MAC Module 12 Eigenvalues and Eigenvectors

Transcription

1 MAC 23 Module 2 Eigenvalues and Eigenvectos

2 Leaning Objectives Upon completing this module, you should be able to:. Solve the eigenvalue poblem by finding the eigenvalues and the coesponding eigenvectos of an n x n matix. Find the algebaic multiplicity and the geometic multiplicity of an eigenvalue. 2. Find a basis fo each eigenspace of an eigenvalue. 3. Detemine whethe a matix A is diagonalizable. 4. Find a matix P, P -, and D that diagonalize A if A is diagonalizable. 5. Find an othogonal matix P with P - = P T and D that diagonalize A if A is symmetic and diagonalizable. 6. Detemine the powe and the eigenvalues of a matix, A k. Rev.F9 2

3 Eigenvalues and Eigenvectos The majo topics in this module: Eigenvalues, Eigenvectos, Eigenspace, Diagonalization and Othogonal Diagonalization Rev.9 3

4 What ae Eigenvalues and Eigenvectos? If A is an n x n matix and λ is a scala fo which Ax = λx has a nontivial solution x Rⁿ, then λ is an eigenvalue of A and x is a coesponding eigenvecto of A. Ax = λx is called the eigenvalue poblem fo A. Note that we can ewite the equation Ax = λx = λi n x as follows: λi n x - Ax = o (λi n - A)x =. x = is the tivial solution. But ou solutions must be nonzeo vectos called eigenvectos that coespond to each of the distinct eigenvalues. Rev.F9 4

5 What ae Eigenvalues and Eigenvectos? Since we seek a nontivial solution to (λi n - A)x = (λi - A)x =, λi - A must be singula to have solutions x. This means that the det(λi - A) =. The det(λi - A) = p(λ) = is the chaacteistic equation, whee det(λi - A) = p(λ) is the chaacteistic polynomial. The deg(p(λ)) = n and the n oots of p(λ), λ, λ 2,,λ n, ae the eigenvalues of A. The polynomial p(λ) always has n oots, so the zeos always exist; but some may be complex and some may be epeated. In ou examples, all of the oots will be eal. Fo each λ i we solve fo x i = p i the coesponding eigenvecto, and Ap i = λ i p i fo each distinct eigenvalue. Rev.F9 5

6 Rev.F9 How to Solve the Eigenvalue Poblem, Ax = λx? Example : Find the eigenvalues and the coesponding eigenvectos of A. 3 2 A 5 Step : Find the chaacteistic equation of A and solve fo its eigenvalues. p() I A ( 3) (2)(5) 2 3 ( 5) ( 2) Thus, the eigenvalues ae 5, 2 2. Each eigenvalue has algebaic multiplicity. Let D which is a diagonal matix. 6

7 How to Solve the Eigenvalue Poblem, Ax = λx?(cont.) Step 2: Use Gaussian elimination with back-substitution to solve the (λi - A) x = fo λ and λ 2. The augmented matix fo the system with λ = -5: 5I A ~ 2 2 The second column is not a leading column, so x 2 = t is a fee vaiable, and x = x 2 = t. Thus, the solution coesponding to λ = -5 is Rev.F9 x x x t t : t 5 2 2,t. 7.

8 How to Solve the Eigenvalue Poblem, Ax = λx? (Cont.) Since t is a fee vaiable, thee ae infinitely many eigenvectos. Fo convenience, we choose t =, and x p as the eigenvecto fo λ = -5. If we want p to be a unit vecto, we will choose t so that / 2 Howeve, t = is fine in this p poblem. / 2. The augmented matix fo the system with λ 2 = 2: 2I A : Rev.F :

9 How to Solve the Eigenvalue Poblem, Ax = λx? (Cont.) Again, the second column is not a leading column, so x 2 = t is a fee vaiable, and x = - 2x 2 / 5 = - 2t / 5. Thus, the solution coesponding to λ = 2 is x x x 2 2 t 5 t t 2 5,t. Fo convenience, we choose t = 5 and x p as the eigenvecto fo λ = 2. Alight, we have finished solving the eigenvalue poblem fo 3 2 A. 5 Rev.F9 9

10 Eigenspaces Fo each eigenvalue λ, thee is an eigenspace E λ with a basis fomed fom the linealy independent eigenvectos fo λ. The dim(e λ ) is the geometic multiplicity of λ, which is the numbe of linealy independent eigenvectos associated with λ. We will see that the geometic multiplicity equals the algebaic multiplicity fo each eigenvalue. B { p } is a basis fo E, the eigenspace of, and B 2 { p 2 } is a basis fo E 2, the eigenspace of 2. So, E span(b ) span({ p }) span E 2 span(b 2 ) span({ p 2 }) span 2 5 and dim(e ) dim(e 2 ). T T, Rev.F9

