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

Linear Algebra Chapter 8: Eigenvalues: Further Applications and Computations Section 8.1. Diagonalization of Quadratic Forms—Proofs of Theorems December 29, 2018 () Linear Algebra December 29, 2018 1 / 8

Table of contents Page 417 Number 4(a) 1 Page 417 Number 4(b) 2 Page 417 Number 12 3 () Linear Algebra December 29, 2018 2 / 8

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 x 2 1 − 2 x 2 2 + x 2 3 + 6 x 2 4 − 2 x 1 x 4 + 6 x 2 x 4 − 8 x 1 x 3 . Solution. We take the coefficient of x i x j where i ≤ j as u ij . All u ij = 0 for i > j . So we have for the nonzero u ij : u 11 = 1, u 22 = − 2, u 33 = 1, u 44 = 6, u 14 = − 2, u 24 = 6, u 13 = − 8. Hence  1 0 − 8 − 2  0 − 2 0 6   U =  .   0 0 1 0  0 0 0 6 � () Linear Algebra December 29, 2018 3 / 8

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 x 2 1 − 2 x 2 2 + x 2 3 + 6 x 2 4 − 2 x 1 x 4 + 6 x 2 x 4 − 8 x 1 x 3 . Solution. We take the coefficient of x i x j where i ≤ j as u ij . All u ij = 0 for i > j . So we have for the nonzero u ij : u 11 = 1, u 22 = − 2, u 33 = 1, u 44 = 6, u 14 = − 2, u 24 = 6, u 13 = − 8. Hence  1 0 − 8 − 2  0 − 2 0 6   U =  .   0 0 1 0  0 0 0 6 � () Linear Algebra December 29, 2018 3 / 8

Page 417 Number 4(b) Page 417 Number 4(b) Page 417 Number 4(b). Find the symmetric coefficient matrix of the quadratic form x 2 1 − 2 x 2 2 + x 2 3 + 6 x 2 4 − 2 x 1 x 4 + 6 x 2 x 4 − 8 x 1 x 3 . Solution. From part (a) we have the upper-triangular coefficient matrix  1 0 − 8 − 2  0 − 2 0 6   U =  .   0 0 1 0  0 0 0 6 We take a ij = a ji = u ij / 2 for i < j to get  1 0 − 4 − 1  0 − 2 0 3   A =  .   − 4 0 1 0  − 1 3 0 6 � () Linear Algebra December 29, 2018 4 / 8

Page 417 Number 4(b) Page 417 Number 4(b) Page 417 Number 4(b). Find the symmetric coefficient matrix of the quadratic form x 2 1 − 2 x 2 2 + x 2 3 + 6 x 2 4 − 2 x 1 x 4 + 6 x 2 x 4 − 8 x 1 x 3 . Solution. From part (a) we have the upper-triangular coefficient matrix  1 0 − 8 − 2  0 − 2 0 6   U =  .   0 0 1 0  0 0 0 6 We take a ij = a ji = u ij / 2 for i < j to get  1 0 − 4 − 1  0 − 2 0 3   A =  .   − 4 0 1 0  − 1 3 0 6 � () Linear Algebra December 29, 2018 4 / 8

Page 417 Number 12 Page 417 Number 12 Page 417 Number 12. Consider the quadratic form x 2 + 2 xy + y 2 . Find an orthogonal substitution that diagonalizes the quadratic form and find the diagonalized form. Solution. For x 2 + 2 xy + y 2 , the upper-triangular coefficient matrix is � 1 � 1 � � 2 1 U = , so the symmetric coefficient matrix is A = . 0 1 1 1 () Linear Algebra December 29, 2018 5 / 8

Page 417 Number 12 Page 417 Number 12 Page 417 Number 12. Consider the quadratic form x 2 + 2 xy + y 2 . Find an orthogonal substitution that diagonalizes the quadratic form and find the diagonalized form. Solution. For x 2 + 2 xy + y 2 , the upper-triangular coefficient matrix is � 1 � 1 � � 2 1 U = , so the symmetric coefficient matrix is A = . We 0 1 1 1 need the eigenvalues of A , so consider � � 1 − λ 1 � = (1 − λ ) 2 − (1) 2 � � det( A − λ I ) = � � 1 1 − λ � 1 − 2 λ + λ 2 − 1 = λ 2 − 2 λ = λ ( λ − 2) , so that the eigenvalues of A are λ 1 = 0 and λ 2 = 2. Now for the eigenvectors. () Linear Algebra December 29, 2018 5 / 8

Page 417 Number 12 Page 417 Number 12 Page 417 Number 12. Consider the quadratic form x 2 + 2 xy + y 2 . Find an orthogonal substitution that diagonalizes the quadratic form and find the diagonalized form. Solution. For x 2 + 2 xy + y 2 , the upper-triangular coefficient matrix is � 1 � 1 � � 2 1 U = , so the symmetric coefficient matrix is A = . We 0 1 1 1 need the eigenvalues of A , so consider � � 1 − λ 1 � = (1 − λ ) 2 − (1) 2 � � det( A − λ I ) = � � 1 1 − λ � 1 − 2 λ + λ 2 − 1 = λ 2 − 2 λ = λ ( λ − 2) , so that the eigenvalues of A are λ 1 = 0 and λ 2 = 2. Now for the eigenvectors. () Linear Algebra December 29, 2018 5 / 8

