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=
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
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
数学人气：840 ℃时间：2020-10-01 02:26:19
优质解答
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
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