Lecture 6. Eigen-analysis

Transcription

1 Lecture 6 Eigen-analysis University of British Columbia, Vancouver Yue-Xian Li March 7

2 6 Definition of eigenvectors and eigenvalues Def: Any n n matrix A defines a LT, A : R n R n A vector v and a scalar λ (real or complex) are called an eigen-pair of A if () v ; () A v λ v, where v is called an eigenvector and λ the corresponding eigenvalue An eigenvector is never a The eigenvalue can be zero An eigenvector is so special, its image is in the same direction as itself with its length changed by a factor λ An eigenvector defines an eigen-direction, thus if v is an eigenvector, then u α v represents the same eigenvector for any scalar α because A u A(α v) αa v αλ v λ(α v) λ u For an n n matrix, often there exist n eigen-pairs {λ i, v i } (i,,, n), especially when no repeated eigenvalues occur

3 Eg 6 Determine if v, and if yes, what is the correspond- 5 5 A ing eigenvalue? is an eigenvector of Ans: A v ( ) Yes, it is an eigenvector with the corresponding eigenvalue -

4 Eg 6 Given that the matrix A [ defines the reflection [ of all vectors in R in the direction defined by a Use the definition above to determine all eigen-pairs of A Ans: An eigenvector is a direction in R that remains unchanged after the LT defined by A Based on our knowledge of reflection, it leaves all vectors in the direction of a unchanged Thus, for any v α a, (α ), A v v, corresponding eigenvalue : λ Let s pick α, thus v a, and verify the result A v [ [ [ [ Thus, {λ, v } {, [ } is one eigen-pair Similarly, for any vector v β a β [, (β ), 4

5 A v v, corresponding eigenvalue : λ Let s pick β, thus v a, and verify the result A v [ [ [ + ( ) [ Therefore, {λ, v } {, [ } is another eigenpair Question: If P a defines a matrix the projects all vector in R in a, what are the eigen-pairs for P a? 5

6 6 Calculating eigen-pairs Finding eigen-pairs of matrix A is to solve A v λ v for all possible pair of λ, v in which v Notice that A v λ v (A λi) v If det(a λi), then v (A λi), cannot be eigenvector! Therefore, det(a λi) is a necessary condition for finding any nonzero vector that satisfies A v λ v For an n n matrix, det(a λi) defines a polynomial of degree n in the unknown λ It is thus called the characteristic equation or polynomial It yields n eigenvalues (counting the repeated ones) Then the corresponding eigenvectors can be solved 6

7 [ a a Eg 6 A a a, characteristic equation? Ans: A λi [ a a a a [ λ λ [ a λ a a λ a det(a λi) a λ a a λ a (a λ)(a λ) a a λ (a +a )λ+a a a a λ Trλ+Det, where { Tr Trace(A) a + a, Det det(a) a a a a For any square matrix, trace is defined as the sum of all diagonal entries Trthe sum of all eigenvalues (elaborated later) Detthe product of all eigenvalues (elaborated later) 7

8 For any matrix, the characteristic equation is λ Trλ + Det Eg 6 A eigenvalues [ is a projection in R Find all Ans: Tr, Det Thus λ Trλ + Det λ λ λ(λ ) λ, λ Alternatively, A λi λ λ ( ) λ ( ) ( ) λ ( ) λ ± λ ±, 8

9 Eg 6 Check if is an eigenvalue of the matrix A Ans: One needs to check if det(a I) A I , yes, it is! Question: How to solve for eigenvectors? Answer: For each eigenvalue λ i (i,, n), solve the homogeneous system (A λ i I) v, for v i 9

10 Back to Eg 6 Find the eigenvector corresponding to each eigenvalue for A [ Ans: We found previously that λ, λ [ v For λ (solve for v ): A λ I A I [ v [ [ v t, v t v [ ()()+() () () t [ For λ (solve for v A λ I A [ [ v v ): ()() () ()() v t, v t v [ [ [ t These results are consistent with the fact that A is the matrix that project all vectors on the 45 line

