Chapter 5 Eigenvalues and Eigenvectors Section 5.1 Eigenvectors - PowerPoint PPT Presentation

Chapter 5 Eigenvalues and Eigenvectors

Section 5.1 Eigenvectors and Eigenvalues

Motivation: Difference equations A Biology Question How to predict a population of rabbits with given dynamics : 1. half of the newborn rabbits survive their first year; 2. of those, half survive their second year; 3. their maximum life span is three years; 4. Each rabbit gets 0 , 6 , 8 baby rabbits in their three years, respectively. Approach: Each year, count the population by age :    = first-year rabbits in year n f n f n    where v n = s n = second-year rabbits in year n s n   t n  = third-year rabbits in year n t n The dynamics say : v n +1 Av n � �� � � �� �         f n +1 6 s n + 8 t n 0 6 8 f n  =  = 1 s n +1 f n / 2 0 0 s n       2 1 s n / 2 0 0 t n +1 t n 2 .

Motivation: Difference equations Continued This is a difference equation : Av n = v n +1 If you know initial population v 0 , what happens in 10 years v 10 ? Plug in a computer: Notice any patterns? v 0 v 10 v 11 1. Each segment of the       1 9459 19222 population essentially doubles 2 2434 4729       every year: Av 11 ≈ 2 v 10 . 3 577 1217 2. The ratios get close to       3 30189 61316 (16 : 4 : 1): 7 7761 15095         16 9 1844 3881 v 11 ≈ (big#) · 4  .        16 16384 32768 1 4 4096 8192       1 1024 2048 New terms coming: eigenvalue, and eigenvector

Motivation: Difference equations Continued (2) We want a formula for vectors v 0 , v 1 , v 2 , . . . , such that Av 0 = v 1 Av 1 = v 2 Av 2 = v 3 . . . We can see that v n = A n v 0 . But multiplying by A each time is inefficient! If v 0 satisfies A v 0 = λ v 0 then v n = A n − 1 ( Av 0 ) = λ A n − 1 v 0 = λ 2 A n − 2 v 0 . . . = λ n v 0 . It is much easier to compute v n = λ 10 v 0 . Example     0 6 8 16 1 A = 0 0 v 0 = 4 Av 0 = 2 v 0 .     2 1 0 0 1 2 Starting with 16 baby rabbits, 4 first-year rabbits, and 1 second-year rabbit: ◮ The population will exactly double every year, ◮ In 10 years, you will have 2 10 · 16 baby rabbits, 2 10 · 4 first-year rabbits, and 2 10 second-year rabbits.

Eigenvectors and Eigenvalues This is the most important definition in the course. Definition Let A be an n × n matrix. 1. An eigenvector of A is a nonzero vector v in R n such that Av = λ v , for some λ in R . In other words, Av is a multiple of v . 2. We say that the number λ is the eigenvalue for v , and v is an eigenvector for λ . 3. Alternatively, λ in R is an eigenvalue of A if the equation Av = λ v has a nontrivial solution . Notes: ◮ Eigenvalues and eigenvectors are only for square matrices. ◮ Eigenvectors are by definition nonzero. Eigenvalues may be equal to zero.

Verifying Eigenvectors Example     0 6 8 16 1 A = 0 0 v = 4     2 1 0 0 1 2 Multiply:       0 6 8 16 32 1  =  = 2 v Av = 0 0 4 8     2 1 0 0 1 2 2 Hence v is an eigenvector of A , with eigenvalue λ = 2. Example � 2 � � 1 � 2 A = v = − 4 8 1 Multiply: � 2 � � 1 � � 4 � 2 Av = = = 4 v − 4 8 1 4 Hence v is an eigenvector of A , with eigenvalue λ = 4.

Poll Poll Which of the vectors � 1 � 1 � � � − 1 � � 2 � � 0 � A. B. C. D. E. 1 − 1 1 1 0 � 1 � 1 are eigenvectors of the matrix ? 1 1 � � � � � � 1 1 1 1 = 2 eigenvector with eigenvalue 2 1 1 1 1 � � � � � � 1 1 1 1 = 0 eigenvector with eigenvalue 0 1 1 − 1 − 1 � � � � � � 1 1 − 1 − 1 = 0 eigenvector with eigenvalue 0 1 1 1 1 � � � � � � 1 1 2 3 = not an eigenvector 1 1 1 3 � � 0 is never an eigenvector 0

Verifying Eigenvalues � 2 � − 4 Question: Is λ = 3 an eigenvalue of A = ? − 1 − 1 In other words, does  Av = 3 v   Av − 3 v = 0 have a nontrivial solution?   ( A − 3 I ) v = 0 We know how to answer that! Row reduction! � 2 � 1 � − 1 � 1 � � � � − 4 0 − 4 4 A − 3 I = − 3 = − 1 − 1 0 1 − 1 − 4 0 0 � v 1 � � − 4 � Parametric vector form : = v 2 . v 2 1 � − 4 � Then: Any nonzero multiple of is an eigenvector with eigenvalue λ = 3 1 Check one of them: � 2 � � − 4 � � − 12 � � − 4 � − 4 ✧ = = 3 . − 1 − 1 1 3 1

