Linear Algebra Chapter 8: Eigenvalues: Further Applications and - - PowerPoint PPT Presentation

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Linear Algebra Chapter 8: Eigenvalues: Further Applications and - - PowerPoint PPT Presentation

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Linear Algebra Chapter 8: Eigenvalues: Further Applications and Computations Section 8.1. Diagonalization of Quadratic FormsProofs of Theorems December 29, 2018 () Linear Algebra December 29, 2018 1 / 8 Table of contents Page 417 Number

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SLIDE 1

Linear Algebra

December 29, 2018 Chapter 8: Eigenvalues: Further Applications and Computations Section 8.1. Diagonalization of Quadratic Forms—Proofs of Theorems

() Linear Algebra December 29, 2018 1 / 8

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Table of contents

1

Page 417 Number 4(a)

2

Page 417 Number 4(b)

3

Page 417 Number 12

() Linear Algebra December 29, 2018 2 / 8

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SLIDE 3

Page 417 Number 4(a)

Page 417 Number 4(a)

Page 417 Number 4(a). Find the upper-triangular coefficient matrix of the quadratic form x2

1 − 2x2 2 + x2 3 + 6x2 4 − 2x1x4 + 6x2x4 − 8x1x3.

  • Solution. We take the coefficient of xixj where i ≤ j as uij. All uij = 0 for

i > j. So we have for the nonzero uij: u11 = 1, u22 = −2, u33 = 1, u44 = 6, u14 = −2, u24 = 6, u13 = −8. Hence U =     1 −8 −2 −2 6 1 6     .

  • ()

Linear Algebra December 29, 2018 3 / 8

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SLIDE 4

Page 417 Number 4(a)

Page 417 Number 4(a)

Page 417 Number 4(a). Find the upper-triangular coefficient matrix of the quadratic form x2

1 − 2x2 2 + x2 3 + 6x2 4 − 2x1x4 + 6x2x4 − 8x1x3.

  • Solution. We take the coefficient of xixj where i ≤ j as uij. All uij = 0 for

i > j. So we have for the nonzero uij: u11 = 1, u22 = −2, u33 = 1, u44 = 6, u14 = −2, u24 = 6, u13 = −8. Hence U =     1 −8 −2 −2 6 1 6     .

  • ()

Linear Algebra December 29, 2018 3 / 8

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SLIDE 5

Page 417 Number 4(b)

Page 417 Number 4(b)

Page 417 Number 4(b). Find the symmetric coefficient matrix of the quadratic form x2

1 − 2x2 2 + x2 3 + 6x2 4 − 2x1x4 + 6x2x4 − 8x1x3.

  • Solution. From part (a) we have the upper-triangular coefficient matrix

U =     1 −8 −2 −2 6 1 6     . We take aij = aji = uij/2 for i < j to get A =     1 −4 −1 −2 3 −4 1 −1 3 6     .

  • ()

Linear Algebra December 29, 2018 4 / 8

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SLIDE 6

Page 417 Number 4(b)

Page 417 Number 4(b)

Page 417 Number 4(b). Find the symmetric coefficient matrix of the quadratic form x2

1 − 2x2 2 + x2 3 + 6x2 4 − 2x1x4 + 6x2x4 − 8x1x3.

  • Solution. From part (a) we have the upper-triangular coefficient matrix

U =     1 −8 −2 −2 6 1 6     . We take aij = aji = uij/2 for i < j to get A =     1 −4 −1 −2 3 −4 1 −1 3 6     .

  • ()

Linear Algebra December 29, 2018 4 / 8

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SLIDE 7

Page 417 Number 12

Page 417 Number 12

Page 417 Number 12. Consider the quadratic form x2 + 2xy + y2. Find an orthogonal substitution that diagonalizes the quadratic form and find the diagonalized form.

  • Solution. For x2 + 2xy + y2, the upper-triangular coefficient matrix is

U = 1 2 1

  • , so the symmetric coefficient matrix is A =

1 1 1 1

  • .

() Linear Algebra December 29, 2018 5 / 8

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SLIDE 8

Page 417 Number 12

Page 417 Number 12

Page 417 Number 12. Consider the quadratic form x2 + 2xy + y2. Find an orthogonal substitution that diagonalizes the quadratic form and find the diagonalized form.

  • Solution. For x2 + 2xy + y2, the upper-triangular coefficient matrix is

U = 1 2 1

  • , so the symmetric coefficient matrix is A =

1 1 1 1

  • . We

need the eigenvalues of A, so consider det(A − λI) =

  • 1 − λ

1 1 1 − λ

  • = (1 − λ)2 − (1)2

1 − 2λ + λ2 − 1 = λ2 − 2λ = λ(λ − 2), so that the eigenvalues of A are λ1 = 0 and λ2 = 2. Now for the eigenvectors.

() Linear Algebra December 29, 2018 5 / 8

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SLIDE 9

Page 417 Number 12

Page 417 Number 12

Page 417 Number 12. Consider the quadratic form x2 + 2xy + y2. Find an orthogonal substitution that diagonalizes the quadratic form and find the diagonalized form.