11 Eg 64 Reflection in line y ( + [ )x is represented by A, calculate all eigen-pairs Ans: Based on geometry, the eigen-pairs are: λ, v [ + ; λ, v [ + Let s now solve these pairs Det Thus, For this matrix, Tr and λ Trλ + Det λ λ ± [ v For λ (solve for v ): v [ () () () () A λ I A I [ ()( [ )() ()() () [ v t, v ( + [ + )t v For λ (solve for v [ v v ):

12 [ + () () () () A λ I A + I + [ + + ()(+ [ )() + + ()() () [ + v t, v (+ [ )t v Note that ( + [ [ + ) [ [ + to what was expected from geometry +, and that These results are identical

13 Eg 65 Find all eigen-pairs of a matrix A Ans: A λi expansion in st column λ λ λ ( λ) 9 λ 5 + λ λ ()() () λ λ 5 + λ λ ( 7) λ λ (λ+) [(λ 9)(λ ) + 5(λ ) (λ+)(λ 6λ+8) (λ + )(λ )(λ 4) λ, λ 4, λ We were lucky here because one of the two minors in the column expansion was! For λ : A I ()()/ ()()/( 5) ()() () ()() () RREF ()()+7() ()() () v t

14 For λ 4: A 4I ()()/( ) ()() () () ()()/( 7) () ()()+() RREF ()()+7() v t For λ : A + I ()()+() RREF ()() () () () ()()/5 (), ()()/( 7) () v t 4

15 Eg 66 Find all eigen-pairs of A Ans: A λi λ 6 λ λ ()()+() λ 6 λ 4 4 λ 4 λ (4 λ) λ 6 λ 4 expansion in rd row ( (4 λ) λ λ 6 λ ) (4 λ) ( λ 7λ + ) (λ 4)(λ 4)(λ ) λ, λ λ 4 Now, for eigenvectors For λ : A I ()() () ()()+() ()() () ()()/( ) RREF ()()+() ()()/ v t 5

16 For λ 4: A 4I ()()+() ()()+() ()()/( ) v, s + t s t s, t s, t, Although the eigenvalue λ, 4 is repeated twice, there still exist two LI eigenvectors corresponding to it 6

17 Question: Is there a formula for the characteristic polynomial det(a λi) for or n n matrices? Asnwer: Yes For matrix A det(a λi) a a a a a a a a a, a λ a a a a λ a a a a λ λ Trλ + c λ Det, where c a a + a a + a a a a a a a a 7

18 Question: Is there a faster/easier way in solving the eigenvectors? Asnwer: Yes Finding nontrivial linear relation(s) (LRs) between columns of A λi Recall that to solve the eigenvector v i corresponding to the eigenvalue λ i, we need to solve the homogeneous linear system (A λ i I) v, that is, we solve for nonzero vectors that are mapped to zero by A λ i I Def: For all A n n, its kernel or nullspace is defined as the set of all vectors that are mapped to zero by matrix A: { } ker(a) : N (A) : x R n : A x Therefore, solving (A λ i I) v for eigenvector v i is equivalent to finding the kernel of A λ i I 8

19 v i ker(a λ i I) Definition of Linear Relation (LR): Given a set of n vectors { c, c,, c n } A LR is defined as an ordered set of scalars {α, α,, α n } that satisfies α c + α c + + α n c n This ordered set of scalars defines a vector α [α α α n T Thus, we often call α a LR of the above set of vectors If α, it defines a trivial LR; otherwise (if α ), α is non-trivial Vectors { c, c,, c n } are LI if the only LR between them is trivial Vectors { c, c,, c n } are LD if there exists as least one non-trivial LR between them 9

20 Eg 67 Show that c, c, c are LD Find a non-trivial LR, α, between them Ans: Notice that c c c, thus, they are LD Because c c + c α is one LR

