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设向量a为n维列向量,a^t*a=1,令H=E-2a*a^t,证明H是正交矩阵
(E—2aa^T)^T怎么求?
人气:146 ℃ 时间:2020-04-10 07:16:01
解答
H^T=(E—2aa^T)^T
=E^T—2(a^T)^T·a^T
=E—2aa^T


H·H^T=(E-2aa^T)·(E-2aa^T)
=E-2aa^T-2aa^T+4aa^T·aa^T
=E-4aa^T+4a·(a^T·a)·a^T
=E-4aa^T+4aa^T
=E


所以,H是正交矩阵.
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