Function Spaces. a x 3. (Letting x = 1 =)) a(0) + b + c (1) = 0. Row reducing the matrix. b 1. e 4 3. e 9. >: (x = 1 =)) a(0) + b + c (1) = 0

Size: px

Start display at page:

Download "Function Spaces. a x 3. (Letting x = 1 =)) a(0) + b + c (1) = 0. Row reducing the matrix. b 1. e 4 3. e 9. >: (x = 1 =)) a(0) + b + c (1) = 0"

Gwen James
5 years ago
Views:

1 unction Spacs Prrquisit: Sction 4.7, Coordinatization n this sction, w apply th tchniqus of Chaptr 4 to vctor spacs whos lmnts ar functions. Th vctor spacs P n and P ar familiar xampls of such spacs. Othr important xampls ar C (R) = {all continuous ral-valud functions on R} and C (R) = {all continuously di rntiabl ral-valud functions on R}. Linar ndpndnc in unction Spacs Proving that a nit subst S of a function spac is linarly indpndnt usually rquirs a modi cation of th stratgy usd in R n. n Exampl Considr th subst S = x x; x x ; sin xo of C (R) : W will show that S is linarly indpndnt using th d nition of linar indpndnc. Lt a; b; and c b ral numbrs such that a x x + b x x + c sin x = for vry valu of x. W must show that a = b = c = : Th abov quation must b satis d for vry valu of x: n particular, it is tru for x = ; x = ; and x = : This yilds th following systm: (Ltting x = =)) a() + b + c () = >< (Ltting x = =)) >: (Ltting x = =)) Row rducing th matrix a(6) + b a(4) + b 4 + c () = 9 + c ( ) = : a b 4 9 c 7 5 to a 4 b c 5 shows that th trivial solution a = b = c = is th only solution to this homognous systm. Hnc, th st S is linarly indpndnt by th d nition of linar indpndnc. Whn proving linar indpndnc using th tchniqu of Exampl, w try to choos nic valus of x to mak computations asir. Evn so, th us of a calculator or computr is oftn dsirabl whn working with function spacs. Othr problms may occur bcaus of th choic of x-valus. Rturning to Exampl, if instad w had pluggd in x = ; x = ; and x = ; w would hav obtaind th systm (x = =)) a() + b + c ( ) = >< (x = =)) a() + b () + c () = ; >: (x = =)) a() + b + c () = which has in nitly many nontrivial solutions. To prov linar indpndnc, w must xamin furthr valus of x, gnrating mor quations for th systm, until th nw systm w obtain has only th trivial solution, as in Exampl. Andrilli/Hckr Elmntary Linar Algbra, 4th d. March 5, Copyright, Elsvir nc. All rights rsrvd.

2 Suppos, howvr, that aftr substituting many valus for x and crating a hug homognous systm, w still hav nontrivial solutions. W cannot conclud that th st of functions is linarly dpndnt, although w may suspct that it is. n gnral, to prov that a st of functions ff ; : : : ; f n g is linarly dpndnt, w must nd ral numbrs a ; : : : ; a n ; not all zro, such that a f (x) + a f (x) + + a n f n (x) = is a functional idntity for vry valu of x; not just thos w hav trid. Exampl Lt S = fsin x; cos x; sin x; cos xg; a subst of C (R). Suppos w attmpt to show that S is linarly indpndnt using th d nition of linar indpndnc. Lt a; b; c; and d rprsnt ral numbrs such that a(sin x) + b(cos x) + c(sin x) + d(cos x) = : Sinc w hav four vctors in S, w substitut four di rnt valus for x into this quation to obtain th following systm: (x = =) ) a() + b () + c () + d() = >< (x = 4 =) ) a() + b () + c + d = : (x = =) ) a() + b ( ) + c () + d() = >: (x = 4 =) ) a( ) + b () + c + d = Sinc th co cint matrix for this homognous systm row rducs to a b c d ; thr ar nontrivial solutions to th systm, such as a = ; b = ; c = ; d = : At this point, w cannot infr that S is linarly indpndnt bcaus w hav nontrivial solutions. W also cannot conclud that S is linarly dpndnt bcaus w hav tstd only a fw valus for x. W could try mor valus, such as x = 6 and x =, but w would still nd that a = ; b = ; c = ; d = satis s ach quation w gnrat. This situation lads us to bliv that th st S is linarly dpndnt. To b crtain, w must chck that th valus a = ; b = ; c = ; and d = yild a functional idntity whn pluggd into th original functional quation. Substituting ths valus yilds (sin x) + ( )(cos x) + ( )(sin x) + ()(cos x) = ; or cos x = cos x sin x; a wll-known trigonomtric idntity. Thus, on vctor in S can b xprssd as a linar combination of th othr vctors in S, and S is linarly dpndnt. Nw Vocabulary C (R) (continuous ral-valud functions on R) C (R) (ral-valud functions on R having a continuous drivativ) function spacs linarly dpndnt st (in a function spac) linarly indpndnt st (in a function spac) Andrilli/Hckr Elmntary Linar Algbra, 4th d. March 5, Copyright, Elsvir nc. All rights rsrvd.