21 Theorem: For all A n n, its kernel or nullspace is ker(a) : N (A) : {LRs of columns of A} Proof: Based on the definition { } ker(a) : N (A) : x R n : A x Let A [ c c c n, where c, c,, c n are column vectors of A Let x x x be a vector in ker(a), thus x n A x [ c c c n x x x n x c +x c + +x n c n x is a LR of columns of A Therefore, for every x ker(a), x defines a LR of columns of A This proves ker(a) : N (A) : {LRs of columns of A}

22 [ Eg 68 ker 4 {[ } { } There exists no nontrivial LR between the columns Column vectors are LI Matrix is invertible (det A ) (This situation would never happen to A λi if λ is a true eigenvalue of A!) [ Eg 69 ker 4 }{{} c c Eg 6 ker [, the kernel is D }{{} c + c + c & c + c + c,, the kernel is D There are infinitely many choices of the two LRs in this case Other combinations could be,,,,

23 Back to Eg 65 Knowing that λ, λ 4, λ 5 5 are eigenvalues of matrix A Find the 7 7 eigenvector corresponding to each eigenvalue Ans: v ker(a I) ker }{{} v ker(a 4I) ker c + c + c }{{} v ker(a + I) ker c + c + c } {{ } c + c + c

24 Eg 6 Find all eigen-pairs of A Ans: A λi 9 λ 6 4 λ 6 6 λ ()()+() linear in (5 + λ) st row 4 λ 6 6 λ ( ) λ 6 (5 + λ) λ λ 5 λ 5 λ 4 λ 6 6 λ expansion in st row (λ + 5)(λ + 5λ + 4) (λ + 5)(λ + 4)(λ + ) (λ + )(λ + 4)(λ + 5) λ, λ 4, λ 5 We were lucky here because entries in the first row shares a common factor! 4

25 For λ : v ker(a + I) ker } {{ } c c + c For λ 4: v ker(a + 4I) ker } {{ } ( c c )+ c For λ 5: v ker(a + 5I) ker } {{ } c c + c If one encounters any difficulty in finding the LR between the columns, do a few row operations and simplify the matrix to a form closer to RREF, one recognizes the LR soon after a couple of row operations are made 5

26 Eg 6 Repeated eigenvalue[ with a missing eigenvector Find all eigen-pairs of A Ans: Tr4, Det4 λ 4λ+4 (λ ) Thus, λ λ λ, (Identical eigenvalue repeated twice!) Algebraic Multiplicity (AM) is defined as the number of times the same eigenvalue occurs as a root of the characteristic equation det(a λi) Thus, AM(λ ) for the matrix in this example [ [ v ker(a I) ker }{{} c + c Geometric Multiplicity (GM) is defined as the number of LI eigenvectors corresponding to a repeated eigenvalue Therefore, GM(λ ) for the matrix in this example For any n n matrices, if GM(λ) < AM(λ) for at least one eigenvalue, the number of LI eigenvectors is less than n 6

27 Eg 6 Repeated eigenvalue without missing eigenvector Find all eigen-pairs of A Ans: A λi λ λ λ expansion in st column ( λ) λ λ ( λ)( λ)( λ) (λ + )(λ ) Thus, λ, λ λ For λ : v ker(a + I) ker } {{ } c c c For λ (AM(λ ) ): v ker(a I) ker }{{} c + c + c & c + c + c 7

28 v, v Therefore, for this case, AM(λ ) GM(λ ) No eigenvector is missing! There exist LI eigenvectors for this matrix 8

29 Back to Eg 66 Find all eigen-pairs of A Ans: We already found that λ, λ, 4 We now solve the corresponding eigenvalues by finding the LRs of the columns of A λi v ker(a I) ker } {{ } c c + c v v, ker(a 4I) ker }{{} c + c + c & c + c c, Notice that the nd eigenvector for the repeated eigenvalue looks different from the one we obtained previously in Eg 66 There exist infinitely many different combinations of a set of two such eigenvectors that are all allowable choices 9