11 How to Detemine if a Matix A is Diagonalizable? If P = [ p p 2 ], then AP = PD even if A is not diagonalizable. Since AP p 2 p () (2) 5() 2(5) p p2 5 2 A p p2 A p Ap PD. Rev.F9

12 How to Detemine if a Matix A is Diagonalizable? So, ou two distinct eigenvalues both have algebaic multiplicity and geometic multiplicity. This ensues that p and p 2 ae not scala multiples of each othe; thus, p and p 2 ae linealy independent eigenvectos of A. Since A is 2 x 2 and thee ae two linealy independent eigenvectos fom the solution of the eigenvalue poblem, A is diagonalizable and P - AP = D. We can now constuct P, P - and D. Let P Then, P Rev.F9 p p2 5 / 7 2 / 7 / 7 / and D

13 How to Detemine if a Matix A is Diagonalizable? (Cont.) Note that, if we multiply both sides on the left by P, then AP A Rev.F9 5 P AP p p / 7 2 / 7 / 7 / 7 A p Ap / 7 2 / 7 / 7 / p p2 2 5 p p / 7 / 7 / 7 4 / () 2(2) 5() 2(5) PD becomes D.

14 How to Detemine if a Matix A is Diagonalizable? (Cont.) Example 2: Find the eigenvalues and eigenvectos fo A. 3 4 A 3 Step : Find the eigenvalues fo A. Recall: The deteminant of a tiangula matix is the poduct of the elements at the diagonal. Thus, the chaacteistic equation of A is λ = has algebaic multiplicity and λ 2 = 3 has algebaic multiplicity 2.. Rev.F9 p() det(i A) I A ( 3) 2 ( ). 4

15 How to Detemine if a Matix A is Diagonalizable? (Cont.) Step 2: Use Gaussian elimination with back-substitution to solve (λi - A) x =. Fo λ =, the augmented matix fo the system is I A : :. Column 3 is not a leading column and x 3 = t is a fee vaiable. The geometic multiplicity of λ = is one, since thee is only one fee vaiable. x 2 = and x = 2x 2 = Rev.F9 5

16 How to Detemine if a Matix A is Diagonalizable? (Cont.) The eigenvecto coesponding to λ = is x x x 2 x 3 t t ou choice fo the eigenvecto.. If we choose t, then p B { p } is a basis fo the eigenspace, E, with dim(e ). The dimension of the eigenspace is because the eigenvalue has only one linealy independent eigenvecto. Thus, the geometic multiplicity is and the algebaic multiplicity is fo λ =. is Rev.F9 6

17 How to Detemine if a Matix A is Diagonalizable? (Cont.) The augmented matix fo the system with λ 2 = 3 is 3I A 4 2 : : : Column is not a leading column and x = t is a fee vaiable. Since thee is only one fee vaiable, the geometic multiplicity of λ 2 is one. Rev.F9 7

18 How to Detemine if a Matix A is Diagonalizable? (Cont.) x 2 = x 3 = and the eigenvecto coesponding to λ 2 = 3 is x t x x 2 t,we choose t, and p 2 is x 3 ou choice fo the eigenvecto. B 2 { p 2 } is a basis fo the eigenspace, E 2, with dim(e 2 ). The dimension of the eigenspace is because the eigenvalue has only one linealy independent eigenvecto. Thus, the geometic multiplicity is while the algebaic multiplicity is 2 fo λ 2 = 3. This means thee will not be enough linealy independent eigenvectos fo A to be diagonalizable. Thus, A is not diagonalizable wheneve the geometic multiplicity is less than the algebaic multiplicity fo any eigenvalue. Rev.F9 8

19 How to Detemine if a Matix A is Diagonalizable? (Cont.) This time, AP A p 2 p2 2 p2 p p2 p2 p p2 p2 A p Ap 2 Ap PD. P does not exist since the columns of P ae not linealy independent. It is not possible to solve fo D P AP, so A is not diagonalizable. Rev.F9 9

20 A is Diagonalizable iff A is Simila to a Diagonal Matix Fo A, an n x n matix, with chaacteistic polynomial oots, 2,..., n, then AP A p p2... p n A p Ap 2... Ap n p 2 p2... n pn PD p p2... p n O n fo eigenvalues λ i of A with coesponding eigenvectos p i. P is invetible iff the eigenvectos that fom its columns ae linealy independent iff dim(e i ) geometic multiplicity algbaic multiplicity fo each distinct i. Rev.F9 2

