设f(x)=a0+a1x+…+amxm是数域K上的一元多项式 设A是数域K上的n级矩阵 定义f(A)

正确答案：因为A～B故存在可逆方阵而P使P^-1AP=B ①当k=0时由A^k=B^k=E从布A^k～B^k；当k为正整数时由①可得：(P^-1AP)^k=B^k即P^-1A^kP=B^k所以A^k～B^k．则f(B)=a₀I+a₀B+…+a_mB_m=a₀P^-1IP+a₁P^-1AP+…+a_mP^-1AP=P^-1(a₀I+a₁A+…+a_mA^m)P=P^-1.f(A)P即f(A)～f(B)．
因为A～B,故存在可逆方阵而P,使P-1AP=B①当k=0时,由Ak=Bk=E,从布Ak～Bk；当k为正整数时,由①可得：(P-1AP)k=Bk,即P-1AkP=Bk,所以Ak～Bk．则f(B)=a0I+a0B+…+amBm=a0P-1IP+a1P-1AP+…+amP-1AP=P-1(a0I+a1A+…+amAm)P=P-1.f(A)P即f(A)～f(B)．