  • Solution. For x2 + 2xy + y2, the upper-triangular coefficient matrix is

U = 1 2 1

  • , so the symmetric coefficient matrix is A =

1 1 1 1

  • . We

need the eigenvalues of A, so consider det(A − λI) =

  • 1 − λ

1 1 1 − λ

  • = (1 − λ)2 − (1)2

1 − 2λ + λ2 − 1 = λ2 − 2λ = λ(λ − 2), so that the eigenvalues of A are λ1 = 0 and λ2 = 2. Now for the eigenvectors.

() Linear Algebra December 29, 2018 5 / 8

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SLIDE 10

Page 417 Number 12

Page 417 Number 12 (continued 1)

Solution (continued). λ1 = 0 We consider the system [A − λ1I | 0] = 1 1 1 1 R2→R2−R1

  • 1

1

  • .

So we need v1 + v2 = 0 0 = 0 or v1 = −v2 v2 = v2

  • r with r = v2 as a free variable,
  • v1 = r

−1 1

  • where r ∈ R, r = 0. We need a unit eigenvector so we take

r = 1/ √ 2 and v1 =

1 √ 2

−1 1

  • =

−1/ √ 2 1/ √ 2

  • .

λ2 = 2 We consider the system [A − λ2I | 0] = −1 1 1 −1 R1→R1+R2

  • 1

−1

  • ()

Linear Algebra December 29, 2018 6 / 8

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SLIDE 11

Page 417 Number 12

Page 417 Number 12 (continued 1)

Solution (continued). λ1 = 0 We consider the system [A − λ1I | 0] = 1 1 1 1 R2→R2−R1

  • 1

1

  • .

So we need v1 + v2 = 0 0 = 0 or v1 = −v2 v2 = v2

  • r with r = v2 as a free variable,
  • v1 = r

−1 1

  • where r ∈ R, r = 0. We need a unit eigenvector so we take

r = 1/ √ 2 and v1 =

1 √ 2

−1 1

  • =

−1/ √ 2 1/ √ 2

  • .

λ2 = 2 We consider the system [A − λ2I | 0] = −1 1 1 −1 R1→R1+R2

  • 1

−1

  • ()

Linear Algebra December 29, 2018 6 / 8

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SLIDE 12

Page 417 Number 12

Page 417 Number 12 (continued 2)

Solution (continued).

R1↔R2

  • 1

−1

  • .

So we need v1 − v2 = 0 0 = 0 or v1 = v2 v2 = v2

  • r with s = v2 as a free variable,
  • v2 = s

1 1

  • where s ∈ R, s = 0. We need a unit eigenvector so we take

2 = 1/ √ 2 and v2 =

1 √ 2

1 1

  • =

1/ √ 2 1/ √ 2

  • . So we consider the
  • rthogonal matrix

C = [ v1 v2] = −1/ √ 2 1/ √ 2 1/ √ 2 1/ √ 2

  • .

() Linear Algebra December 29, 2018 7 / 8

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SLIDE 13

Page 417 Number 12

Page 417 Number 12 (continued 2)

Solution (continued).

R1↔R2

  • 1

−1

  • .

So we need v1 − v2 = 0 0 = 0 or v1 = v2 v2 = v2

  • r with s = v2 as a free variable,
  • v2 = s

1 1

  • where s ∈ R, s = 0. We need a unit eigenvector so we take

2 = 1/ √ 2 and v2 =

1 √ 2

1 1

  • =

1/ √ 2 1/ √ 2

  • . So we consider the
  • rthogonal matrix

C = [ v1 v2] = −1/ √ 2 1/ √ 2 1/ √ 2 1/ √ 2

  • .

() Linear Algebra December 29, 2018 7 / 8

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SLIDE 14

Page 417 Number 12

Page 417 Number 12 (continued 3)

Solution (continued). The orthogonal substitution is

  • x =

x y

  • = C

t = −1/ √ 2 1/ √ 2 1/ √ 2 1/ √ 2 t1 t2

  • =

(−1/ √ 2)t1 + (1/ √ 2)t2 (1/ √ 2)t1 + (1/ √ 2)t2

  • .

The diagonal form is then

  • xTA

x = λ1t2

1 + λ2t2 2 = 0t2 1 + 2t2 2 =

2t2

2.

  • ()

Linear Algebra December 29, 2018 8 / 8

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SLIDE 15

Page 417 Number 12

Page 417 Number 12 (continued 3)

Solution (continued). The orthogonal substitution is

  • x =

x y

  • = C

t = −1/ √ 2 1/ √ 2 1/ √ 2 1/ √ 2 t1 t2

  • =

(−1/ √ 2)t1 + (1/ √ 2)t2 (1/ √ 2)t1 + (1/ √ 2)t2

  • .

The diagonal form is then

  • xTA

x = λ1t2

1 + λ2t2 2 = 0t2 1 + 2t2 2 =

2t2

2.

  • ()

Linear Algebra December 29, 2018 8 / 8