Post on 06-Jan-2016
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Math 307Spring, 2003
Hentzel
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
Monday, Apr 21 Chapter 7.3 Page 324 Problems 16,20.34,36
Main Idea: You want lots of eigen vectors. You might not get all you want.
Key Words: Algebraic Multiplicity, Geometric Multiplicity, Eigen Space
Goal: Learn to expect additional eigen vectors for multiple roots, but accept graciously the possibility that they do not exist.
Previous Assignment
Friday, April 18 Chapter 7.2
Page 310 Problems 6,8,10,20
Page 310 Problem 6Find all real eigenvalues, with their algebraic
multiplicities.
A = |1 2 | |3 4 |Det [A-x I ] = Det | 1-x 2 | = (1-x)(4-x) – 6 | 3 4-x |
= x2 -5 x – 2
5 +/- Sqrt [33] x = ----------------- 2 x = -0.372281 x = 5.37228
Page 310 Problem 8
Find all real eigenvalues, with their algebraic
multiplicities.
| -1 -1 -1 |
A = | -1 -1 -1 |
| -1 -1 -1 |
| -1-x -1 -1 |
Det [A-xI] = | -1 -1-x -1 |
| -1 -1 -1-x |
| -1-x -1 -1 |
Det[A-xI] = | x -x 0 |
| -1 -1 -1-x |
| -1-x -1 -1 |
Det [A-xI] = x | 1 -1 0 |
| -1 -1 -1-x |
| -2-x -1 -1 |Det [A-xI] = x | 0 -1 0 | | -2 -1 -1-x |
| -2-x -1 |Det [A-xI] = -x | | | -2 -1-x | Det [A-xI] = -x [ (-2-x)(-1-x) - 2 ]
De t[A-xI] = -x [ 2+2x+x+x2 - 2 ]
Det [A-xI] = -x [ x2 + 3x ]
Det [A-xI] = -x2 [ x + 3 ]
The eigen values are 0,0,-3.
Page 310 Problem 10
Find all real eigenvalues, with their algebraic
multiplicities.
| -3 0 4 |
A = | 0 -1 0 |
| -2 7 3 |
| -3-x 0 4 |
Det [A - x I ] = | 0 -1-x 0 |
| -2 7 3-x |
| -3-x 0 4 |
Det [A - x I ] = (-1-x)| 0 1 0 |
| -2 7 3-x |
| -3-x 4 |
Det [A - x I ] = (-1-x)| |
| -2 3-x |
Det [A - x I ] = (-1-x)( -9 + 3 x - 3 x + x2 + 8 )
Det [A - x I ] = (-1-x)( x2 -1) = -(x+1)2 (x-1)
Eigen values are -1,-1,1
Page 310 Problem 20
Consider a 2x2 matrix A with two distinct real eigenvalues c1 and c2. Express Det [A] in
terms of c1 and c2. Do the same for the
trace of A.
A = | a b |
| c d |
Det | a-x b | = (a-x)(d-x) - bc
| c d-x |
Det[A-xI] = x2 -(a+d)x + (ad-bc)
= x2 - trace[A] x + Det[A].
= (x-c1)(x-c2) = x2 - (c1+c2) x + c 1c2
So c1+c2 = trace[A] and c1c2 = Det[A].
New Material:
The Characteristic polynomial of a matrix A is
Det[A-xI].
Theorem. The Characteristic polynomial is
invariant under similarity.
The Characteristic Polynomial of P -1 A P =
Det [ P -1 A P - x I ] =
Det [ P -1 (A-x I) P ] =
Det [ P -1 Det[A-x I] Det[P] =
Det [ P -1 P] Det[A-xI] =
Det [A-xI] =
The Characteristic Polynomial of A.
Corollary:
Similar matrices have the same eigen values. If
c, V;
are an eigen value and eigen vector of A, then
c, P -1 V;
are an eigen value and vector of
P -1 A P.
Proof: P -1 A P P -1 V =
P -1 A V =
P -1 c V =
c P -1 V.
Definition: The Algebraic multiplicity is the number of times c appears as a root of the characteristic polynomial.
The Geometric multiplicity of the eigen value c is the number of linearly independent eigen vectors with eigen value c.
Find the eigen values and eigen vectors of
| 0 1 0 |
A = | 0 0 1 |
| 0 0 0 |
| 0-x 1 0 |
Det [A - x I ] = | 0 0-x 1 | = -x3
| 0 0 0-x |
The eigen values are 0,0,0.
To find the corresponding eigen vectors.
| 0 1 0 | | 1 |
| 0 0 1 | has null space | 0 |
| 0 0 0 | | 0 |
So 0 has algebraic multiplicity 3 and geometric multiplicity 1 for A.
Page 321 Example 7. Solve the recursive dependence relation
X(t+1) = A x(t)
| 750 | | 0 19 12 |
Where X(0) = | 200 | and A =(1/20) | 16 0 0 |
| 200 | | 0 10 0 |
The Eigen Values and Vectors are:
| 9 | | 2 | | 5 |
-3/5 |-12 | -2/5 | -4 | 1 | 4 |
| 10 | | 5 | | 2 |
| 9 2 5 |
P = | -12 -4 4 |.
| 10 5 2 |
| -3/5 0 0 |
P -1 .A.P = | 0 -2/5 0 |
| 0 0 1 |
| 750 | | 50 |
Solve P X = | 200 | X = | -100 |
| 200 | | 100 |
| 9 | | 2 | | 5 |
Xo = 50 |-12 | - 100 |-4 | + 100 | 4 |
| 10 | | 5 | | 2 |
| 9 | | 2 | | 5 |
Xn = 50 (-3/5)n |-12 | - 100 (-2/5) n |-4 | +100| 4 |
| 10 | | 5 | | 2 |
| 500 |
The limiting situation is | 400 |
| 200 |
Theorem: Eigen Vectors for distinct Eigen Values are linearly independent.
Proof: Suppose that ci, Vi are eigen values
and eigen vectors for distinct eigen values . Suppose that
a1 V1 + a2 V2 + … + an Vn = 0 then
a1 c1n V1 + a2 c2
n V2+ … + an cnn Vn = 0
for all n. This can only happen when all ai are
zero.