Application of Jordan Canonical Form

Transcription

1 CHAPTER 6 Applicatio of Jorda Caoical Form Notatios R is the set of real umbers C is the set of complex umbers Q is the set of ratioal umbers Z is the set of itegers N is the set of o-egative itegers Z + is the set of positive itegers Re(z, Im(z, z ad z are the real part, imagiary part, cojugate ad modulus of a complex umber z M m (F is the set of m matrices with etries i F F(x is the set of polyomials i x with coefficiets i x Row(A, Col(A ad Nul(A are the row, colum ad ull spaces of a matrix A, respectively rak(a is the rak of a matrix A I [ e e 2 e ] isthe idetitymatrix,wheree,e 2,,e are the colum vectors of I K(T, R(T ad rak(t are the kerel, rage ad rak of a liear trasformatio T gcd(f (x,f 2 (x,,f m (x is the greatest commo divisor of polyomials f (x,f 2 (x,,f m (x lcm(f (x,f 2 (x,,f m (x is the least commo multiple of polyomials f (x,f 2 (x,,f m (x Computatio of A I may applicatios, we eed to compute the -th power of a give square matrix A (cf [P] Here by computig A, we mea to give a closed formula for A that depeds oly o for a give A Our mai tools are Jorda Caoical Form ad characteristic/miimal polyomials There are essetially two approaches to the problem The first method uses directly the JCF of A For every complex square matrix A There exists a ivertible matrix P such that P AP J is a Jorda matrix The A PJP ad (6 A PJ P 97

2 98 6 APPLICATION OF JORDAN CANONICAL FORM It suffices to figure out the -th power of a Jorda matrix, which comes dow to computig the -th power of the Jorda block J λ,m : ( Jλ,m (λi +J 0,m λ k J0,m k k k0 m ( (62 λ k J0,m k k k0 ( ( ( λ I + λ J 0,m ++ λ m+ J0,m m 0 m by Biomial Theorem, sice J0,m k 0 for k m If we use the otatio E ij for the m m matrix with at i-th row ad j-th colum ad 0 everywhere else, the (63 J 0,m E 2 +E E m,m m i J k 0,m E,k+ +E 2,k+2 + +E m k,m m Jλ,m k0 m k0 ( λ k J0,m k k m k i m k0 ( k ( λ k E i,i+k k i j m E i,i+ m k i E i,i+k ( m k λ k E i,i+k i ( j i λ +i j E ij So Jλ,m is a upper triagular matrix whose etry at i-th row ad j-th colum is ( j i λ +i j for i j That is, ( ( 0 λ ( λ ( 2 λ 2 m λ m+ ( ( (64 Jλ,m 0 λ ( λ m 2 λ m+2 ( ( 0 λ λ λ For example, (65 J λ,2 [ λ λ] [( 0 λ ( ( 0 Example 6 Let us compute A for [ ] 8 4 (66 A 9 4 ( 0 ] [ λ λ λ λ ] λ The characteristic polyomial of A is (x 2 2 ad { (67 dimnul(a 2I k : k 0,,2, } {0,,2,2,}

3 COMPUTATION OF A 99 So the Youg tableau of A is give i Figure 7 It suffices to choose a vector v Nul(A 2I ad the [ (68 P 2 AP J 2,2 for P 2] [ (A 2Iv v ] We may choose v e The [ ] 6 P 9 0 (69 A PJ 2,2 P ad [ ] [ A PJ2,2P ][ [ ][ 6 2 (2 ][ ] [ (3+2 (2 + ] 9(2 ( 32 A equivalet way to compute A is to compute A v by decomposig v to a sum of geeralized eigevectors of A The colum vectors of the ivertible matrix P i (6 form a basis B of C which is a disjoit uio (60 B B B 2 B l of cycles of geeralized eigevectors Suppose that each B i is geerated by the vector v i correspodig to eigevalue λ i ad B i m i To compute A v, it suffices to write v as a liear combiatio of B: ] (6 l v c ij (A λ i I j v i i j0 l A v A c ij (A λ i I j v i i j0 where we may take j from 0 to for simplicity sice (A λ i I j v i 0 for j m i The term A (A λ i I j v i ca be computed by expadig A as (62 ( A (λ i I +(A λ i I λi k (A λ i I k k k0 ( λ k i (A λ i I k k k0