30 It is important to notice some special cases: () For A [ e c e, {, e } and {, e } are eigen-pairs () For A [ e e e, {, e }, {, e }, {, e } are eigen-pairs () For A [ e e e, {, e } is an eigen-pair (4) For A 7 [ 5 e c c, λ,, 5,, 5 5 are eigenvalues, { 5, e } is an eigen-pair (5) For A [ c c 6 e, λ,,,, are eigenvalues, {6, e } is an eigen-pair

31 Eg 64 [ Complex eigenvalues Find all eigen-pairs of A (This is rotation CC by π ) Ans: Tr, Det λ λ + λ λ, ±i For λ i: [ i v ker(a ii) ker i }{{} c +i c [ i For λ i λ : v v [ i [ i

32 Eg 65 Find all eigen-pairs of A Ans: A λi λ λ 4 4 λ ()() () λ λ λ λ (λ ) λ λ expan in rd row (λ ) ( λ 4 4 λ λ ) (λ )[λ 6λ + (λ )[(λ ) + 4 Thus, λ, (λ ) + 4 λ, ± i For λ : v ker(a I) ker } {{ } c c c For λ i:

33 v ker(a ( i)i) ker ker + i i i + i ker + i i i + i + i i + i ()/, ()/ ()/ } {{ } (+i) c c c ker ( + i) ()() () + i + i i + i + i For λ + i λ : v v + i i

34 Eg 66 Find all eigen-pairs of A Ans: A λi expansion in st column λ λ λ λ λ λ ()()+() ( λ) λ λ λ λ λ ( λ)[λ + ( λ) ( λ)(λ λ + ) Thus, λ, λ λ + (λ ) + λ, ± i For λ : v ker(a I) ker } {{ } c + c + c For λ + i: v ker(a ( + i)i) ker i i i ()( i)() 4

35 ker ( i) i i i ()(+i)() ker i i i ()() () ()()+() ker ()() ()) ker i ( i) i i i i } {{ } i c + c +(+i) c i + i For λ i λ : v v i + i i i 5

36 Summary: For any n n matrix A, A v λ v or (A λi) v define eigen-pairs {λ i, v i }, (i,,, n) Tr(A)a + a + + a nn λ + λ + + λ n Det(A)λ λ λ n (Another method for calculating det(a)) 4 det(a λi) for matrices is: λ Trλ + Det 5 det(a λi) for matrices is: λ Trλ + c λ Det, where c a a + a a + a a a a a a a a 6 det(a λi) for n n matrices is: λ n Trλ n + + ( ) n Det 7 v j ker(a λ j I)LR(s) of columns of A λ j I 8 If the j th column of A is α e j, then {α, e j } is an eigenpair 9 If there exist n distinct eigenvalues, ie λ i λ j if i j, then there exist n LI eigenvectors If GM(λ) < AM(λ) for at least one repeated eigenvalue, then there exist less than n LI eigenvectors 6

37 6 Some applications of eigen-analysis 6 Evaluate A k v If v is an eigenvector of A corresponding to eigenvalue λ, then the answer is easy A k v λ k v What about any vector v that is not an eigenvector? Try to express v as a LC of eigenvectors of A Eg 6 For A A v [, v [ 4, calculate Ans: Tr-4, Det λ +4λ+ (λ+)(λ+) λ, λ [ v ker(a + I) ker }{{} c + c [ 7

38 [ v ker(a + I) ker }{{} c c [ Notice that v [ 4 [ [ v v, thus A v A ( v v ) A v A v ( ) v ( ) v [ [ [ This is accomplishable only when A has n LI eigenvectors that span R n thus guaranteeing that any vector x R n can be expressed as a LC of the eigenvectors of A 8

39 6 Knowing all eigen-pairs, find the matrix Eg 6 Let A be a matrix that reflects vectors in R in the line at an angle π Find A (This is not the easiest method Using diagonalization, this becomes much easier as we shall demonstrate later!) Ans: We know that the eigen-pairs of A are λ, v [ ; λ, v [ Notice that [ e 4 [ + 4 [ 4 v + 4 v ; e [ 4 [ 4 [ 4 v 4 v Thus, A e A( 4 v + 4 v ) 4 A v A v 4 v 4 v

