Matricial Potentiation

Transcription

1 Matrcal Potetato By Ezo March* ad Mart Mates** Abstract I ths short ote we troduce the potetato of matrces of the same sze. We study some smple propertes ad some example. * Emertus Professor UNSL, Sa Lus Argeta. Fouder ad Frst Drector of the IMASL, UNSL CONICET. ** UTN. (ex. Superor Researcher CONICET.

2 Itroducto We troduce the matrx potetato. The problem was pose byguamett. Cosder two matrcesa, of szemxad B, mx. We wsh to defe the potetato For ths purpose. We tae the logarthm C = A B l C = B l A, Ths s vald f CadA are ot sgular. From ow o we assome that whe we tae al the argumet s ot sgular. The l C s well defed for the matrces Aad B real or complex. I the secod case we have that t s a multvalued fucho. Ths assumg that he l A s a covergg sequece. Cosder the matrx D = A I where I stads for the detty matrx. The l A = l (D + I = D D + D3 3 D4 4 + D5 ⁱ+¹D + ( 5 From here, t s medated that f m = C = eb l A ca be well defed. The oly codto that s ecessary the covergece of l A, or (A I (A I3 l C = B l A = B [(A I + 3 = B [ ( +1 A I ( ] = B [ ( +1 ( j A ( I j ] =1 =1 If A Is dagoalzable the A I = Q +1 Λ Q 1 where the matrx Λ s dagoal wth all the egevalues the ma dagoal, ad Q s formed by the lgevectors. Therefore l C = B l A = B Q [ ( +1 =1 Λ ]Q 1 or

3 C = e B l A = e B Q [ ( +1 =1 Λ ] Q 1 = 1 [B Q K (A Q 1 ] t t! t=0 where K (A = ( +1 =1 Λ We have l A = ( +1 ( j ( j A j o=1 ad t easy to see that A s dagoalzable the followg way Therefore A = Q (Λ + I Q 1 l A = ( +1 ( j ( j Q (Λ + Ij Q 1 =1 = ( +1 =1 Q { ( j ( (Λ + Ij} Q j 1 As a example f A s dagoal: A = dag(x 1, x,, x m them A j = dag(x 1, x,, x m. replacg we see that l A s also dagoal l A (r, r = ( +1 ( ( j (λ r + 1 j = (λ r ( =1 =1 = (λ r + 1 ( = lm λ s =1 l A (r, s = r s =1 theml A s dagoal ad t s covergg λ 1 0 l λ 1 0 l ( λ = ( l λ 0 λ m 0 l λ m Next case, we have that A s dagoalzable A = P Λ P 1

4 l Them the A = ( +1 p +1 Λ p 1 = p +1 ( =1 (Λ p 1 = p 1 [ ( +1 =1 ButQ = p 1 =1 =1 (Λ ] p 1 q 11 q 1 QΛ = λ 1 [ q 1 ] + λ [ q ] + = λ r q m1 q m r=1 q 1r [ q r ] q r therefore ad replacg to equato (1,t tws out that (l A (r, s = ( +1 =1 = ( +1 =1 QΛ P (r, s = λ q r p s ( (λ 1 q r p s ( (λ 1 q r p s = q r [ ( +1 =1 = [p +1 (l Λ p 1 ] (r, s (λ 1 ] p s = q r (l λ p s As a cocluso we have for A = p +1 Λ p 1 dagoalzable wth the egevalues for Fub0 < λ r < cocet l A = p +1 (l Λ p 1 ad l λ 1 c l Λ = (( 0 l λ The we have proved the followg result. Theorem: Gve two dagoalzables matrces mx: A ad B, where A I has egevaluesx r : 0 < x r, the the matrx C = A B

5 s well defes ad t has the form (BA t C = t! t=0 where A = l A Propertes Ths we have bee successful the defto of matrx potetato. I ths secto we are gog to for dagoalzable matrces some the frst oe, already proved s A B exp(b l A Now we study We have wherd = A B. (A B C (A B C = D C = exp(c l D = exp(c l A B = exp(c B l A = A CB O the other had A B A C = exp(b l Aexp(C l A = exp(b l A + C l A = exp((b + C l A = A B+C We follow wth exp(l A = A We ow e x = x 1 a f A h dagoalzable we have l A = P l Λ p 1 where the dagoal matrx Λ s formed by the egevalues by the egevector as colums. The

