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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
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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.
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Previous Assignment Friday, April 18 Chapter 7.2 Page 310 Problems 6,8,10,20
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Page 310 Problem 6 Find 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 | = x 2 -5 x – 2 5 +/- Sqrt [33] x = ----------------- 2 x = -0.372281 x = 5.37228
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Page 310 Problem 8 Find all real eigenvalues, with their algebraic multiplicities. | -1 -1 -1 | A = | -1 -1 -1 | | -1 -1 -1 |
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| -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 |
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| -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+x 2 - 2 ] Det [A-xI] = -x [ x 2 + 3x ] Det [A-xI] = -x 2 [ x + 3 ] The eigen values are 0,0,-3.
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Page 310 Problem 10 Find all real eigenvalues, with their algebraic multiplicities. | -3 0 4 | A = | 0 -1 0 | | -2 7 3 |
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| -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 |
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Det [A - x I ] = (-1-x)( -9 + 3 x - 3 x + x 2 + 8 ) Det [A - x I ] = (-1-x)( x 2 -1) = -(x+1) 2 (x-1) Eigen values are -1,-1,1
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Page 310 Problem 20 Consider a 2x2 matrix A with two distinct real eigenvalues c 1 and c 2. Express Det [A] in terms of c 1 and c 2. 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 |
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Det[A-xI] = x 2 -(a+d)x + (ad-bc) = x 2 - trace[A] x + Det[A]. = (x-c 1 )(x-c 2 ) = x 2 - (c 1 +c 2 ) x + c 1 c 2 So c 1 +c 2 = trace[A] and c 1 c 2 = Det[A].
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New Material: The Characteristic polynomial of a matrix A is Det[A-xI].
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Theorem. The Characteristic polynomial is invariant under similarity.
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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.
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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.
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Proof: P -1 A P P -1 V = P -1 A V = P -1 c V = c P -1 V.
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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.
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Find the eigen values and eigen vectors of | 0 1 0 | A = | 0 0 1 | | 0 0 0 |
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| 0-x 1 0 | Det [A - x I ] = | 0 0-x 1 | = -x 3 | 0 0 0-x | The eigen values are 0,0,0.
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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.
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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 |
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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 |
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| -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 |
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| 9 | | 2 | | 5 | Xo = 50 |-12 | - 100 |-4 | + 100 | 4 | | 10 | | 5 | | 2 |
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| 9 | | 2 | | 5 | X n = 50 (-3/5) n |-12 | - 100 (-2/5) n |-4 | +100| 4 | | 10 | | 5 | | 2 | | 500 | The limiting situation is | 400 | | 200 |
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Theorem: Eigen Vectors for distinct Eigen Values are linearly independent.
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Proof: Suppose that c i, V i are eigen values and eigen vectors for distinct eigen values. Suppose that a 1 V 1 + a 2 V 2 + … + a n V n = 0 then a 1 c 1 n V 1 + a 2 c 2 n V 2 + … + a n c n n V n = 0 for all n. This can only happen when all a i are zero.
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