Chapter 5 Eigenvalues and Eigenvectors
Section 5. Eigenvectors and Eigenvalues
Motivation: Difference equations A Biology Question How to predict a population of rabbits with given dynamics:. 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: f n f n = first-year rabbits in year n v n = s n where s n = second-year rabbits in year n t n = third-year rabbits in year n The dynamics say: v n+ t n { }} { {}} { f n+ 6s n + 8t n 0 6 8 f n s n+ = f n/2 = 0 0 s 2 n t n+ s n/2 0 t n Av n 2 0.
Motivation: Difference equations Continued This is a difference equation: Av n = v n+ If you know initial population v 0, what happens in 0 years v 0? Plug in a computer: v 0 v 0 v 2 9459 2434 9222 4729 3 577 27 3 7 3089 776 636 5095 9 844 388 6 4 6384 4096 024 32768 892 2048 Notice any patterns?. Each segment of the population essentially doubles every year: Av 2v 0. 2. The ratios get close to (6 : 4 : ): 6 v (big#) 4. New terms coming: eigenvalue, and eigenvector
Motivation: Difference equations Continued (2) We want a formula for vectors v 0, v, v 2,..., such that Av 0 = v Av = 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 v0 = λv 0 then v n = A n (Av 0) = λa n v 0 = λ 2 A n 2 v 0... = λ n v 0. It is much easier to compute v n = λ 0 v 0. Example 0 6 8 6 A = 0 0 2 v 0 = 4 Av 0 = 2v 0. 0 2 0 Starting with 6 baby rabbits, 4 first-year rabbits, and second-year rabbit: The population will exactly double every year, In 0 years, you will have 2 0 6 baby rabbits, 2 0 4 first-year rabbits, and 2 0 second-year rabbits.
Eigenvectors and Eigenvalues Definition Let A be an n n matrix.. An eigenvector of A is a nonzero vector v in R n such that Av = λv, 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 = λv 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 Multiply: 0 6 8 6 A = 0 0 2 v = 4 0 2 0 0 6 8 6 32 Av = 0 0 4 2 = 8 = 2v 0 2 2 0 Hence v is an eigenvector of A, with eigenvalue λ = 2. Example Multiply: A = Av = ( 2 ) 2 4 8 ( 2 2 4 8 ) ( ) = v = ( ) ( ) 4 = 4v 4 Hence v is an eigenvector of A, with eigenvalue λ = 4.
Poll Poll Which of the vectors ( ) ( ) A. B. C. are eigenvectors of the matrix ( ) D. ( )? ( ) 2 E. ( ) 0 0 ( ) ( ) ( = 2 ) ( ) ( ) ( ) = 0 ( ) ( ) ( ) = 0 ( ) ( ) ( 2 3 = 3) ( 0 0) eigenvector with eigenvalue 2 eigenvector with eigenvalue 0 eigenvector with eigenvalue 0 not an eigenvector is never an eigenvector
Verifying Eigenvalues ( ) 2 4 Question: Is λ = 3 an eigenvalue of A =? In other words, does Av = 3v Av 3v = 0 have a nontrivial solution? (A 3I )v = 0 We know how to answer that! Row reduction! ( ) ( ) ( ) ( ) 2 4 0 4 4 A 3I = 3 = 0 4 0 0 ( ) ( ) v 4 Parametric vector form: = v v 2. 2 ( ) 4 Then: Any nonzero multiple of is an eigenvector with eigenvalue λ = 3 Check one of them: ( ) ( ) ( ) ( ) 2 4 4 2 4 = = 3. 3 "
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: λ-eigenspace = { v in R n Av = λv } = { v in R n (A λi )v = 0 } = Nul ( A λi ). 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 6 A = 2 6. 2 8 2 6 A 2I = 2 6 2 6 row reduce parametric vector form basis 3 2 0 0 0 0 0 0 v 3 2 v 2 = v 2 + v 3 0 v 3 0 3 2, 0 0
Eigenspaces Picture This is how eigenvalues and eigenvectors make matrices easier to understand. 2 What does this 2-eigenspace look like? A basis is, 0 3 0. Av v v Av For any v in the 2-eigenspace, Av = 2v 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 v Av v is an eigenvector 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! v Which vectors don t move off their line v is an eigenvector with eigenvalue. Av 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! w Aw Which vectors don t move off their line w is an eigenvector with eigenvalue. 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 u 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! 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 -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 ( )-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.. λ is an eigenvalue of A if and only if (A λi )x = 0 has a nontrivial solution, if and only if Nul(A λi ) {0}. 2. Finding a basis for the λ-eigenspace of A means finding a basis for Nul(A λi ) as usual, through the general solution to (A λi )x = 0 (parametric vector form). 3. The eigenvectors with eigenvalue λ are the nonzero elements of Nul(A λi ) that is, the nontrivial solutions to (A λi )x = 0.
Some facts you can work out yourself Fact A is invertible if and only if 0 is not an eigenvalue of A. Fact 2 If v, v 2,..., v k are eigenvectors of A with distinct eigenvalues λ,..., λ k, then {v, v 2,..., v k } is linearly independent. Consequence of Fact 2 An n n matrix has at most n distinct eigenvalues. Why Fact? 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, then v 2 is contained in the λ -eigenspace. This is not true as v 2 does not have the same eigenvalue as v.
The Eigenvalues of a Triangular Matrix are the Diagonal Entries If we know λ is eigenvalue: easy to find eigenvectors (row reduction). And to find all eigenvalues? Will need to compute a determinant. Finding λ that has a non-trivial solution to (A λi )v = 0 boils down to finding λ that makes det(a λi ) = 0. Theorem The eigenvalues of a triangular matrix are the diagonal entries. Example 3 4 Find all eigenvalues of A = 0 2. 0 0 3 3 4 3 λ 4 A λi = 0 2 λi 3 = 0 λ 2 0 0 3 0 0 3 λ Since det(a λi ) = (3 λ) 2 ( λ), eigenvalues are λ = 3 and.