40 4 [ 4 [ [ ; A e A( 4 v 4 v ) 4 A v 4 A v 4 v + 4 v [ + [ 4 4 [ Therefore, [ A [A e A e This is accomplishable only when A has n LI eigenvectors that span R n thus guaranteeing that any basis vector in R n can be expressed as a LC of the eigenvectors of A 4

41 6 Equilibrium (steady-state) probability vector Eg 6 Let P [ be a positive (aka regular) Markov[ matrix for a -state random walk process and that x() (a) Is there a probability vector x e ( e for equilibrium) such that P x e x e? (b) Starting from x(), what happens to x(n) after many, many steps (ie as n )? Ans: (a) If P x e x e is possible for some x e, then x e must be an eigenvector corresponding to eigenvalue λ Since Tr, Det λ λ (λ )(λ+ ), thus λ, λ [ v ker(a I) ker }{{ 4 } c + c [ 4

42 v ker(a + I) ker [ 4 (b) Notice that [ x() }{{} c c [ Therefore, x e v [ + [ [ v + v Thus, x(n) P n x() p n ( v + v ) P n v + P n v () n [ + ( ) n [ n [ x e 4

43 Question: Does x e always exist? Answer: λ always exists for all Markov matrix P AM(λ ) is guaranteed for all positive (regular) Markov matrices However, if P is not positive, AM(λ ) GM(λ ) > can occur When this occurs, there would be more than one LI eigenvector for λ Thus, for different x(), x(n) P n x() can approach different equilibria as n Proof: λ is always an eigenvalue Notice that (i) P T and P have identical eigenvalues because P λi P T λi (P λi) T (ii) v [ T is the eigenvector of P T corresponding to λ : P T v p p p n p p p n p n p n p nn n j p j n j p j n j p jn v 4

44 Theorem: λ is always an eigenvalue for all Markov matrices If P is positive/regular, the λ has only one corresponding eigenvector v This is the equilibrium probability vector for P All other eigenvalues satisfy λ j <, (j,, n) and the all components of the corresponding eigenvectors sum to zero 44

45 64 Matrix similarity, diagonalization, and calculating A k Def: Two n n matrices A, B are similar to each other if there exists an invertible matrix T such that B T AT If A and B are similar, then rank(a)rank(b) det(a)det(b) Tr(A)Tr(B) det(a λi) and det(b λi) are identical Identical eigenvalues (but not necessarily eigenvectors) For an arbitrary matrix, A k is hard to compute when k is large However, for a diagonal matrix a a a nn k a k a k a k nn If matrix A is not diagonal but is similar to a diagonal matrix, A T DT (D is diagonal), then 45

46 A k (T DT ) k T D k T T λ k λ k λ k n T Theorem: For all n n matrices A, if there exist n LI eigenvectors (this happens either when all eigenvalues are distinct or for each repeated eigenvalue λ, we have AM(λ) GM(λ), ie no eigenvector is missing), then A is similar to a diagonal matrix and is called diagonalizable And that A T DT T λ λ λ n T T A d T, where the diagonal entries λ, λ,, λ n are the eigenvalues of A and that T is the matrix whose columns are the corresponding eigenvectors T [ v v v n Proof: Let {λ j, v j }, (j,,, n) be the eigen-pairs of 46

47 A Since the n eigenvectors are LI, then T [ v v v n must be an invertible matrix Then, AT A[ v v v n [A v A v A v n [λ v λ v λ n v n [ v v v n λ λ λ n T A d A T A d T [ Eg 64 For A, we found in Eg 6 that { [ } { [ },,, are the eigen-pairs Calculate A Ans: Based on the theorem, [ [ [ [ λ A T T λ Notice that T [ [ Therefore, 47

48 [ [ [ A [ [ [ ( ) ( ) [ [ [ + + Eg 65 For A that,, eigen-pairs Calculate A Ans: Based on the theorem, A T A λ λ λ A T,,, we found in Eg 6 ( ) () (), are the () + () 48