4 00 6 APPLICATION OF JORDAN CANONICAL FORM Therefore, (63 A v l A c ij (A λ i I j v i i i j0 ( l ( k l k0 i j,k 0 c ij λ k i λ k i (A λ i I k c ij (A λ i I j v i j0 ( (A λ i I j+k v i k Lettig j +k a ad k a j, we ca rewrite (63 as (64 A v l i j,k 0 i a0 c ij λ k i l a l i m i a0 j0 a j0 ( (A λ i I j+k v i k ( c ij λ +j a i (A λ i I a v i a j ( c ij λ +j a i (A λ i I a v i a j sice (A λ i I a v i 0 for a m i So we may use (64 to compute A v for all v To compute A itself, we observe that (65 A A I A [ e e 2 e m ] [ A e A e 2 A e m ] ad apply the above algorithm to each A e i Example 62 Let us redo Example 6 by computig A e i via writig e i as a liear combiatio of B {(A 2Iv,v}, where we choose v e The (66 e e e [ + 0] [ 6 9 9] 2 3 e + 9 (A 2Ie

5 COMPUTATION OF A 0 ad hece A e A e (2I +(A 2I e ( 2 k (A 2I k e k k0 ( 2 k (A 2I k e 2 e +(2 (A 2Ie k k0 [ ] [ [ ] 2 +(2 0 6 (3+2 9] 9(2 A e 2 A ( 2 3 e + 9 (A 2Ie (67 (2I +(A 2I ( 2 3 e + 9 (A 2Ie ( ( ( 2 k (A 2I k 2 k 3 e + 9 (A 2Ie k0 ( 2 I +(2 (A 2I ( 2 3 e + 9 (A 2Ie ( e + 9 (2 (A 2Ie 3 Therefore, 2+ 3 [ 0] + ( 2 9 (2 3 [ 6 9] [ (2 + ] ( 32 (68 A [ [ A e A ] (3+2 (2 e + ] 2 9(2 ( 32 The mai complexity of the above algorithm of computig A via JCF lies i the fidig of the ivertible matrix P, which is a time-cosumig process Our secodmethodsidesteps thewholeissue ofp, butit itroduces its ow complexity The mai idea is that if we have a ozero polyomial f(x such that f(a 0, we ca divide x by f(x to obtai (69 x q(xf(x+r(x where q(x ad r(x is the quotiet ad remaider of the divisio of x by f(x such that degr(x < degf(x; the (620 A q(af(a+r(a r(a sice f(a 0 Of course, q(x is irrelevat here The key of the above approach is the determiatio of the polyomial r(x We may take f(x to be ay ozero polyomial satisfyig f(a 0, ie, ay ozero polyomial that is divisible by the miimal polyomial of

6 02 6 APPLICATION OF JORDAN CANONICAL FORM A But the lesser the degree of f(x, the better the above algorithm works IfthemiimalpolyomialofAiskow, wetakef(xtobethemiimalpolyomial Otherwise, we usually take f(x to be the characteristic polyomial of A by Cayley-Hamilto To determie r(x, we write (69 as (62 x r(x q(x+ f(x f(x Let f(x (x λ d (x λ 2 d 2 (x λ m dm We may write r(x/f(x as partial fractios: (622 x r(x q(x+ f(x f(x q(x+ m d i i j c ij (x λ i j for some umbers c ij To determie c ij for each i, we multiply both sides of (622 by (x λ i d i ad take derivatives at λ i of order k 0,,,d i i sequece: ( x (x λ i d i (k (k d i (623 f(x λi j c ij (x λ i d i j λi where (624 h i (x q(x+ l i + ( h i (x(x λ i d i (k λi d l j Sice h i (x is C at λ i, it is easy to see that c lj (x λ l j (625 ( hi (x(x λ i d i (k λi 0 for all k < d i So ( x (x λ i d i (k (626 f(x λi (k d i c ij (x λ i d i j j λi (k!c i,di k for k 0,,,d i ad hece ( x (x λ i d i (627 c ij (d i j! f(x (di j λi I complex aalysis, c ij is the residue of the meromorphic fuctio (628 z (z λ i j f(z

