Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall

Math 307Spring, 2003

Hentzel

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

Instructor: Irvin Roy HentzelOffice 432 Carver

Phone 515-294-8141E-mail: [email protected]

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

10 Problems 6,8,10,20

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

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.

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

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.

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.

Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall

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