B 则 00 00 A2 A3=A4 =B3=B4= 00 +(>)A=|+(e-1)A= 01 =H+∑)B=1+(-1B=°1-e n 01 2e+1 0 可见ee≠ee A+B/20 A+B)2=2 00 00|=2A+B),(A+B)3=2(A+B), B+=|+C∑2)(A+B)=|+(a2-1)A+B)=/0 2 所以,e≠e ≠eBe [定理]若AB=BA,则eA=e′eB=ee [证明] eeB=(+A+A2+…)(|+B+B2+…) =|+(A+B)+(A2+2AB+B2)+(A3+3A3+3AB2+B3) 2! =|+(A+B)+(A+B)2+(A+B)+…=eA 3! (A+B)2=(A+B)(A+B)=A+AB+BA +B=A2+2AB+B2 A+B)3=…=A3+3AB+3AB2+B      1 1 A = 0 0 ,       1 -1 B = 0 0 , 则       2 3 4 1 1 A = = A = A = 0 0       2 3 4 1 -1 B = = B = B = 0 0         A n=1 1 e e - 1 e =I+( )A =I+(e - 1)A = n! 0 1         B n=1 1 e 1- e e =I+( )B =I+(e - 1)B = n! 0 1   =     2 2 A B e 2e - e - 1 e e 0 1   =     2 2 B A e e - 2e +1 e e 0 1 可见 A B B A e e ≠e e       2 0 A + B = 0 0 ,       2 2 0 (A + B) = 2 = 2(A + B) 0 0 , 3 2 (A + B) = 2 (A + B),         2 A+B n-1 2 n=1 1 1 e 0 e =I+( 2 )(A + B)=I+ (e - 1)(A + B)= n! 2 0 1 所以, A+B A B e ≠e e , A+B B A e ≠e e [定理] 若 AB = BA , 则 A+B A B B A e = e e = e e [证明]: A B 2 2 1 1 e e =(I+ A + A + )(I+ B + B + ) 2! 2! 1 1 2 2 3 2 2 3 =I+(A + B)+ (A + 2AB + B )+ (A + 3A B + 3AB + B )+ 2! 3! 1 1 2 3 A+B =I+(A + B)+ (A + B) + (A + B) + = e 2! 3! 2 2 2 2 2 (A + B) =(A + B)(A + B)= A + AB + BA + B = A + 2AB + B 3 3 2 2 3 (A + B) = = A + 3A B + 3AB + B