7 at z λ i COMPUTATION OF A 03 Example 63 For example, suppose that f(x (x 2 (x 2 3 ad we write x (x (629 2 (x 2 3 q(x+ c x + c 2 (x 2 + c 2 x 2 + c 22 (x c 23 (x 2 3 To determie c ad c 2, we multiply both sides of (629 by (x 2 ad take derivatives at of order k 0,: ( x (k (x 2 3 ( (k x c (x +c 2 x (630 + ( h (x(x 2 (k x }{{} 0 where (63 h (x q(x+ c 2 x 2 + c 22 (x c 23 (x 2 3 So for k 0,, we obtai ( x 3 c 2 (x 2 x (632 ( x c (x 2 x 3 3 To determie c 2, c 22 ad c 23, we multiply both sides of (629 by (x 2 3 ad take derivatives at 2 of order k 0,,2: where ( x (k (x 2 ( c 2 (x 2 2 (k x2 +c 22 (x 2+c 23 x2 (633 + ( h 2 (x(x 2 3 (k x2 }{{} 0 where (634 h 2 (x q(x+ c x + c 2 (x 2 So for k 0,,2, we obtai ( x 2 c 23 2 (x x2 ( x (635 c 22 (x x2 2 ( 42 c 2 ( x 2 (x x2 2 (

8 04 6 APPLICATION OF JORDAN CANONICAL FORM Therefore, (636 ad hece (637 If f(a 0, the (638 x +3 (x 2 q(x (x 2 3 x (x 2 + ( x 2 + ( 42 2 (x (x 2 3 x q(xf(x (+3(x (x 2 3 (x 2 3 +( (x 2 (x 2 2 +( 42 (x 2 (x 2+2 (x 2 A (+3(A I(A 2I 3 (A 2I 3 +( (A I 2 (A 2I 2 +( 42 (A I 2 (A 2I+2 (A I 2 Example 64 Let us redo Example 6 usig the characteristic polyomial of A Let f(x det(xi A (x 2 2 ad let (639 for some umbers c ad c 2 The x (x 2 2 q(x+ c x 2 + c 2 (x 2 2 (640 c (x x2 (2 ad c 2 x x2 2 Therefore, (64 x q(xf(x+(2 (x 2+2 A (2 (A 2I+2 I (2 A+( 2 I [ ] [ ] ( ( [ (3+2 (2 + ] 9(2 ( 32 Example 65 Let us compute A for (642 A Let f(x det(xi A (x 2 (x 4 ad let (643 x (x 2 (x 4 q(x+ a x + a 2 (x 2 + b x 4

9 COMPUTATION OF A 05 for some umbers a,a 2 ad b The ( x a x 4 (644 x 3 ( x 9, a 2 x 4 x 3 ad ( x 2 b (x x4 4 9 Therefore, ( x q(xf(x ( A ( Actually, A is diagoalizable sice (x (x 4 3 (x (x 2 (A I(A 4I 3 (A 4I+ 4 9 (A I2 (646 dimnul(a I+dimNul(A 4I 2+ 3 ad the miimal polyomial of A is (x (x 4 So we may choose f(x (x (x 4 ad the above computatio ca be simplified: (647 ad hece (648 x (x (x 4 q(x+ a x + b x 4 q(x 3(x + 4 3(x 4 x q(xf(x 3 (x (x A 3 (A 4I+ 4 3 (A I As a applicatio of computatio of A, we study liear recurrece A liear recurrece is a sequece {a 0,a,,} satisfyig the coditio (649 a +l c a +l +c 2 a +l 2 ++c l a +b 2