21 A is Diagonalizable iff A is Simila to a Diagonal Matix (Cont.) This gives us n linealy independent eigenvectos fo P, so P - exists. Theefoe, A is diagonalizable since P AP P PD D O n. The squae matices S and T ae simila iff thee exists a nonsingula P such that S = P - TP o PSP - = T. Since A is simila to a diagonal matix, A is diagonalizable. Rev.F9 2

22 Anothe Example Example 4: Solve the eigenvalue poblem Ax = λx and find the eigenspace, algebaic multiplicity, and geometic multiplicity fo each eigenvalue. A Step : Wite down the chaacteistic equation of A and solve fo its eigenvalues. p() I A ()( ) Rev.F9 22

23 p() Anothe Example (Cont.) ( )( 4)( 5) 8( ) ( 2 5 4)( 5) 8( ) ( ) ( )( 2 2) ( )( 2)( ) ( 2)( ) 2. So the eigenvalues ae 2, 2. Since the facto (λ - 2) is fist powe, λ = 2 is not a epeated oot. λ = 2 has an algebaic multiplicity of. On the othe hand, the facto (λ +) is squaed, λ 2 = - is a epeated oot, and it has an algebaic multiplicity of 2. Rev.F9 23

24 Anothe Example (Cont.) Step 2: Use Gaussian elimination with back-substitution to solve (λi - A) x = fo λ and λ 2. Fo λ = 2, the augmented matix fo the system is 2I A ~ / ~ : / 2 3 / 2 / 2. In this case, x 3 =, x 2 =, and x = -/2() + = + =. Rev.F9 24

25 25 Rev.F9 Anothe Example (Cont.) Thus, the eigenvecto coesponding to λ = 2 is x x x 2 x 3,. If we choose p, then B is a basis fo the eigenspace of 2. E span({ p }) and dim(e ), so the geometic multiplicity is. A x 2 x o (2I A) x

26 Anothe Example (Cont.) Fo λ 2 = -, the augmented matix fo the system is ()I A ~ ~ x 3 = t, x 2 = s, and x = -s + 2t. Thus, the solution has two linealy independent eigenvectos fo λ 2 = - with x x x 2 x 3 s 2t s t s t 2,s,t. Rev.F9 26

27 27 Rev.F9 Anothe Example (Cont.) If we choose p 2, and p 3 2, then B 2, 2 is a basis fo E 2 span({ p 2, p 3 }) and dim(e 2 ) 2, so the geometic multiplicity is 2.

28 Anothe Example (Cont.) Since the geometic multiplicity is equal to the algebaic multiplicity fo each distinct eigenvalue, we found thee linealy independent eigenvectos. The matix A is diagonalizable since P = [p p 2 p 3 ] is nonsingula. Thus, we have AP PD as follows : Rev.F9 28

29 We can find P - as follows: Anothe Example (Cont.) P I : : : 2. So, P 2. Rev.F9 29

30 Anothe Example (Cont.) Note that A and D ae simila matices. AP PD gives us A APP PDP. Thus, PDP A Rev.F9 3

31 Anothe Example (Cont.) Also, D = P - AP = D So, A and D ae simila with D = P - AP and A = PD P -. Rev.F9 3

32 The Othogonal Diagonalization of a Symmetic Matix If P is an othogonal matix, its invese is its tanspose, P - = P T. Since P T P T p T p 2 p p2 p3 p T 3 p p p p 2 p p 3 p 2 p p2 p 2 p2 p 3 p 3 p p3 p 2 p3 p 3 p i p j T T T p p p p2 p p3 p T T T 2 p p2 p2 p2 p3 T T p 3 p p3 p2 p T 3p 3 I because p i p j fo i j and p i p j fo i j,2,3. So, P P T. Rev.F9 32

33 The Othogonal Diagonalization of a Symmetic Matix (Cont.) A is a symmetic matix if A = A T. Let A be diagonalizable so that A = PDP -. But A = A T and A T (PDP ) T (PDP T ) T (P T ) T D T P T PDP T A. This shows that fo a symmetic matix A to be diagonalizable, P must be othogonal. If P - P T, then A A T. The eigenvectos of A ae mutually othogonal but not othonomal. This means that the eigenvectos must be scaled to unit vectos so that P is othogonal and composed of othonomal columns. Rev.F9 33

34 The Othogonal Diagonalization of a Symmetic Matix (Cont.) Example 5: Detemine if the symmetic matix A is diagonalizable; if it is, then find the othogonal matix P that othogonally diagonalizes the symmetic matix A. Let A () 3 ( 2) 5 5 2, then det(i A) ( 2)( 5)2 ( 2) ( 4)( 2 4 2) ( 4)( 2)( 6) Thus, 4, 2 2, 3 6. Since we have thee distinct eigenvalues, we will see that we ae guaanteed to have thee linealy independent eigenvectos. Rev.F9 34

