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=

问题描述:

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
1个回答 分类:数学 2014-11-12

问题解答:

我来补答
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,
再问: 省略号省略的项都是0 嘛?
再答: 是啊!0*0=0
 
 
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