Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Matrix Calculations: Diagonalisation, Orthogonality, and Applications A. Kissinger Institute for Computing and Information Sciences Radboud University Nijmegen Version: autumn 2018 A. Kissinger Version: autumn 2018 Matrix Calculations 1 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Last time • Vectors look different in different bases, e.g. for: � 1 �� 1 � �� �� 1 � � 1 �� B = , C = , 1 − 1 1 2 • we have: � 1 � 2 � � 1 � � 2 = = 0 − 1 1 S C 2 B A. Kissinger Version: autumn 2018 Matrix Calculations 2 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Last time � 1 �� 1 � �� �� 1 � � 1 �� B = , C = , 1 − 1 1 2 • We can transform bases using basis transformation matrices. Going to standard basis is easy (basis elements are columns): � 1 � � 1 � 1 1 ❚ B⇒S = ❚ C⇒S = 1 − 1 1 2 • ...coming back means taking the inverse: � 1 � ❚ S⇒B = ( ❚ B⇒S ) − 1 = 1 1 1 − 1 2 � 2 � − 1 ❚ S⇒C = ( ❚ C⇒S ) − 1 = − 1 1 A. Kissinger Version: autumn 2018 Matrix Calculations 3 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Last time • The change of basis of a vector is computed by applying the matrix. For example, changing from S to B is: ✈ ′ = ❚ S⇒B · ✈ • The change of basis for a matrix is computed by surrounding it with basis-change matrices. • Changing from a matrix ❆ in S to a matrix ❆ ′ in B is: ❆ ′ = ❚ S⇒B · ❆ · ❚ B⇒S • (Memory aid: look at the first matrix after the equals sign to see what basis transformation you are doing.) A. Kissinger Version: autumn 2018 Matrix Calculations 4 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up • Many linear maps have their ‘own’ basis, their eigenbasis , which has the property that all basis elements ✈ ∈ B do this: ❆ · ✈ = λ ✈ • λ is called an eigenvalue, ✈ is called an eigenvector. • Eigenvalues are computed by solving: det( ❆ − λ ■ ) = 0 A. Kissinger Version: autumn 2018 Matrix Calculations 5 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Outline Eigenvectors and diagonalisation Inner products and orthogonality Wrapping up A. Kissinger Version: autumn 2018 Matrix Calculations 6 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Computing eigenvectors • For an n × n matrix, the equation det( ❆ − λ ■ ) = 0 has n solutions, which we’ll write as: λ 1 , λ 2 , . . . , λ n • (e.g. a 2 × 2 matrix involves solving a quadratic equation, which has 2 solutions λ 1 and λ 2 ) • For each of these solutions, we get a homogeneous system: ( ❆ − λ i ■ ) · ✈ i = 0 � �� � matrix • Solving this homogeneous system gives us the associated eigenvector ✈ i for the eigenvalue λ i A. Kissinger Version: autumn 2018 Matrix Calculations 8 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Example • This matrix: � 1 � − 2 ❆ = 0 − 1 • Has characteristic polynomial: � − λ + 1 � − 2 = λ 2 − 1 det 0 − λ − 1 • The equation λ 2 − 1 = 0 has 2 solutions : λ 1 = 1 and λ 2 = − 1. A. Kissinger Version: autumn 2018 Matrix Calculations 9 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Example • For λ 1 = 1, we get a homogeneous system: ( ❆ − λ 1 · ■ ) · ✈ 1 = 0 • Computing ( ❆ − (1) · ■ ): � 1 � � 1 � � 0 � − 2 0 − 2 − (1) · = 0 − 1 0 1 0 − 2 • So, we need to find a non-zero solution for: � 0 � − 2 · ✈ 1 = 0 0 − 2 (just like in lecture 2) � 0 � • This works: ✈ 1 = 1 A. Kissinger Version: autumn 2018 Matrix Calculations 10 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Example • For λ 2 = − 1, we get another homogeneous system: ( ❆ − λ 2 · ■ ) · ✈ 2 = 0 • Computing ( ❆ − ( − 1) · ■ ): � 1 � � 1 � � 2 � − 2 0 − 2 − ( − 1) · = 0 − 1 0 1 0 0 • So, we need to find a non-zero solution for: � 2 � − 2 · ✈ 2 = 0 0 0 � 1 � • This works: ✈ 2 = 1 A. Kissinger Version: autumn 2018 Matrix Calculations 11 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Example So, for the matrix ❆ , we computed 2 eigenvalue/eigenvector pairs: � 0 � λ 1 = 1 , ✈ 1 = 1 and � 1 � λ 2 = − 1 , ✈ 2 = 1 A. Kissinger Version: autumn 2018 Matrix Calculations 12 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Theorem If the eigenvalues of a matrix ❆ are all different, then their associated eigenvectors form a basis. Proof. We need to prove the ✈ i are all linearly independent. Then suppose (for contradiction) that ✈ 1 , . . . , ✈ n are linearly dependent , i.e.: c 1 ✈ 1 + c 2 ✈ 2 + . . . + c n ✈ n = 0 for k non-zero coefficients. Then, using that they are eigvectors: ❆ · ( c 1 ✈ 1 + . . . + c n ✈ n ) = 0 = ⇒ λ 1 c 1 ✈ 1 + . . . + λ n c n ✈ n = 0 1 Suppose c j � = 0, then subtract λ j times 2nd equation from the 1st equation: c 1 ✈ 1 + c 2 ✈ 2 + . . . + c n ✈ n − 1 λ j ( λ 1 c 1 ✈ 1 + . . . + λ n c n ✈ n ) = 0 This has k − 1 non-zero coefficients (because all the λ i ’s are distinct). Repeat until we have just 1 non-zero coefficient, and we have: c j ✈ k = 0 = ⇒ ✈ k = 0 but eigenvectors are always non-zero, so this is a contradiction. A. Kissinger Version: autumn 2018 Matrix Calculations 13 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Changing basis • Once we have a basis of eigenvectors B = { ✈ 1 , ✈ 2 , . . . , ✈ n } , translating to B gives us a diagonal matrix, whose diagonal entries are the eigenvalues: λ 1 0 0 0 0 λ 2 0 0 ❚ S⇒B · ❆ · ❚ B⇒S = ❉ where ❉ = 0 0 · · · 0 0 0 0 λ n • Going the other direction, we can always write ❆ in terms of a diagonal matrix: ❆ = ❚ B⇒S · ❉ · ❚ S⇒B A. Kissinger Version: autumn 2018 Matrix Calculations 14 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Definition For a matrix ❆ with eigenvalues λ 1 , . . . , λ n and eigenvectors B = { ✈ 1 , . . . , ✈ n } , decomposing ❆ as: λ 1 0 0 0 0 λ 2 0 0 ❆ = ❚ B⇒S · · ❚ S⇒B 0 0 · · · 0 0 0 0 λ n is called diagonalising the matrix ❆ . A. Kissinger Version: autumn 2018 Matrix Calculations 15 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Summary: diagonalising a matrix (study this slide!) We diagonalise a matrix ❆ as follows: 1 Compute each eigenvalue λ 1 , λ 2 , . . . , λ n by solving the characteristic polynomial 2 For each eigenvalue, compute the associated eigenvector ✈ i by solving the homogenious system ( ❆ − λ i ■ ) · ✈ i = 0 . 3 Write down ❆ as the product of three matrices: ❆ = ❚ B⇒S · ❉ · ❚ S⇒B where: • ❚ B⇒S has the eigenvectors ✈ 1 , . . . , ✈ n (in order!) as its columns • ❉ has the eigenvalues (in the same order!) down its diagonal, and zeroes everywhere else • ❚ S⇒B is the inverse of ❚ B⇒S . A. Kissinger Version: autumn 2018 Matrix Calculations 16 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Example: political swingers, part I • We take an extremely crude view on politics and distinguish only left and right wing political supporters • We study changes in political views, per year • Suppose we observe, for each year: • 80% of lefties remain lefties and 20% become righties • 90% of righties remain righties, and 10% become lefties Questions . . . • start with a population L = 100 , R = 150, and compute the number of lefties and righties after one year; • similarly, after 2 years, and 3 years, . . . • We can represent these computations conveniently using matrix multiplication. A. Kissinger Version: autumn 2018 Matrix Calculations 17 / 56
Eigenvectors and diagonalisation Inner products and orthogonality Radboud University Nijmegen Wrapping up Political swingers, part II • So if we start with a population L = 100 , R = 150, then after one year we have: • lefties: 0 . 8 · 100 + 0 . 1 · 150 = 80 + 15 = 95 • righties: 0 . 2 · 100 + 0 . 9 · 150 = 20 + 135 = 155 � � � � L 100 • If = , then after one year we have: R 150 � 95 � 100 � � 0 . 8 � � 100 � � 0 . 1 P · = · = 150 0 . 2 0 . 9 150 155 • After two years we have: � 95 � 95 � 91 . 5 � � 0 . 8 � � � 0 . 1 P · = · = 155 0 . 2 0 . 9 155 158 . 5 A. Kissinger Version: autumn 2018 Matrix Calculations 18 / 56
Recommend
More recommend