10 06 6 APPLICATION OF JORDAN CANONICAL FORM forall, wherel,c,c 2,,c l,barecostats Ifb 0, itishomogeeous If b 0, it is o-homogeeous For example, a geometric sequece/progressio is a homogeeous liear recurrece give by (650 a + ra ad a arithmetic sequece/progressio is a o-homogeeous liear recurrece give by (65 a + a +b The most famous liear recurrece (other tha geometric/arithmetic is the Fiboacci sequece (652 a +2 a + +a (653 Solvig a liear recurrece ivolves fidig a closed formula for a To solve (649 for b 0, we write a +l c a +l +c 2 a +l 2 ++c l a a +l a +l a +l 2 a +l 2 a + a + Let us put both sides of (653 i the matrix form: c a c 2 c l c l +l a +l ( a + }{{} u + } {{ } A a +l a +l 2 a }{{} u This turs the liear recurrece (649 ito a geometric sequece {u } of vectors (655 u + Au for u i C l Obviously, a +l a +l 2 a (656 u Au A 2 u 2 A u 0 for u 0 a l a l 2 So it comes dow to the computatio of A with A give i (654 a 0

11 COMPUTATION OF A 07 Whe b 0, it suffices to add oe more compoet to the vectors u : c a c 2 c l c l b +l a +l a +l a +l 2 (657 a a }{{} }{{} u + }{{} u A The (658 u A u 0 for u 0 a l a l 2 The characteristic polyomial of A for b 0 ca be computed by expadig the determiat of xi A by the first row: (659 x c c 2 c l c l x 0 0 det(xi A det x a 0 x x (x c det ( c 2det x x x + +( l+2 ( c l det x x l c x l c 2 x l 2 c l x x x Similarly, we ca compute the characteristic polyomial of A for b 0: (660 det(xi A (x l c x l c 2 x l 2 c l (x where A is give i (657

12 08 6 APPLICATION OF JORDAN CANONICAL FORM Regardless of b 0 or b 0, we eed to compute A for the correspodig A Note that we are oly iterested i the compoet a i u so we oly eed to kow the l-th row of A Let J P AP be the JCF of A for some ivertible matrix P The (66 u A u 0 PJ P u 0 which is (662 if b 0 ad (663 a +l a +l 2 a a +l a +l 2 a PJ P PJ P a l a l 2 a 0 a l a l 2 if b 0, where a 0,a,,a l are the first l terms of the sequece which are usually give For every Jorda block J 0,m, J0,m is give by (64 We observe that the etries of J0,m, as fuctios of, are i the vector space of { ( ( } Spa λ, λ,, λ m+ m (664 {( } Spa λ k : k 0,,2,,m k For each k, ( k is a polyomial i of degree k, ie, ( ( ( k + { (665 Spa,, 2,, k} k k! Therefore, the space (664 is the same as { ( ( } Spa λ, λ,, λ m+ m (666 Spa { λ,λ,, m λ } { } Spa k λ : k 0,,2,,m which cotais all the etries of J λ,m Suppose that (667 det(xi A (x λ d (x λ 2 d 2 (x λ m dm a 0

13 COMPUTATION OF A 09 with λ,λ 2,,λ m distict Every Jorda block of J is i the form of J λi,a for some a d i So the etries of J, as fuctios of, are i the space { Spa λ,λ,, d λ, (668 λ 2,λ 2,, d2 λ 2, λ m,λ m,, dm λ m } Clearly, the etries of PJ P are also i the above space sice P is a costat matrix It follows that a lies i (668, ie, (669 a m i d i j0 c ij j λ i for some c ij C These c ij ca be determied the by the values of a 0,a,,a l I summary, we ca solve a liear recurrece i the followig steps: ( Determie the characteristic polyomial of the recurrece give by (659 or 660 (2 Fid all the roots of the characteristic polyomial as i (667 (3 Write a as a liear combiatio of the fuctios (670 λ,λ,, d λ, λ 2,λ 2,, d 2 λ 2, λ m,λ m,, dm λ m as (669 (4 Determie the coefficiets c ij of this liear combiatio (669 by settig 0,,,l Alteratively, we may compute a by computig A directly usig its characteristic polyomial Example 66 Let us solve the Fiboacci sequece {a : 0,,} give by a +2 a + +a for 0, a 0 0 ad a We ca write (67 It follows that (672 { a+2 a + +a a + a + [ ] a+ a [ ] a+2 a + }{{} u + [ ] [ a [ ][ ] a+ a }{{}}{{} A u a 0 ]