49 Eg 6 (Revisted!) Let A be a matrix that reflects vectors in R in the line at an angle π Find A Ans: We know that the eigen-pairs of A are λ, v [ ; λ, v [ Using the diagonalization theorem: A T A d T [ [ [ Notice that T is orthogonal, thus T T T A [ [ [ [ 49

50 Eg 66 Given a probability transition matrix P (a) Find all eigen-pairs (b) Find the equilibrium probability vector, x e (c) Show that T P T D (d) Calculate P n [ 5 5 (e) Calculate P and show that P x() x e for all x() 5 4 5, Ans: (a) We know that λ and that λ + λ Tr 6 5 Thus, λ 5 For the corresponding eigenvectors, [ v ker(a I) ker 5 5 }{{ 5 } v ker(a 5 I) ker [ 5 5 c + c }{{} c c [ 4 4 [ (b) Since λ is not repeated and has only one corresponding eigenvector v, thus x e v 5 [ 4 4

51 (c) (d) (e) T [ v v T P T [ 4 4 [ P n T D n T [ 4 4 [ 5 5 [ [ T ( ) [ 4 [ 5 [ 4 4 (5) n (5) n P lim n P n lim n [ [ [ n ( ) n 5 [ (5) n 4 + 4(5) n [ 4 + 4(5) n 4 4(5) n 4 4(5) n 4 + 4(5) n For any x [x x T, x + x, [ 4 4(5) n 4 4(5) n [ [ 4 [ [ [ P x 4 4 x 4 (x [ + x ) 4 4 x 4 (x 4 x + x ) e 4 Therefore, P defines a projection that projects all x R satisfying x + x onto x e! 5

52 64 Systems of linear differential equations (DEs) with constant coefficients Definition of Differential Equation (DE): a DE is an equation that relates at least one derivative of an unknown function x(t) to other quantities that often include x(t) itself and/or its other derivatives Eg 64 x (t) kx(t), (k const, st order) is a DE describing an exponential function x(t) x()e kt Eg 64 my (t) + cy (t) + ky(t), (m, c, k consts, nd order) is a DE describing the displacement of a mass, y(t), relative to the equilibrium location that is attached to a spring and a damper 5

53 y(t) m smg/k Eg 64 LI (t)+ri (t)+ I(t), C (R, L, C consts, nd order) is a DE describing the current in a RLC circuit (see figure), where R (in Ohms) is the resistance of the resistor, C (in Farads) is the capacitance of the capacitor, L (in Henrys) is the inductance of the inductor L I(t) C R 5

54 Solving a DE is in essence finding the unknown function given the relation between its derivative and some other quantities It can be regarded as doing anti-derivation in a special form 54

55 The nd-order DEs can be reduced to a system of two storder DEs Eg 644 Turn mx (t)+cx (t)+kx(t) into a system of two st-order DEs Ans: Let v(t) x (t), thus v (t) x (t) Therefore, { x (t) v(t), v (t) k m x(t) c m v(t) Eg 645 Turn LI (t)+ri (t)+ CI(t) into a system of two st-order DEs Ans: Let Z(t) I (t), thus Z (t) I (t) Therefore, { I (t) Z(t), Z (t) LC I(t) R L Z(t) 55

56 A general system of two linear DEs with constant coefficients is expressed as { x (t) a x (t) + a x (t), x (t) a x (t) + a x (t), (64) x (t), x (t) are the unknown functions to be solved, a ij, (i, j, ) are known scalars Def: A vector function and its derivative are defined as x(t) [ x (t) x (t), d x(t) dt x (t) [ x (t) x (t) Therefore, the linear system in Eq(64) can be written in matrix form [ [ [ x (t) a a x x (t) (t) a a x (t) x (t) A x(t), A is a constant matrix Although we shall focus on systems involving two DEs here, results discussed here can be generalized to systems of n linear DEs expressed in the form x A x 56

