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let-give-A-1-1-2-1-find-e-A-and-e-tA-




Question Number 28258 by abdo imad last updated on 22/Jan/18
let give A  = ( 1       1 )                             (  2    −1)    find  e^(A )     and   e^(−tA) .
letgiveA=(11)(21)findeAandetA.
Commented by abdo imad last updated on 23/Jan/18
e^A = Σ_(n=0) ^∝   (A^n /(n!))   let  find  A^n  the carscteristic polynomial  of  A  is P_c (A)=det(A −XI)= determinant (((1−X       1)),((2           −1−X)))  =−(1−X^2 ) −2= X^2 −3 =(X−(√3))(X +(√(3)))  so the proper values of A?are λ_1 =(√3)  and λ_2 =−(√3)  V(λ_1 ) =ker(A −λ_1 I)  u(x,y)∈V(λ_1 )⇔ (A −λ_1 I)u=0   (((1−(√3)        1)),((2        −2−(√3))) )   ((x),(y) ) = (((o )),(o) )   after cslculus we find  V( λ_1 )= D_e_(1 )    with e_1  ((1),((−(1−(√3)))) )    V(λ_2 )=ker(A −λ_(2 ) I)  u(x,y)∈V(λ_2 )  ⇔ (A −λ_2 I)u=0  ⇔ (((1+(√3)       1)),((2         −1+(√3))) )   ((x),(y) ) = ((0),(0) ) after calculus we find  V(λ_2 )= D_e_2       with  e_2  ((1),((−(1+(√(3))))) )   so the diagonal  mstrix is D=  ((((√3)         0 )),((0          −(√3_ ))) )  and  the inversible  matrix id  P=    (((1                                  1)),((−(1−(√3)       −(1+(√3)  )    )) )  we have  P^(−1)    =((tran(com P))/(detP))  .   com(P)= (−1)^(i+j)  a_(ij)   we find  P^(−1) = (1/(2(√3)))  (((1+(√3)                 1)),((−(1−(√(3)))       −1)) )   we have A =PDP^(−1)    ⇒  A^n = P D^n P^(−1)   after all calculus we find      A^n    =(1/(2(√3)))        (     ((√3)  )^n (1+(√3))−(−(√3))^n (1−(√3))           ((√3))^n  −(−(√3))^n     )                            (   2 ((√3))^n    −2(−(√3))^n                 −(1−(√3))((√3))^n   +(1+(√3)) (−(√3))^n      )   .                       we have  e^A   = Σ_(n=0) ^∝   (A^n /(n!))  and  e^(tA)   = Σ_(n=0) ^∝   A^n    (t^n /(n!)) .
eA=n=0Ann!letfindAnthecarscteristicpolynomialofAisPc(A)=det(AXI)=|1X121X|=(1X2)2=X23=(X3)(X+3)sothepropervaluesofA?areλ1=3andλ2=3V(λ1)=ker(Aλ1I)u(x,y)V(λ1)(Aλ1I)u=0(131223)(xy)=(oo)aftercslculuswefindV(λ1)=De1withe1(1(13))V(λ2)=ker(Aλ2I)u(x,y)V(λ2)(Aλ2I)u=0(1+3121+3)(xy)=(00)aftercalculuswefindV(λ2)=De2withe2(1(1+3))sothediagonalmstrixisD=(3003)andtheinversiblematrixidP=(11(13(1+3))wehaveP1=tran(comP)detP.com(P)=(1)i+jaijwefindP1=123(1+31(13)1)wehaveA=PDP1An=PDnP1afterallcalculuswefindAn=123((3)n(1+3)(3)n(13)(3)n(3)n)(2(3)n2(3)n(13)(3)n+(1+3)(3)n).wehaveeA=n=0Ann!andetA=n=0Antnn!.

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