14 0 6 APPLICATION OF JORDAN CANONICAL FORM The characteristic polyomial of the recurrece is (673 det(xi A x 2 x (x λ (x λ 2 for λ ad λ So a, as a fuctio of, lies i the space spaed by λ ad λ 2 Therefore, ( (674 a c λ +c 2 λ + ( c +c for some costats c ad c 2 Settig 0,, we have { c +c 2 0 c 5 (675 λ c +λ 2 c 2 c 2 5 ad hece ( (676 a 5 + ( Alteratively, we may compute A directly via its characteristic polyomial f(x x 2 x That is, we write (677 x f(x q(x+ b x λ + b 2 x λ 2 for some polyomial q(x ad some costats b ad b 2 The (678 b x x λ 2 λ ad b 2 x λ λ λ 2 x λ λ 2 λ2 λ 2 λ By (677, we obtai (679 x q(xf(x+b (x λ 2 +b 2 (x λ ad hece (680 A q(af(a+b (A λ 2 I+b 2 (A λ I b (A λ 2 I+b 2 (A λ I (b +b 2 A (b λ 2 +b 2 λ I [ ] [ (b +b 2 (b λ 2 +b 2 λ ] [ ] b +b 2 b λ 2 b 2 λ

15 Therefore, (68 COMPUTATION OF A [ ] [ ] a+ A a a a 0 [ ][ ] b +b 2 b λ 2 b 2 λ 0 [ ] b +b 2 So a b +b 2 with b ad b 2 give i (678, which also gives us (676 Example 67 Let us solve the recurrece {a : 0,,2,} give by (682 a +2 3a + 2a +5 for 2 ad a 0 a We write a +2 3a + 2a +5 (683 a + a + a +2 a }{{}}{{} u + A It follows that (684 a + a a a 0 The characteristic polyomial of the recurrece is a + a }{{} u (685 det(xi A (x 3 3x+2(x (x 2 (x 2 So a, as a fuctio of, lies i the space spaed by, ( ad 2 Therefore, (686 a c +c 2 +c 3 (2 for some costats c,c 2 ad c 3 Settig 0,,2, we have c +c 3 c 4 (687 c +c 2 +2c 3 c 2 5 c +2c 2 +4c 3 6 c 3 5 ad hece (688 a 4 5+5(2 Alteratively, we may compute A directly via its characteristic polyomial f(x (x 2 (x 2 That is, we write (689 x f(x q(x+ b x + b 2 (x 2 + b 2 x 2

16 2 6 APPLICATION OF JORDAN CANONICAL FORM for some polyomial q(x ad some costats b ij The ( x b, b 2 x 2 x x 2 ad (690 x b 2 (x By (689, we obtai (69 x q(xf(x+b (x (x 2+b 2 (x 2+b 2 (x 2 ad hece (692 Therefore, (693 A q(af(a+b (A I(A 2I+b 2 (A 2I+b 2 (A I 2 b (A I(A 2I+b 2 (A 2I+b 2 (A I 2 b b b b b b 2 5 b 2 +b 2 2b 2 b 2 5(b +b 2 a + a A a a 0 b 2 +b 2 2b 2 b 2 5(b +b 2 5b b 2 +5b 2 a 5b b 2 +5b (2 2 Solutios of Liear ODEs with Costat Coefficiets As aother applicatio of JCF ad miimal polyomial, let us cosider the solutios of the followig system of first order liear ordiary differetial

17 2 SOLUTIONS OF LINEAR ODES WITH CONSTANT COEFFICIENTS 3 equatios (ODEs with costat coefficiets: dx dt a x +a 2 x 2 ++a x +b dx 2 (694 dt a 2x +a 22 x 2 ++a 2 x +b 2 dx dt a x +a 2 x 2 ++a x +b where x,x 2,,x are ukow fuctios i t ad a ij ad b i are costats Usig matrix otatios, we may put it i the form (695 dx dt Ax+b or x Ax+b for A [ a ij ],x x x 2 ad b b 2 We call the system homogeeous if b 0 ad o-homogeeous otherwise Note that we ca always covert a o-homogeeous system of liear ODEs to a homogeeous oe by itroducig a extra ukow fuctio x + ad turs (694 to (696 x dx dt a x +a 2 x 2 ++a x +b x + dx 2 dt a 2x +a 22 x 2 ++a 2 x +b 2 x + dx a x +a 2 x 2 ++a x +b x + dt dx + 0 dt which is homogeeous For a solutio of (696 give by (697 (x,x 2,,x,x + (f (t,f 2 (t,,f (t,f + (t f + (t c must be a costat If f + (t, the (698 (x,x 2,,x (f (t,f 2 (t,,f (t is a solutio of (694 More precisely, { x : dx } dt Ax+b (699 { [ ] d x x : dt x + [ A b 0 0 ][ x x + ] } ad x + b b

