Table of Contents
CE 5143 Homework 5 Due 10/11/07

— David Wagner 2007/10/03 11:47

1. Problem 4.24

Using Eqs. (4.31) and (4.32), convert the following eigenvalue problem into a standard eigenvalue problem:

delim{[}{A}{]} vec{X} = lambda delim{[}{B}{]} vec{X} = delim{[}{matrix{2}{2}{6 {-2} {-2} 3}}{]} vec{X}=lambda delim{[}{matrix{2}{2}{8 2 2 4}}{]} vec{X}

Note both [A] and [B] are symmetric and positive definite.

delim{[}{B}{]} = delim{[}{U}{]}^T delim{[}{U}{]} by Choleski's Method [pg.178]

  1. u_{11}=sqrt{a_{11}}=sqrt{6} approx2.4495
  2. u_{12}={u_{12}}/{u_{11}} = -2/sqrt{6} approx-0.8165
  3. u_{22}=sqrt{a_22-{u_{12}}^2} }=sqrt{3-0.8165^2}approx2.3333
  4. u_{21}=0

delim{[}{U}{]} =delim{[}{matrix{2}{2}{2.4495 {-0.8165} 0 2.333}}{]}

  1. lambda_{ii}=1/{u_{ii}}
  2. lambda_{12}=-{{-0.8165*0.4286}/{2.4495}}approx0.1429

delim{[}{U}{]}^-1=delim{[}{matrix{2}{2}{0.4082 0.1429 0 0.4286}}{]}

(delim{[}{U}{]}^T )^{-1}=(delim{[}{U}{]}^{-1} )^T=delim{[}{matrix{2}{2}{0.4082 0 0.1429 0.4286}}{]}

(4.32)delim{[}{D}{]}=(delim{[}{U}{]}^T )^{-1} delim{[}{A}{]} delim{[}{U}{]}^{-1}= delim{[}{matrix{2}{2}{0.4082 0 0.1429 0.4286}}{]} delim{[}{matrix{2}{2}{6 {-2} {-2} 3}}{]} delim{[}{matrix{2}{2}{0.4082 0.1429 0 0.4286}}{]}= delim{[}{matrix{2}{2}{0.4082 0 0.1429 0.4286}}{]} delim{[}{matrix{2}{2}{2.4492 {-8571} {-0.8164} 12858}}{]}= delim{[}{matrix{2}{2}{1.0000 {-3498} 0.0000 4286}}{]}

(4.31)delim{[}{D}{]}vec{Y}=lambda vec{Y}= delim{[}{matrix{2}{2}{1.0000 {-3498} 0.0000 4286}}{]}vec{Y} = lambda vec{Y}

2. Problem 4.25

Derive the characteristic polynomial…by using Faddeev-Leverrier method. [text pg. 287]

delim{[}{A}{]} = delim{[}{matrix{4}{4}{ 5 7 6 5 7 10 8 7 6 8 10 9 5 7 9 10 }}{]}

Note the matrix is symmetric.