57 Theorem: For linear DE systems of the form x A x or d x A x, where A is a constant matrix (can be dt generalized to n n matrices) The following statements are true (I) If λ λ or if λ λ λ but GM(λ) AM(λ) so that {λ, v } and {λ, v } are two LI eigen-pairs, then x (t) e λ t v, x (t) e λ t v are the two LI solutions because det[ v v (II) The general solution (ie the solution that includes all possible solutions) of the system is a LC of the two x(t) c x (t) + c x (t) c e λ t v + c e λ t v, where c, c are arbitrary constants [ x () (III) Given the initial condition x(), one can x () uniquely determine the constants c, c by substituting it into the general solution x() c e v +c e v [ v v [ c c T c c T x() The theorem basically concludes that solving the system x A x is equivalent to finding the eigen-pairs of A 57

58 Proof of (I): Show that if {λ, v } and {λ, v } are two LI eigen-pairs of A, then x (t) e λ t v, x (t) e λ t v are two LI solution of the system x A x Substitute x j e λ jt v j into the system and remember that A v j λ j vj: lhs x ( e λ jt v j ) ( e λ j t ) vj λ j e λ jt v j, rhs A x A ( e λ jt v j ) e λ j t A v j e λ jt λ j v j lhs rhs x j e λ jt v j is a solution for j, Also, det[ x (t) x (t) det[e λ t v e λ t v e (λ +λ )t det[ v v Therefore, the two solutions are LI whenever the two eigenvectors are LI 58

59 { x Eg 646 For the linear system (t) x (t) + x (t), x (t) x (t) x (t), (a) Express the system in matrix form (b) Find the general solution (c) If x (), x (), find the unique solution (d) As t, what happens to the solution in (b)? Ans: (a) Let x(t) [ x (t) x (t) x A x,, A [, then is the system in matrixform (b) Tr-4, Det4- λ + 4λ + (λ + )(λ + ) λ, λ [ v ker(a + I) ker }{{} c + c [ v ker(a + I) ker }{{} c c [ [ 59

60 Based on the theorem, the general solution is x(t) c e λ t v + c e λ t v c e t [ + c e t [ (c) Substitute x() in (b), [ x () x () [ into the solution x() c e [ +c e [ c [ +c [ [ c [ c c [ v v x() [ [ [ [ [ [ (d) As t, e t, e t, thus x(t) c e t [ + c e t [ t 6

61 Summary: Solving a homogeneous linear system of two st-order DEs requires finding two LI solutions x (t), x (t) and the general solution is a LC of the two x(t) c x (t) + c x (t) For a linear system x (t) A x(t), where A has two LI eigen-pairs {λ j, v j }, (j, ), x (t) e λ t v, x (t) e λ t v (c, c arbitrary consts) Questions and answers: () Why is a linear combination of solutions also a solution to all homogeneous linear equations? Answer: Principle of superposition Let L x be any homogeneous linear system, where L is any linear operator L is a matrix if it represents a matrix equation, L d dt A if it represents a st-order system DEs If 6

62 x j (j,,, n) are all solutions to the system, ie they satisfy L x j, Then, a LC of them must also be a solution (j,,, n) x c x + c x + + c n x n Proof: Since L is linear, L x L(c x +c x + +c n x n ) c L x +c L x + +c n L x n () Why the general solution of a linear system of two storder DEs involves two LI solutions? Answer: Based on the general theory of differential equations, all possible solutions of a linear system x (t) A x(t), (A is n n) is a vector in an n D functional space Thus, any set of n LI vectors can form a basis of that space such that each solution vector can be expressed as a LC of these basis vectors 6

63 Eg 647 For x A x, A [, (a) General solution in both complex and real form (b) Describe what happens to the solution as t Ans: (a) Tr-, Det+9 λ + λ + (λ + ) + 9 λ, ± i [ i v ker(a ( + i)i) ker i }{{} c +i c [ [ v v i i [ i Now, we obtained two LI solutions in complex form z (t) e λ t v e ( +i)t [ i [, z (t) z e ( i)t i x(t) c z (t)+c z (t) c e ( +i)t [ i +c e ( i)t [ i 6