3 Highlights unction spacs ar vctor spacs whos lmnts ar functions. Exampls of function spacs ar P n, P, C (R), and C (R): A st of functions ff ; : : : ; f n g (in a function spac) is linarly indpndnt if thr ar n di rnt valus of x so that th rsulting n quations of th form a f (x) + a f (x) + + a n f n (x) = form a systm having only th trivial solution a = a = = a n = : A st of functions ff ; : : : ; f n g (in a function spac) is linarly dpndnt if th quation a f (x) + a f (x) + + a n f n (x) = has a nontrivial solution for a ; a ; : : : ; a n for vry possibl valu of x. EXERCSES. n ach part of this xrcis, dtrmin whthr th givn subst S of C (R) is linarly indpndnt. f S is linarly indpndnt, prov that it is. f S is linarly dpndnt, solv for a functional idntity that xprsss on function in S as a linar combination of th othrs. a) S = x ; x ; x b) S = fsin x; sin x; sin x; sin 4xg c) S = (5x ) = + x ; (x + ) = + x ; 7x x + 7x 5 = x 4 + x + d) S = fsin x; sin(x + ); sin(x + ); sin(x + )g. Rcall that a function f(x) C (R) is vn if f(x) = f( x) for all x R and is odd if f(x) = f( x) for all x R. Suppos w want to prov that a nit subst S of C (R) is linarly indpndnt by th mthod of Exampl. a) Suppos that vry lmnt of S is an odd function of x (as in Exampl ). Explain why w would not want to substitut both and for x into th appropriat functional quation. Also xplain why x = would b a poor choic. b) Suppos that vry lmnt of S is an vn function. Would w want to substitut both and for x into th appropriat functional quation? Why? How about x =?. Lt S b th subst fcos(x + ); cos(x + ); cos(x + )g of C (R): a) Show that span(s) has fcos x; sin xg for a basis. (Hint: Th idntity cos ( + ) = cos cos sin sin is usful.) b) Us part (a) to prov that S is linarly dpndnt. 4. or ach givn subst S of C (R), nd a subst B of S that is a basis for V = span(s). a) S = fsin x; cos x; sin x; cos x; sin x cos x; g b) S = f x ; ; x g c) S = fsin(x + ); cos(x + ); sin(x + ); cos(x + )g 5. n ach part of this xrcis, lt B rprsnt an ordrd basis for a subspac V of C (R) and nd [v] B for th givn v V: Andrilli/Hckr Elmntary Linar Algbra, 4th d. March 5, Copyright, Elsvir nc. All rights rsrvd.

4 4 a) B = ( x ; x ; x ); v = 5 x 7 x b) B = (sin x; cos x; sin x); v = c) B = (sin (x + ) ; sin (x + )) ; v = cos x 6. Tru or als: a) A subst ff ; f g of nonzro functions in C (R) is linarly dpndnt if and only if f is a nonzro constant multipl of f. b) Th st fx ; x ; x 4 ; x 5 g is a linarly indpndnt subst of C (R). c) Lt f ; f ; f C (R). f plugging valus for x into af (x) + bf (x) + cf (x) = lads to a = b = c =, thn f, f, and f ar linarly dpndnt. d) Lt f ; f ; f C (R). f plugging di rnt valus for x into af (x) + bf (x) + cf (x) = dos not allow us to conclud that a = b = c =, thn f, f, and f ar linarly dpndnt. Andrilli/Hckr Elmntary Linar Algbra, 4th d. March 5, Copyright, Elsvir nc. All rights rsrvd.

5 5 Answrs to Slctd Exrciss () (a) Linarly indpndnt; to prov that it is, substitut th valus x =, x =, x =, and follow th mthod of Exampl (c) Linarly dpndnt (a =, b =, c = ) (4) (a) B = fsin(x); cos(x); sin xg (c) B = fsin(x + ); cos(x + )g (5) (a) [v] B = [5; ; 7] cos (c) [v] B = [ sin ; cos sin ] [:4945; :64]. (f your answr is mor complicatd than this, compar numrical approximations.) (6) (a) T (b) T (c) (d) Andrilli/Hckr Elmntary Linar Algbra, 4th d. March 5, Copyright, Elsvir nc. All rights rsrvd.

The Matrix Exponential

The Matrix Exponential Th Matrix Exponntial (with xrciss) by D. Klain Vrsion 207.0.05 Corrctions and commnts ar wlcom. Th Matrix Exponntial For ach n n complx matrix A, dfin th xponntial of A to b th matrix A A k I + A + k!