设n维行向量α=（1/2,0,...,0,1/2）,矩阵A=E-α'α,B =E+2α'α,则AB=AB = (E-α'α)(E+_数学解答

设n维行向量α=（1/2,0,...,0,1/2）,矩阵A=E-α'α,B =E+2α'α,则AB=
AB
= (E-α'α)(E+2α'α)
= E + α'α - 2α'αα'α
= E + α'α - 2α'(αα')α
= E + α'α - 2α' (1/2)α//这步不太懂
= E

人气：350 ℃　时间：2020-10-01 02:26:19

解答

因 aa'=(1/2)(1/2)+(1/2)(1/2)=1/2,则E+a‘a-2a'(aa')a=E+a'a-a’a=E,省略号省略的项都是0 嘛？是啊！0*0=0