b|=
|c|
|d|
| -2|
|0|
b | 1| + d | 0 | +
| 0|
|-2 |
| 0|
|1|
• | -2 b + 5 b |
• | -2 d +10 d |
| 5|
| 0|
| 10 |
| 0|
• Check.
• | -2 b + 5
• | -2 d +10
b || 1 | | 5|
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
• Det[A-xI =(x-3) | 1 -1
•
| -1 -1
-1 |
0 |
2-x |
•
| 2-x -1
• Det[A-xI =(x-3) |-1+x 0
•
|-3+x 0
-1 |
1 |
3-x |
• Det[A-xI =(x-3) |-1+x
•
|-3+x
• Det[A-xI =(x-3)^2 |-1+x
•
| 1
• Det[A-xI =(x-3)^2 (-x)
1 |
3-x |
1 |
-1 |
• 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 |
|000|
|000|
• [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|
|300|
|030|
|003|
• 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=
•
• F^n =
•
P | a 0 | P^(-1)
|0b|
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
a -b
------a-b
1+n 1+n
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