18 4 6 APPLICATION OF JORDAN CANONICAL FORM Therefore, to solve the o-homogeeous system (694, it suffices to solve (696 ad choose its solutios with x + as the solutios of the origial system So we oly eed to solve a homogeeous system x Ax The solutio of x Ax satisfyig the iitial coditio x(t 0 v is very explicitly give by (600 x e (t t 0A v where e B exp(b is defied by (60 e B m0 m! Bm for a square matrix B So it comes dow to the computatio of e (t t 0A, which ca be achieved by the JCF or the miimal polyomial of A It is easy to see that e A has the followig properties: (602 e λi e λ I e D e λ e λ 2 e A (e A e A+B e A e B if AB BA e A v e λ v if Av λv P e A P exp(p AP λ for D λ 2 e λ λ Agai, there are two ways to compute the solutio e (t t 0A v of x Ax The first approach uses the JCF of A ad the secod approach uses the miimal or characteristic polyomial of A For simplicity, let us assume that t 0 0 We may always replace t by t t 0 i e ta v to obtai e (t t 0A v Suppose that P AP J is the JCF of A for a ivertible matrix P Let B be the ordered basis of C cosistig of the colum vectors of P As i (60, B B B 2 B l is a disjoit uio of cycles of geeralized eigevectors We write v as a liear combiatio of vectors i B as i (6: (603 v l c ij (A λ i I j v i i j0 where v,v 2,,v l are the vectors geeratig the cycles B,B 2,,B l i B

19 2 SOLUTIONS OF LINEAR ODES WITH CONSTANT COEFFICIENTS 5 We compute e ta v by writig e ta e λiti e t(a λii for each i: l e ta v e ta c ij (A λ i I j v i (604 i i j0 l e λiti e t(a λ ii c ij (A λ i I j v i i j0 l e λit e t(a λ ii c ij (A λ i I j v i l i i e λ it ( k0 j,k 0 j0 t k k! (A λ ii k c ij (A λ i I j v i j0 l e λ c it ij t k (A λ i I j+k v i k! Eve that we sum from j,k 0 to, there are oly fiitely may terms amog(a λ i I j+k v i thatareozerosice(a λ i I m i v i 0form i B i I the special case that A is diagoalizable, m i for all i That is, (A λ i Iv i 0 ad the l e ta v c i0 e λit v i c e λt v +c 2 e λ2t v 2 ++c l e λlt v l (605 i for v c v +c 2 v 2 ++c l v l where we let c i c i0 ad v,v 2,,v l are eigevectors of A that spa C Example 68 Let us solve the system of liear ODEs give by dx (606 dt 8x { 4x 2 x (0 with dx 2 dt 9x x 4x 2 (0 2 We put the system i the matrix form x Ax with [ ] 8 4 (607 A 9 4 which is the same 2 2 matrix i Example 6 From the computatio there, we kow that [ (608 P 2 AP J 2,2 for P 2] [ [ ] ] 6 (A 2Ie e 9 0 The solutio of (606 is give by [ ] [ ] (609 x e ta x (0 e x 2 (0 ta }{{} v