  1. [P_1]=[A]=delim{[}{matrix{4}{4}{ 5 7 6 5 7 10 8 7 6 8 10 9 5 7 9 10 }}{]}, p_1 = trace[P_1]=35;
  2. [P_2]=delim{[}{matrix{4}{4}{ 5 7 6 5 7 10 8 7 6 8 10 9 5 7 9 10 }}{]}delim{[}{matrix{4}{4}{ {-30} 7 6 5 7 {-25} 8 7 6 8 {-25} 9 5 7 9 {-25} }}{]}=delim{[}{matrix{4}{4}{ {-40} {-57} {-19} 3 {-57} {-88} {-15} 2 {-19} {-15} {-69} {-49} 3 2 {-49} {-95} }}{]}, p_2={1/2}trace[P_2]=-292/2=-146
  3. [P_3]=delim{[}{matrix{4}{4}{ 5 7 6 5 7 10 8 7 6 8 10 9 5 7 9 10 }}{]}delim{[}{matrix{4}{4}{ 106 {-57} {-19} 3 {-57} 58 {-15} 2 {-19} {-15} 77 {-49} 3 2 {-49} 51 }}{]}=delim{[}{matrix{4}{4}{ 32 41 17 {-10} 41 75 {-10} 6 17 {-10} 95 3 {-10} 6 3 98 }}{]}, p_3={1/3}trace[P_3]=300/3=100
  4. [P_4]=delim{[}{matrix{4}{4}{ 5 7 6 5 7 10 8 7 6 8 10 9 5 7 9 10 }}{]}delim{[}{matrix{4}{4}{ {-68} 41 17 {-10} 41 {-25} {-10} 6 17 {-10} {-5} 3 {-10} 6 3 {-2} }}{]}=delim{[}{matrix{4}{4}{ {-1} 0 0 0 0 {-1} 0 0 0 0 {-1} 0 0 0 0 {-1} }}{]}, p_4={1/4}trace[P_4]=-4/4=-1
lambda^4 -35 lambda^3 +146 lambda^2 -100 lambda +1 = 0

3. Problem 4.27

Find the eigenvalues and eigenvectors of the matrix

[A]=delim{[}{ matrix{2}{2}{1 0.5000 0.5000 0.3333} }{]}

Note this matrix is symmetric and positive definite.

[P_1]=[A]=delim{[}{ matrix{2}{2}{1 0.5000 0.5000 0.3333} }{]}, p_1=trace[P_1]=1.3333.

[P_2]=delim{[}{ matrix{2}{2}{1 0.5000 0.5000 0.3333} }{]} delim{[}{ matrix{2}{2}{{-0.3333} 0.5000 0.5000 {-1}} }{]}= delim{[}{ matrix{2}{2}{{-0.0833} 0.0000 0.0000 {-0.0833} } }{]}, p_2={1/2}trace[P_1]=-0.1667/2=-0.0833.

lambda^2 -1.3333 lambda +0.0833 = 0

lambda={-b pm sqrt{b^2 - 4ac}}/{2a} = {1.333 pm sqrt{1.3333^2-4*1*0.0833}}/{2}= {1.3333 pm sqrt{1.4444}}/{2}= {1.3333 pm 1.2018}/2= (0.1315,2.5351)

lambda_1=0.1315, lambda_2=2.5351

delim{[}{ matrix{2}{2}{1 0.5000 0.5000 0.3333} }{]} delim{lbrace}{ matrix{2}{1}{x_1 x_2} }{rbrace}= 0.1315 delim{lbrace}{ matrix{2}{1}{x_1 x_2} }{rbrace}

delim{[}{ matrix{2}{2}{0.8685 0.5000 0.5000 0.2018} }{]} delim{lbrace}{ matrix{2}{1}{x_1 x_2} }{rbrace}= delim{lbrace}{ matrix{2}{1}{0 0} }{rbrace}

  • 0.8685 x_1 +0.5x_2=0
  • 0.5 x_1 + 0.2018 x_2 =0

delim{[}{ matrix{2}{2}{1 0.5000 0.5000 0.3333} }{]} delim{lbrace}{ matrix{2}{1}{x_1 x_2} }{rbrace}= 2.5351 delim{lbrace}{ matrix{2}{1}{x_1 x_2} }{rbrace}

delim{[}{ matrix{2}{2}{{-1.5351} 0.5000 0.5000 {-2.2018}} }{]} delim{lbrace}{ matrix{2}{1}{x_1 x_2} }{rbrace}= delim{lbrace}{ matrix{2}{1}{0 0} }{rbrace}

  • -1.5351 x_1 +0.5x_2=0
  • 0.5 x_1 -2.2018 x_2 =0

4. Problem 4.32

  • m1 = 136 lb⋅sec²/in, m2 = 66 lb⋅sec²/in
  • k1 = 30,700 lb/in, k2 = 44,300 lb/in

Problem 5

Derive [A] = [L][L]T [See handout 3.50]

Discussion


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