Page 417 Number 12 Page 417 Number 12 (continued 1) Solution (continued). λ 1 = 0 We consider the system � 1 � 1 � R 2 → R 2 − R 1 � 1 0 1 0 [ A − λ 1 I | � � 0] = . 1 1 0 0 0 0 So we need v 1 + v 2 = 0 0 = 0 or v 1 = − v 2 or with r = v 2 as a free variable, v 2 = v 2 � − 1 � where r ∈ R , r � = 0. We need a unit eigenvector so we take v 1 = r � 1 � − 1 � − 1 / √ √ � � 2 1 √ 2 and � r = 1 / v 1 = = . √ 1 2 1 / 2 λ 2 = 2 We consider the system � − 1 � 0 � R 1 → R 1 + R 2 � 1 0 0 0 [ A − λ 2 I | � � 0] = 1 − 1 0 1 − 1 0 () Linear Algebra December 29, 2018 6 / 8

Page 417 Number 12 Page 417 Number 12 (continued 1) Solution (continued). λ 1 = 0 We consider the system � 1 � 1 � R 2 → R 2 − R 1 � 1 0 1 0 [ A − λ 1 I | � � 0] = . 1 1 0 0 0 0 So we need v 1 + v 2 = 0 0 = 0 or v 1 = − v 2 or with r = v 2 as a free variable, v 2 = v 2 � − 1 � where r ∈ R , r � = 0. We need a unit eigenvector so we take v 1 = r � 1 � − 1 � − 1 / √ √ � � 2 1 √ 2 and � r = 1 / v 1 = = . √ 1 2 1 / 2 λ 2 = 2 We consider the system � − 1 � 0 � R 1 → R 1 + R 2 � 1 0 0 0 [ A − λ 2 I | � � 0] = 1 − 1 0 1 − 1 0 () Linear Algebra December 29, 2018 6 / 8

Page 417 Number 12 Page 417 Number 12 (continued 2) Solution (continued). � 1 R 1 ↔ R 2 − 1 0 � � . 0 0 0 So we need v 1 − v 2 = 0 0 = 0 or v 1 = v 2 or with s = v 2 as a free variable, v 2 = v 2 � 1 � where s ∈ R , s � = 0. We need a unit eigenvector so we take � v 2 = s 1 √ � 1 � 1 / √ � � 2 1 √ 2 = 1 / 2 and � v 2 = = . So we consider the √ 2 1 1 / 2 orthogonal matrix � − 1 / √ √ � 2 1 / 2 √ √ C = [ � v 1 � v 2 ] = . 1 / 2 1 / 2 () Linear Algebra December 29, 2018 7 / 8

Page 417 Number 12 Page 417 Number 12 (continued 2) Solution (continued). � 1 R 1 ↔ R 2 − 1 0 � � . 0 0 0 So we need v 1 − v 2 = 0 0 = 0 or v 1 = v 2 or with s = v 2 as a free variable, v 2 = v 2 � 1 � where s ∈ R , s � = 0. We need a unit eigenvector so we take � v 2 = s 1 √ � 1 � 1 / √ � � 2 1 √ 2 = 1 / 2 and � v 2 = = . So we consider the √ 2 1 1 / 2 orthogonal matrix � − 1 / √ √ � 2 1 / 2 √ √ C = [ � v 1 � v 2 ] = . 1 / 2 1 / 2 () Linear Algebra December 29, 2018 7 / 8

Page 417 Number 12 Page 417 Number 12 (continued 3) Solution (continued). The orthogonal substitution is � x � − 1 / √ √ � � t 1 � � 2 1 / 2 = C � √ √ x = � t = y 1 / 2 1 / 2 t 2 � ( − 1 / √ √ � 2) t 1 + (1 / 2) t 2 √ √ = . (1 / 2) t 1 + (1 / 2) t 2 The diagonal form is then x = λ 1 t 2 1 + λ 2 t 2 2 = 0 t 2 1 + 2 t 2 x T A � 2 t 2 � 2 = 2 . � () Linear Algebra December 29, 2018 8 / 8

Page 417 Number 12 Page 417 Number 12 (continued 3) Solution (continued). The orthogonal substitution is � x � − 1 / √ √ � � t 1 � � 2 1 / 2 = C � √ √ x = � t = y 1 / 2 1 / 2 t 2 � ( − 1 / √ √ � 2) t 1 + (1 / 2) t 2 √ √ = . (1 / 2) t 1 + (1 / 2) t 2 The diagonal form is then x = λ 1 t 2 1 + λ 2 t 2 2 = 0 t 2 1 + 2 t 2 x T A � 2 t 2 � 2 = 2 . � () Linear Algebra December 29, 2018 8 / 8