If the characteristic polynomial of a linear operator TTT on a complex inner product space VVV is
fT(x)=x4−2x2+1f_T(x) = x^4 - 2x^2 + 1fT(x)=x4−2x2+1,
then the rank(S)\operatorname{rank}(S)rank(S) of the linear operator
S=T6−2T5−T4+4T3−T2−2TS = T^6 - 2T^5 - T^4 + 4T^3 - T^2 - 2TS=T6−2T5−T4+4T3−T2−2T
in terms of rank(T)\operatorname{rank}(T)rank(T) is ___.
參考答案與解析
rank(S)=rank(T)\operatorname{rank}(S)=\operatorname{rank}(T)rank(S)=rank(T)（皆為滿秩，等於VVV的維度4）。特徵多項式fT(x)=x4−2x2+1=(x−1)2(x+1)2f_T(x)=x^4-2x^2+1=(x-1)^2(x+1)^2fT(x)=x4−2x2+1=(x−1)2(x+1)2，故TTT的特徵值為1,1,−1,−11,1,-1,-11,1,−1,−1（皆非0，TTT滿秩=4）。將p(x)=x6−2x5−x4+4x3−x2−2xp(x)=x^6-2x^5-x^4+4x^3-x^2-2xp(x)=x6−2x5−x4+4x3−x2−2x代入：p(1)=−1≠0p(1)=-1\ne0p(1)=−1=0，p(−1)=−1≠0p(-1)=-1\ne0p(−1)=−1=0，故SSS的特徵值為−1,−1,−1,−1-1,-1,-1,-1−1,−1,−1,−1（皆非0，SSS也滿秩=4）。因此rank(S)=rank(T)\operatorname{rank}(S)=\operatorname{rank}(T)rank(S)=rank(T)（兩者都等於dimV=4\dim V=4dimV=4）。