Eigenspaces Definition Let A be an n × n matrix and let λ be an eigenvalue of A . The λ -eigenspace of A is the set of all eigenvectors of A with eigenvalue λ , plus the zero vector: v in R n | Av = λ v � � λ -eigenspace = v in R n | ( A − λ I ) v = 0 � � = � � = Nul A − λ I . The λ -eigenspace is a subspace of R n . How to find a basis? Parametric vector form!

Eigenspaces Example Find a basis for the 2-eigenspace of   λ 4 − 1 6  . A = 2 1 6  2 − 1 8     − 1 2 − 1 6 1 3 row reduce 2 A − 2 I = 2 − 1 6 0 0 0     2 − 1 6 0 0 0 parametric vector       1 v 1 − 3 form 2  = v 2  + v 3 v 2 1 0     v 3 0 1       1 − 3 basis   2  , 1 0      0 1

Eigenspaces Picture This is how eigenvalues and eigenvectors make matrices easier to understand.       1 − 3   2 What does this 2-eigenspace look like? A basis is 1  , 0  .     0 1 v v Av Av For any v in the 2-eigenspace, Av = 2 v by definition. This means, on its 2-eigenspace, A acts by scaling by 2.

Geometrically Eigenvectors An eigenvector of a matrix A is a nonzero vector v such that: ◮ Av is a multiple of v , which means ◮ Av is on the same line as v . Aw w Av v is an eigenvector v w is not an eigenvector

Eigenspaces Geometry; example Let T : R 2 → R 2 be reflection over the line L defined by y = − x , and let A be the matrix for T . Question: Eigenvalues and eigenspaces of A ? No computations! Which vectors don’t move off their line v v is an eigenvector with eigenvalue − 1. L Av

Eigenspaces Geometry; example Let T : R 2 → R 2 be reflection over the line L defined by y = − x , and let A be the matrix for T . Question: Eigenvalues and eigenspaces of A ? No computations! Which vectors don’t move off their line wAw w is an eigenvector with eigenvalue 1. L

Eigenspaces Geometry; example Let T : R 2 → R 2 be reflection over the line L defined by y = − x , and let A be the matrix for T . Question: Eigenvalues and eigenspaces of A ? No computations! Which vectors don’t move off their line u is not an eigenvector. Au L u

Eigenspaces Geometry; example Let T : R 2 → R 2 be reflection over the line L defined by y = − x , and let A be the matrix for T . Question: Eigenvalues and eigenspaces of A ? No computations! z Which vectors don’t move off their line Neither is z . Az L

Eigenspaces Geometry; example Let T : R 2 → R 2 be reflection over the line L defined by y = − x , and let A be the matrix for T . Question: Eigenvalues and eigenspaces of A ? No computations! Which vectors don’t move off their line The 1-eigenspace is L (all the vectors x where Ax = x ). L

Eigenspaces Geometry; example Let T : R 2 → R 2 be reflection over the line L defined by y = − x , and let A be the matrix for T . Question: Eigenvalues and eigenspaces of A ? No computations! Which vectors don’t move off their line The ( − 1)-eigenspace is the line y = x (all the vectors x where Ax = − x ). L

Eigenspaces Summary Let A be an n × n matrix and let λ be a number. 1. λ is an eigenvalue of A if and only if ( A − λ I ) x = 0 has a nontrivial solution , if and only if Nul( A − λ I ) � = { 0 } . 2. Finding a basis for the λ -eigenspace of A means finding a basis for Nul( A − λ I ) as usual, through the general solution to ( A − λ I ) x = 0 (parametric vector form). 3. The eigenvectors with eigenvalue λ are the nonzero elements of Nul( A − λ I ) that is, the nontrivial solutions to ( A − λ I ) x = 0.

Some facts you can work out yourself Fact 1 A is invertible if and only if 0 is not an eigenvalue of A . Fact 2 If v 1 , v 2 , . . . , v k are eigenvectors of A with distinct eigenvalues λ 1 , . . . , λ k , then { v 1 , v 2 , . . . , v k } is linearly independent . Consequence of Fact 2 An n × n matrix has at most n distinct eigenvalues. Why Fact 1? 0 is an eigenvalue of A ⇐ ⇒ Ax = 0 has a nontrivial solution ⇐ ⇒ A is not invertible. Why Fact 2 (for two vectors)? If v 2 is a multiple of v 1 , then v 2 is contained in the λ 1 -eigenspace. This is not true as v 2 does not have the same eigenvalue as v 1 .