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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)=a0I+a1A+…+amAm.显
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)=a0I+a1A+…+amAm.显然,(A)仍是数域K上的一个n级矩阵,称,(A)是矩阵A的多项式.证明:如果A~B,则f(A)~f(B).
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因为A~B故存在可逆方阵而P使P-1AP=B ①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).
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).
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