35 The Othogonal Diagonalization of a Symmetic Matix (Cont.) Since λ = 4, λ 2 = -2, and λ 3 = 6, ae distinct eigenvalues, each of the eigenvalues has algebaic multiplicity. An eigenvalue must have geometic multiplicity of at least one. Othewise, we will have the tivial solution. Thus, we have thee linealy independent eigenvectos. We will use Gaussian elimination with back-substitution as follows: Rev.F9 35

36 The Othogonal Diagonalization of a Symmetic Matix (Cont.) Rev.F9 Fo 4, I A 6 : : x 2 s, x 3, x s. x x x 2 x 3 s o p / 2 / 2. 36

37 The Othogonal Diagonalization of a Symmetic Matix (Cont.) Rev.F9 Fo 2 2, 2 I A 7 7 : 7 x 3 s, x 2, x. x x x 2 x 3 s o p 2. 37

38 The Othogonal Diagonalization of a Symmetic Matix (Cont.) Fo 3 6, 3 I A 8 : x 2 s, x 3, x s. x x x 2 x 3 s o p 3 / 2 / 2. Rev.F9 38

39 The Othogonal Diagonalization of a Symmetic Matix (Cont.) As we can see the eigenvectos of A ae distinct, so {p, p 2, p 3 } is linealy independent, P - exists fo P =[p p 2 p 3 ] and AP PD PDP. Thus A is diagonalizable. Since A = A T (A is a symmetic matix) and P is othogonal with appoximate scaling of p, p 2, p 3, P - = P T. PP PP T / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 I. Rev.F9 39

40 The Othogonal Diagonalization of a Symmetic Matix (Cont.) As we can see the eigenvectos of A ae distinct, so {p, p 2, p 3 } is linealy independent, P - exists fo P =[p p 2 p 3 ] and AP PD PDP. Thus A is diagonalizable. Since A = A T (A is a symmetic matix) and P is othogonal with appoximate scaling of p, p 2, p 3, P - = P T. PP PP T / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 I. Rev.F9 4

41 The Othogonal Diagonalization of a Symmetic Matix (Cont.) Note that A and D ae simila matices. PD P - = PDP T / 2 / 2 / 2 / / 2 / 2 / 2 / 2 4 / 2 6 / 2 4 / 2 6 / 2 2 / 2 / 2 / 2 / A. Rev.F9 4

42 The Othogonal Diagonalization of a Symmetic Matix (Cont.) Also, D = P - AP = P T AP / 2 / 2 / 2 / / 2 / 2 / 2 / 2 4 / 2 4 / / 2 6 / 2 / 2 / 2 / 2 / So, A and D ae simila with D = P T AP and A = PDP T. Rev.F9 42

43 The Othogonal Diagonalization of a Symmetic Matix (Cont.) A T = (PD P - ) T = (PD P T ) T = (P T ) T D T P T = P D T P T T / 2 / 2 4 / 2 / 2 / 2 / 2 2 / 2 / 2 6 / 2 / 2 / 2 / 2 Rev.F / 2 / 2 / 2 / 2 4 / 2 6 / 2 / 2 / / 2 6 / / 2 / 2 2 = A. This shows that if A is a symmetic matix, P must be othogonal with P - = P T. T 43

44 How to Detemine the Powe and the Eigenvalues of a Matix, A k? Fom Example, the diagonal matix fo matix A is : D and A 3 PDP PDP PDP PD 3 P 2 5 (5) 3 64 / / / 7 29 / 7 (2) 3 P AP D A PDP, 5 / 7 2 / 7 / 7 / / 7 2 / 7 / 7 / 7. Fo A 3,the eigenvalues, ae 3 25 aand In geneal, the powe of a matix, A k PD k P. and the eigenvalues ae i k, whee i is on the main diagonal of D. Rev.F9 44

45 We have leaned to: What have we leaned?. Solve the eigenvalue poblem by finding the eigenvalues and the coesponding eigenvectos of an n x n matix. Find the algebaic multiplicity and the geometic multiplicity of an eigenvalue. 2. Find a basis fo each eigenspace of an eigenvalue. 3. Detemine whethe a matix A is diagonalizable. 4. Find a matix P, P -, and D that diagonalize A if A is diagonalizable. 5. Find an othogonal matix P with P - = P T and D that diagonalize A if A is symmetic and diagonalizable. 6. Detemine the powe and the eigenvalues of a matix, A k. Rev.F9 45

46 Cedit Some of these slides have been adapted/modified in pat/whole fom the following textbook: Anton, Howad: Elementay Linea Algeba with Applications, 9th Edition Rev.F9 46