20 6 6 APPLICATION OF JORDAN CANONICAL FORM We write v as a liear combiatio of e ad (A 2Ie : [ (60 v ] [ + 3 0] [ 6 9 9] 3 e + 9 (A 2Ie The (6 So the solutio is (62 e ta v e ta ( 3 e + 9 (A 2Ie ( e 2tI+t(A 2I 3 e + 9 (A 2Ie ( e 2t e t(a 2I 3 e + 9 (A 2Ie ( ( e 2t t k k! (A 2Ik 3 e + 9 (A 2Ie k0 ( e 2t (+t(a 2I 3 e + 9 (A 2Ie ( ( e 2t 3 e t (A 2Ie 3 [ ( e2t +e 3 0] 2t 9 + t [ [ ] 6 (+2te 2t 3 9] (+3te 2t { x (+2te 2t x 2 (+3te 2t The difficulty with the above approach is the fidig of the matrix P, just as i the computatio of A Our secod approach will avoid this issue As before, we choose a ozero polyomial f(x such that f(a 0 ad try to divide e tx by f(x We may take f(x to be the characteristic or miimal polyomial of A Although e tx is ot a polyomial, we ca still divide e tx by f(x such that (63 e tx q(xf(x+r(x where r(x is a polyomial of degr(x < degf(x ad q(x is a etire fuctio, ie, a fuctio that has complex derivatives everywhere for all x C Whe we replace x by A, we still have (64 e ta q(af(a+r(a r(a So it suffices to fid r(x As before, it ca be computed usig partial fractios: (65 e tx r(x q(x+ f(x f(x q(x+ m d i i j c ij (x λ i j

21 2 SOLUTIONS OF LINEAR ODES WITH CONSTANT COEFFICIENTS 7 for f(x (x λ d (x λ 2 d 2 (x λ m dm with λ,λ 2,,λ m distict As before, to determie c ij for each i, we multiply both sides of (65 by (x λ i d i ad take derivatives at λ i of order k 0,,,d i, which yields (66 c ij (d i j! ( e tx (x λ i d i f(x (di j λi Example 69 Let us redo Example 68 usig the characteristic polyomial of A, which is f(x (x 2 2 Suppose that (67 e tx (x 2 2 q(x+ c x 2 + c 2 (x 2 2 for some etire fuctio q(x ad some costats c ad c 2 We multiply both sides by (x 2 2 ad take derivatives of order k 0, at 2: (68 c (e tx x2 te 2t ad c 2 (e tx e 2t x2 Therefore, (69 e tx q(x(x 2 2 +c (x 2+c 2 q(x(x 2 2 +te 2t (x 2+e 2t Replacig x by A, we obtai (620 e ta q(a(a 2I 2 +te 2t (A 2I+e 2t I te 2t (A 2I+e 2t The the solutio of (606 is [ ] x e x ta v te 2t (A 2Iv+e 2t v 2 [ ] [ ] (62 te 2t 6 4 +e 9 6][ 2t [ ] (+2te 2t (+3te 2t which agrees with (62 Example 60 Let us solve the system of liear ODEs give by dx dt 2x 2 x 3 x (0 0 dx (622 2 dt x +3x 2 +x 3 with x 2 (0 dx x 3 (0 3 dt x +x 2 +2x 3 We put it i the matrix form x Ax for 0 2 (623 A 3 2

22 8 6 APPLICATION OF JORDAN CANONICAL FORM The characteristic polyomial of A is f(x (x (x 2 2 Suppose that (624 e tx (x (x 2 2 q(x+ a x + b x 2 + b 2 (x 2 2 for some etire fuctio q(x ad some costats a,b ad b 2 The ( e tx 2 a e (x 2 x t, ( e tx (625 b te 2t e 2t, ad x ( x2 e tx b 2 e 2t Therefore, (626 x x2 e tx q(xf(x+a(x 2 2 +b (x (x 2+b 2 (x e ta q(af(a+a(a 2I 2 +b (A I(A 2I+b 2 (A I a(a 2I 2 +b (A I(A 2I+b 2 (A I e t (A 2I 2 +(t e 2t (A I(A 2I+e 2t (A I Therefore, the solutio of (622 is x x 2 e ta x (0 x 2 (0 x 3 x 3 (0 }{{} v e t (A 2I 2 v+(t e 2t (A I(A 2Iv (627 +e 2t (A Iv e t (t e 2t e 2t 2 0 et (+2te 2t (+2te 2t e t +2e 2t