64 Noe, how to find the general solution in real form? Notice that x (t) Re{ z (t)} z (t) + z (t), x (t) Im{ z (t)} z (t) z (t) i are also two LI solutions of the system and are realvalued Therefore, the general solution in real form is x(t) c x (t) + c x (t) c Re{ z (t)} + c Im{ z (t)} Using Euler s formula z (t) e ( +i)t [ i e t e it [ i e t [cos(t)+i sin(t) [ i e t [ cos(t) + i sin(t) i cos(t) sin(t) e t [ cos(t) sin(t) + ie t [ sin(t) cos(t) [ cos(t) x (t) Re{ z (t)} e t sin(t) [ sin(t) x (t) Im{ z (t)} e t cos(t) 64

65 The general solution in real form is x(t) c x (t)+c x (t) c e t [ cos(t) sin(t) +c e t [ sin(t) cos(t) (b) As t, e t, thus x(t) c e t [ cos(t) sin(t) + c e t [ sin(t) cos(t) t The above result is summarized in the following theorem 65

66 Theorem: For x A x (A is and const), if λ, α±βi, (α, β R) with corresponding eigenvectors v, v, then the general solution in complex form is x(t) c e λ t v + c e λ t v ; and the general solution in real form is x(t) c Re { e λ t v } + c Im { e λ t v } Proof: Since λ λ, v, v are LI Thus, x(t) c e λ t v + c e λ t v is the general solution Also because Re { e λ t v } Im { e λ t v } i [e λt v + e λ t v, [e λ t v e λ t v both are LCs of solutions, based on Principle of Superposition, they must also be solutions to the system It is not hard to show (not included here) that the two are LI Thus, x(t) c Re { e λ t v } +c Im { e λ t v } is also the general solution 66

67 Eg 648 For x A x, A [ 5 (a) Find the general solution in complex and real forms [ (b) Given the initial condition, x(), find the unique solution for this initial value problem (IVP) (c) Roughly sketch two components of the solution vector in (b), determine what happens as t, Ans: (a) Tr-, Det λ + λ + (λ + ) + 9 λ, ± i [ i v ker(a ( +i)i) ker 5 i }{{} ( i) c +5 c v v [ i 5 [ + i 5 [ i 5 Now, we obtained two LI solutions in complex form 67

68 z (t) e λ t v e ( +i)t [ i 5, z (t) z e ( i)t [ + i 5 x(t) c z (t)+c z (t) c e ( +i)t [ i 5 +c e ( i)t [ + i 5 To find the general solution in real form, one needs the real and imaginary parts of the complex solutions x (t) Re{ z (t)} z (t) + z (t), x (t) Im{ z (t)} z (t) z (t) i Using Euler s formula z (t) e ( +i)t [ i 5 e t e it [ i e t [ cos t + sin t + i(sin t cos t) 5 cos t + 5i sin t e t [ cos t + sin t 5 cos t [ i e t [cos t+i sin t 5 + ie t [ sin t cos t 5 sin t 68

69 x (t) Re{ z (t)} e t [ cos t + sin t 5 cos t x (t) Im{ z (t)} e t [ sin t cos t 5 sin t The general solution in real form is x(t) c x (t) + c x (t) c e t [ cos t + sin t 5 cos t + c e t [ sin t cos t 5 sin t (b) Substituting the initial condition (IC) into the general solution x() c [ 5 + c [ [ [ 5 [ c c [ [ c c 5 [ 5 [ [ x(t) e t [ sin t cos t 5 sin t 69

70 (c) Since x(t) [ x (t) x (t) e t [ sin t cos t 5 sin t [ ( ) e t cos t sin t 5 sin t e t [ cos t sin t 5 sin t e t [ cos(t + θ) 5 sin t where θ tan The sketches are given in the figures below As t, e t, thus x(t) e t [ cos(t + θ) 5 sin t t [ 7