6 l (exp A = l ( P Λ p 1 = l (P Λ!! =0 =0 P 1 = ( 1 (P 1 exp(λ P = ( 1 = P 1 ( ( 1 (expλ P = P 1 Λ P = A P 1 (expλ P ad ths way we have proved the property. O the other had, we have, aother basc property, for dagoalzable matrces amely. Others propertes are Cosder exp(a + B = exp A exp B e (A+B (A + B ( = = t At B t = At B t!! t! ( t! =0 =0 t=0 Ad ow by Fub property ad the stadardamegamet = At B t t! ( t! =0 =t ad by a varable chage t = j ad t = the = A B j = ( A ( Bj =! j! j! =0 =t ad s ths way the property s proved. O the other had, we ow prove =0 l (AB = l A + l B always the case that AB = BA, or they commets let =0 t=0 exp A exp B l (AB = C exp C = exp(l AB = AB O the other had f we call

7 D = l A + l B the exp D = exp(l A + l B = exp(l A. exp(l B = AB the C = D. Now we preset A B A C = A B+C Let Next A B A C = exp(b l A. exp(c l A = exp(b l A + C l A = exp((b + C l A = exp l (A B+C = A B+C cosder the frst term. Callg D = A B the (A B C = A B+C (A B C = D C = exp(c l D = exp(c l A B = exp(c B l A = A CB Next we cosder a property about the determg amely det(l A = det(l Λ Whe A = P 1 Λ P whch s medate by the decomposto. Cosder the trasperse of A A t, the Let A = P 1 Λ P the O the other had exp(a t = (exp A t A t = (P 1 Λ P t = P t Λ (P 1 t = P t Λ (P t 1 exp(a t = (At 1 = P t ( Λ (p t 1 = (P 1 Λ!!! =0 =0 =0 P t = (e A t Now we cosder aother property, amely det exp A = exp tra Λ Let

8 e A = P 1 ( Λ =0 P! det(e A = det ( Λ = det(e Λ = e λ! =0 tra A = e where tra, dcate the trace of the matrz. Now we cosder a example. j=1 = e λ 1e λ e λ = e λ 1+λ +λ tra Λ = e ( 1 =1 ( 1 ( =1 ( 1 =1 ( 1 ( 1 ( 1 =1 0 l 1 = l C = ( 0 l 1 Ej. 0 l 1 ( 0 l 1 I + l C 1! (l C = ( + ( l C! + ( l C 3 3! 0 l 1 ( 0 l = ( 1 ( ( H = ( H 0 l 1 0 l 1 0 l 1 C = e l C = I + ( 0 l 1 + (

9 1 (l 1 =1 1 0 (l 1 ( 1 =1 = ( 1 e l = e l 1 = ( 0 1 Example We wsh to comput C = A B Cosder as example A = ( adb = (1 oug l C = B l A = B l P Λ P 1 = B P l Λ P 1 them l l Λ = ( 0 l 1 = ( 0 l ad where P = ( Therefore 0 l 16 l A = P l Λ P 1 = ( 0 l hece l C = B l A = ( l 16 0 l 16 ( = ( 0 l 0 l 64

10 from here C = e B l A = ( B l A = 1!! I=0 = l 0 3 ( 1 ( l 0 3 l C = ( 0 3 e 6 l 0 e 3 l

11 Bblografía Burde R.L. ad J. DougasFares: AálssNumérco. Cogage 001. Poole D: Algebra leal. Thompso 004. Waye L. Wso: Ivestgacó de Operacoes 4ta Ed. Thompso 004. The authors would le to tha the Isaac Newto Isttute for Mathematcal Sceces for support ad hosptalty durg the programme Dscrete Aalyss whe wor o ths paper was udertae. Ths wor was supported by EPSRC Grat Number EP/K0308/1